标题: Quantitative Methods 【Reading 11】Sample [打印本页]
作者: Mechanic 时间: 2012-3-22 16:08 标题: [2012 L1] Quantitative Methods 【Session 3 - Reading 11】Sample
Which of the following statements about testing a hypothesis using a Z-test is least accurate? A)
| If the calculated Z-statistic lies outside the critical Z-statistic range, the null hypothesis can be rejected. |
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B)
| The calculated Z-statistic determines the appropriate significance level to use. |
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C)
| The confidence interval for a two-tailed test of a population mean at the 5% level of significance is that the sample mean falls between ±1.96 σ/√n of the null hypothesis value. |
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The significance level is chosen before the test so the calculated Z-statistic can be compared to an appropriate critical value.
作者: Mechanic 时间: 2012-3-22 16:08
Susan Bellows is comparing the return on equity for two industries. She is convinced that the return on equity for the discount retail industry (DR) is greater than that of the luxury retail (LR) industry. What are the hypotheses for a test of her comparison of return on equity?A)
| H0: µDR = µLR versus Ha: µDR ≠ µLR. |
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B)
| H0: µDR ≤ µLR versus Ha: µDR > µLR. |
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C)
| H0: µDR = µLR versus Ha: µDR < µLR. |
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The alternative hypothesis is determined by the theory or the belief. The researcher specifies the null as the hypothesis that she wishes to reject (in favor of the alternative). Note that this is a one-sided alternative because of the “greater than” belief.
作者: Mechanic 时间: 2012-3-22 16:09
In the process of hypothesis testing, what is the proper order for these steps?A)
| State the hypotheses. Specify the level of significance. Collect the sample and calculate the test statistics. Make a decision. |
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B)
| Collect the sample and calculate the sample statistics. State the hypotheses. Specify the level of significance. Make a decision. |
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C)
| Specify the level of significance. State the hypotheses. Make a decision. Collect the sample and calculate the sample statistics. |
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The hypotheses must be established first. Then the test statistic is chosen and the level of significance is determined. Following these steps, the sample is collected, the test statistic is calculated, and the decision is made.
作者: Mechanic 时间: 2012-3-22 16:09
The first step in the process of hypothesis testing is: A)
| to state the hypotheses. |
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B)
| selecting the test statistic. |
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C)
| the collection of the sample. |
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The researcher must state the hypotheses prior to the collection and analysis of the data. More importantly, it is necessary to know the hypotheses before selecting the appropriate test statistic.
作者: Mechanic 时间: 2012-3-22 16:09
Which of the following statements least describes the procedure for testing a hypothesis? A)
| Compute the sample value of the test statistic, set up a rejection (critical) region, and make a decision. |
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B)
| Develop a hypothesis, compute the test statistic, and make a decision. |
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C)
| Select the level of significance, formulate the decision rule, and make a decision. |
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Depending upon the author there can be as many as seven steps in hypothesis testing which are:- Stating the hypotheses.
- Identifying the test statistic and its probability distribution.
- Specifying the significance level.
- Stating the decision rule.
- Collecting the data and performing the calculations.
- Making the statistical decision.
- Making the economic or investment decision.
作者: Mechanic 时间: 2012-3-22 16:10
Which of the following is the correct sequence of events for testing a hypothesis? A)
| State the hypothesis, select the level of significance, formulate the decision rule, compute the test statistic, and make a decision. |
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B)
| State the hypothesis, select the level of significance, compute the test statistic, formulate the decision rule, and make a decision. |
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C)
| State the hypothesis, formulate the decision rule, select the level of significance, compute the test statistic, and make a decision. |
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Depending upon the author there can be as many as seven steps in hypothesis testing which are:- Stating the hypotheses.
- Identifying the test statistic and its probability distribution.
- Specifying the significance level.
- Stating the decision rule.
- Collecting the data and performing the calculations.
- Making the statistical decision.
- Making the economic or investment decision.
作者: Mechanic 时间: 2012-3-22 16:10
Which of the following statements about hypothesis testing is most accurate?A)
| If you can disprove the null hypothesis, then you have proven the alternative hypothesis. |
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B)
| The power of a test is one minus the probability of a Type I error. |
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C)
| The probability of a Type I error is equal to the significance level of the test. |
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The probability of getting a test statistic outside the critical value(s) when the null is true is the level of significance and is the probability of a Type I error. The power of a test is 1 minus the probability of a Type II error. Hypothesis testing does not prove a hypothesis, we either reject the null or fail to reject it.
作者: Mechanic 时间: 2012-3-22 16:10
An analyst conducts a two-tailed z-test to determine if small cap returns are significantly different from 10%. The sample size was 200. The computed z-statistic is 2.3. Using a 5% level of significance, which statement is most accurate?A)
| You cannot determine what to do with the information given. |
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B)
| Reject the null hypothesis and conclude that small cap returns are significantly different from 10%. |
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C)
| Fail to reject the null hypothesis and conclude that small cap returns are close enough to 10% that we cannot say they are significantly different from 10%. |
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At the 5% level of significance the critical z-statistic for a two-tailed test is 1.96 (assuming a large sample size). The null hypothesis is H0: x = 10%. The alternative hypothesis is HA: x ≠ 10%. Because the computed z-statistic is greater than the critical z-statistic (2.33 > 1.96), we reject the null hypothesis and we conclude that small cap returns are significantly different than 10%.
作者: Mechanic 时间: 2012-3-22 16:11
An analyst conducts a two-tailed test to determine if mean earnings estimates are significantly different from reported earnings. The sample size is greater than 25 and the computed test statistic is 1.25. Using a 5% significance level, which of the following statements is most accurate? A)
| The analyst should reject the null hypothesis and conclude that the earnings estimates are significantly different from reported earnings. |
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B)
| The analyst should fail to reject the null hypothesis and conclude that the earnings estimates are not significantly different from reported earnings. |
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C)
| To test the null hypothesis, the analyst must determine the exact sample size and calculate the degrees of freedom for the test. |
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The null hypothesis is that earnings estimates are equal to reported earnings. To reject the null hypothesis, the calculated test statistic must fall outside the two critical values. IF the analyst tests the null hypothesis with a z-statistic, the crtical values at a 5% confidence level are ±1.96. Because the calculated test statistic, 1.25, lies between the two critical values, the analyst should fail to reject the null hypothesis and conclude that earnings estimates are not significantly different from reported earnings. If the analyst uses a t-statistic, the upper critical value will be even greater than 1.96, never less, so even without the exact degrees of freedom the analyst knows any t-test would fail to reject the null.
作者: Mechanic 时间: 2012-3-22 16:12
Given the following hypothesis:- The null hypothesis is H0 : µ = 5
- The alternative is H1 : µ ≠ 5
- The mean of a sample of 17 is 7
- The population standard deviation is 2.0
What is the calculated z-statistic?
The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean − hypothesized mean) / (population standard deviation / (sample size)1/2 = (X − μ) / (σ / n1/2) = (7 − 5) / (2 / 171/2) = (2) / (2 / 4.1231) = 4.12.
作者: Mechanic 时间: 2012-3-22 16:12
What kind of test is being used for the following hypothesis and what would a z-statistic of 1.68 tell us about a hypothesis with the appropriate test and a level of significance of 5%, respectively?
H0: B ≤ 0
HA: B > 0
A)
| One-tailed test; fail to reject the null. |
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B)
| Two-tailed test; fail to reject the null. |
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C)
| One-tailed test; reject the null. |
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The way the alternative hypothesis is written you are only looking at the right side of the distribution. You are only interested in showing that B is greater than 0. You don't care if it is less than zero. For a one-tailed test at the 5% level of significance, the critical z value is 1.645. Since the test statistic of 1.68 is greater than the critical value we would reject the null hypothesis.
作者: Mechanic 时间: 2012-3-22 16:12
In a two-tailed test of a hypothesis concerning whether a population mean is zero, Jack Olson computes a t-statistic of 2.7 based on a sample of 20 observations where the distribution is normal. If a 5% significance level is chosen, Olson should: A)
| reject the null hypothesis and conclude that the population mean is not significantly different from zero. |
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B)
| fail to reject the null hypothesis that the population mean is not significantly different from zero. |
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C)
| reject the null hypothesis and conclude that the population mean is significantly different from zero. |
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At a 5% significance level, the critical t-statistic using the Student’s t-distribution table for a two-tailed test and 19 degrees of freedom (sample size of 20 less 1) is ± 2.093 (with a large sample size the critical z-statistic of 1.960 may be used). Because the critical t-statistic of 2.093 is to the left of the calculated t-statistic of 2.7, meaning that the calculated t-statistic is in the rejection range, we reject the null hypothesis and we conclude that the population mean is significantly different from zero.
作者: Mechanic 时间: 2012-3-22 16:13
In order to test whether the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken and the sample value of the computed test statistic, tn-1 = 3.4. The null and alternative hypotheses are: A)
| H0: µ = 100; Ha: µ ≠ 100. |
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B)
| H0: µ ≤ 100; Ha: µ > 100. |
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C)
| H0: X ≤ 100; Ha: X > 100. |
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The null hypothesis is that the theoretical mean is not significantly different from zero. The alternative hypothesis is that the theoretical mean is greater than zero.
作者: Mechanic 时间: 2012-3-22 16:13
In order to test if the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken and the sample value of the computed test statistic, tn-1 = 1.2. If you choose a 5% significance level you should: A)
| fail to reject the null hypothesis and conclude that the population mean is not greater than 100. |
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B)
| reject the null hypothesis and conclude that the population mean is greater than 100. |
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C)
| fail to reject the null hypothesis and conclude that the population mean is greater than 100. |
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At a 5% significance level, the critical t-statistic using the Student’s t distribution table for a one-tailed test and 29 degrees of freedom (sample size of 30 less 1) is 1.699 (with a large sample size the critical z-statistic of 1.645 may be used). Because the critical t-statistic is greater than the calculated t-statistic, meaning that the calculated t-statistic is not in the rejection range, we fail to reject the null hypothesis and we conclude that the population mean is not significantly greater than 100.
作者: Mechanic 时间: 2012-3-22 16:13
If the null hypothesis is H0: ρ ≤ 0, what is the appropriate alternative hypothesis?
The alternative hypothesis must include the possible outcomes the null does not.
作者: Mechanic 时间: 2012-3-22 16:14
Jo Su believes that there should be a negative relation between returns and systematic risk. She intends to collect data on returns and systematic risk to test this theory. What is the appropriate alternative hypothesis?
The alternative hypothesis is determined by the theory or the belief. The researcher specifies the null as the hypothesis that she wishes to reject (in favor of the alternative). The theory in this case is that the correlation is negative.
作者: Mechanic 时间: 2012-3-22 16:14
Jill Woodall believes that the average return on equity in the retail industry, µ, is less than 15%. What are the null (H0) and alternative (Ha) hypotheses for her study? A)
| H0: µ ≤ 0.15 versus Ha: µ > 0.15. |
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B)
| H0: µ < 0.15 versus Ha: µ ≥ 0.15. |
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C)
| H0: µ ≥ 0.15 versus Ha: µ < 0.15. |
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This is a one-sided alternative because of the "less than" belief.
作者: Mechanic 时间: 2012-3-22 16:14
Brian Ci believes that the average return on equity in the airline industry, µ, is less than 5%. What are the appropriate null (H0) and alternative (Ha) hypotheses to test this belief? A)
| H0: µ ≥ 0.05 versus Ha: µ < 0.05. |
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B)
| H0: µ < 0.05 versus Ha: µ ≥ 0.05. |
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C)
| H0: µ < 0.05 versus Ha: µ > 0.05. |
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The alternative hypothesis is determined by the theory or the belief. The researcher specifies the null as the hypothesis that he wishes to reject (in favor of the alternative). Note that this is a one-sided alternative because of the "less than" belief.
作者: Mechanic 时间: 2012-3-22 16:15
George Appleton believes that the average return on equity in the amusement industry, µ, is greater than 10%. What is the null (H0) and alternative (Ha) hypothesis for his study? A)
| H0: > 0.10 versus Ha: ≤ 0.10. |
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B)
| H0: > 0.10 versus Ha: < 0.10. |
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C)
| H0: ≤ 0.10 versus Ha: > 0.10. |
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The alternative hypothesis is determined by the theory or the belief. The researcher specifies the null as the hypothesis that he wishes to reject (in favor of the alternative). Note that this is a one-sided alternative because of the "greater than" belief.
作者: Mechanic 时间: 2012-3-22 16:15
A researcher is testing the hypothesis that a population mean is equal to zero. From a sample with 64 observations, the researcher calculates a sample mean of -2.5 and a sample standard deviation of 8.0. At which levels of significance should the researcher reject the hypothesis? | [td=1,1,120]1% significance | 5% significance | 10% significance |
A)
| | Fail to reject | Fail to reject | Reject |
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B)
| | Reject | Fail to reject | Fail to reject |
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C)
| | Fail to reject | Reject | Reject |
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This is a two-tailed test. With a sample size greater than 30, using a z-test is acceptable. The test statistic = 
= −2.5. For a two-tailed z-test, the critical values are ±1.645 for a 10% significance level, ±1.96 for a 5% significance level, and ±2.58 for a 1% significance level. The researcher should reject the hypothesis at the 10% and 5% significance levels, but fail to reject the hypothesis at the 1% significance level.
Using Student's t-distribution, the critical values for 60 degrees of freedom (the closest available in a typical table) are ±1.671 for a 10% significance level, ±2.00 for a 5% significance level, and ±2.66 for a 1% significance level. The researcher should reject the hypothesis at the 10% and 5% significance levels, but fail to reject the hypothesis at the 1% significance level.
作者: Mechanic 时间: 2012-3-22 16:15
James Ambercrombie believes that the average return on equity in the utility industry, µ, is greater than 10%. What are the null (H0) and alternative (Ha) hypotheses for his study? A)
| H0: µ ≤ 0.10 versus Ha: µ > 0.10. |
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B)
| H0: µ < 0.10 versus Ha: µ > 0.10. |
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C)
| H0: µ > 0.10 versus Ha: µ < 0.10. |
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This is a one-sided alternative because of the "greater than" belief.
作者: Mechanic 时间: 2012-3-22 16:16
Which one of the following is the most appropriate set of hypotheses to use when a researcher is trying to demonstrate that a return is greater than the risk-free rate? The null hypothesis is framed as a: A)
| less than statement and the alternative hypothesis is framed as a greater than or equal to statement. |
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B)
| less than or equal to statement and the alternative hypothesis is framed as a greater than statement. |
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C)
| greater than statement and the alternative hypothesis is framed as a less than or equal to statement. |
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If a researcher is trying to show that a return is greater than the risk-free rate then this should be the alternative hypothesis. The null hypothesis would then take the form of a less than or equal to statement.
作者: Mechanic 时间: 2012-3-22 16:16
Which one of the following best characterizes the alternative hypothesis? The alternative hypothesis is usually the: |
B)
| hypothesis to be proved through statistical testing. |
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C)
| hypothesis that is accepted after a statistical test is conducted. |
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The alternative hypothesis is typically the hypothesis that a researcher hopes to support after a statistical test is carried out. We can reject or fail to reject the null, not 'prove' a hypothesis.
作者: Mechanic 时间: 2012-3-22 16:16
Jill Woodall believes that the average return on equity in the retail industry, µ, is less than 15%. What is null (H0) and alternative (Ha) hypothesis for her study?A)
| H0: µ ≥ 0.15 versus Ha: µ < 0.15. |
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B)
| H0: µ < 0.15 versus Ha: µ = 0.15. |
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C)
| H0: µ = 0.15 versus Ha: µ ≠ 0.15. |
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This is a one-sided alternative because of the “less than” belief. We expect to reject the null.
作者: Mechanic 时间: 2012-3-22 16:16
James Ambercrombie believes that the average return on equity in the utility industry, µ, is greater than 10%. What is null (H0) and alternative (Ha) hypothesis for his study?A)
| H0: µ = 0.10 versus Ha: µ ≠ 0.10. |
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B)
| H0: µ ≥ 0.10 versus Ha: µ < 0.10. |
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C)
| H0: µ ≤ 0.10 versus Ha: µ > 0.10. |
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This is a one-sided alternative because of the “greater than” belief. We expect to reject the null.
作者: Mechanic 时间: 2012-3-22 16:17
What is the most common formulation of null and alternative hypotheses? A)
| Less than for the null and greater than for the alternative. |
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B)
| Equal to for the null and not equal to for the alternative. |
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C)
| Greater than or equal to for the null and less than for the alternative. |
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The most common set of hypotheses will take the form of an equal to statement for the null and a not equal to statement for the alternative.
作者: JonnyKay 时间: 2012-3-22 16:18
Robert Patterson, an options trader, believes that the return on options trading is higher on Mondays than on other days. In order to test his theory, he formulates a null hypothesis. Which of the following would be an appropriate null hypothesis? Returns on Mondays are: A)
| not greater than returns on other days. |
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B)
| greater than returns on other days. |
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C)
| less than returns on other days. |
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An appropriate null hypothesis is one that the researcher wants to reject. If Patterson believes that the returns on Mondays are greater than on other days, he would like to reject the hypothesis that the opposite is true–that returns on Mondays are not greater than returns on other days.
作者: JonnyKay 时间: 2012-3-22 16:19
For a two-tailed test of hypothesis involving a z-distributed test statistic and a 5% level of significance, a calculated z-statistic of 1.5 indicates that:A)
| the null hypothesis cannot be rejected. |
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B)
| the null hypothesis is rejected. |
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C)
| the test is inconclusive. |
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For a two-tailed test at a 5% level of significance the calculated z-statistic would have to be greater than the critical z value of 1.96 for the null hypothesis to be rejected.
作者: JonnyKay 时间: 2012-3-22 16:19
A pitching machine is calibrated to deliver a fastball at a speed of 98 miles per hour. Every day, a technician samples the speed of twenty-five fastballs in order to determine if the machine needs adjustment. Today, the sample showed a mean speed of 99 miles per hour with a standard deviation of 1.75 miles per hour. Assume the population is normally distributed. At a 95% confidence level, what is the t-value in relation to the critical value? A)
| The critical value exceeds the t-value by 1.3 standard deviations. |
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B)
| The t-value exceeds the critical value by 1.5 standard deviations. |
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C)
| The t-value exceeds the critical value by 0.8 standard deviations. |
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t = (99 – 98) / (1.75 / √25) = 2.86. The critical value for a two-tailed test at the 95% confidence level with 24 degrees of freedom is ±2.06 standard deviations. Therefore, the t-value exceeds the critical value by 0.8 standard deviations.
作者: JonnyKay 时间: 2012-3-22 16:20
Ron Jacobi, manager with the Toulee Department of Natural Resources, is responsible for setting catch-and-release limits for Lake Norby, a large and popular fishing lake. For the last two months he has been sampling to determine whether the average length of Northern Pike in the lake exceeds 18 inches (using a significance level of 0.05). Assume that the p-value is 0.08. In concluding that the average size of the fish exceeds 18 inches, Jacobi: |
B)
| makes a Type II error. |
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This statement is an example of a Type I error, or rejection of a hypothesis when it is actually true (also known as the significance level of the test). Here, Ho: μ = 18 inches and Ha: μ > 18 inches. When the p-value is greater than the significance level (0.08 > 0.05), we should fail to reject the null hypothesis. Since Jacobi rejected Ho when it was true, he made a Type 1 error.
The other statements are incorrect. Type II errors occur when you fail to reject a hypothesis when it is actually false (also known as the power of the test).
作者: JonnyKay 时间: 2012-3-22 16:20
Kyra Mosby, M.D., has a patient who is complaining of severe abdominal pain. Based on an examination and the results from laboratory tests, Mosby states the following diagnosis hypothesis: Ho: Appendicitis, HA: Not Appendicitis. Dr. Mosby removes the patient’s appendix and the patient still complains of pain. Subsequent tests show that the gall bladder was causing the problem. By taking out the patient’s appendix, Dr. Mosby:
This statement is an example of a Type II error, which occurs when you fail to reject a hypothesis when it is actually false (also known as the power of the test).
The other statements are incorrect. A Type I error is the rejection of a hypothesis when it is actually true (also known as the significance level of the test).
作者: JonnyKay 时间: 2012-3-22 16:20
Which of the following statements about hypothesis testing is least accurate? A)
| The null hypothesis is a statement about the value of a population parameter. |
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B)
| If the alternative hypothesis is Ha: µ > µ0, a two-tailed test is appropriate. |
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C)
| A Type II error is failing to reject a false null hypothesis. |
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The hypotheses are always stated in terms of a population parameter. Type I and Type II are the two types of errors you can make – reject a null hypothesis that is true or fail to reject a null hypothesis that is false. The alternative may be one-sided (in which case a > or < sign is used) or two-sided (in which case a ≠ is used).
作者: JonnyKay 时间: 2012-3-22 16:20
Which of the following statements about hypothesis testing is most accurate? A Type I error is the probability of:A)
| failing to reject a false hypothesis. |
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B)
| rejecting a true alternative hypothesis. |
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C)
| rejecting a true null hypothesis. |
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The Type I error is the error of rejecting the null hypothesis when, in fact, the null is true.
作者: JonnyKay 时间: 2012-3-22 16:21
Which of the following statements about hypothesis testing is least accurate?A)
| A Type I error is the probability of rejecting the null hypothesis when the null hypothesis is false. |
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B)
| The significance level is the probability of making a Type I error. |
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C)
| A Type II error is the probability of failing to reject a null hypothesis that is not true. |
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A Type I error is the probability of rejecting the null hypothesis when the null hypothesis is true.
作者: JonnyKay 时间: 2012-3-22 16:21
John Jenkins, CFA, is performing a study on the behavior of the mean P/E ratio for a sample of small-cap companies. Which of the following statements is most accurate?A)
| One minus the confidence level of the test represents the probability of making a Type II error. |
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B)
| The significance level of the test represents the probability of making a Type I error. |
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C)
| A Type I error represents the failure to reject the null hypothesis when it is, in truth, false. |
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A Type I error is the rejection of the null when the null is actually true. The significance level of the test (alpha) (which is one minus the confidence level) is the probability of making a Type I error. A Type II error is the failure to reject the null when it is actually false.
作者: JonnyKay 时间: 2012-3-22 16:22
A Type II error: A)
| fails to reject a false null hypothesis. |
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B)
| fails to reject a true null hypothesis. |
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C)
| rejects a true null hypothesis. |
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A Type II error is defined as accepting the null hypothesis when it is actually false. The chance of making a Type II error is called beta risk.
作者: JonnyKay 时间: 2012-3-22 16:22
If we fail to reject the null hypothesis when it is false, what type of error has occured?
A Type II error is defined as failing to reject the null hypothesis when it is actually false.
作者: JonnyKay 时间: 2012-3-22 16:22
Which of the following statements regarding hypothesis testing is least accurate? A)
| The significance level is the risk of making a type I error. |
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B)
| A type I error is acceptance of a hypothesis that is actually false. |
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C)
| A type II error is the acceptance of a hypothesis that is actually false. |
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A type I error is the rejection of a hypothesis that is actually true.
作者: JonnyKay 时间: 2012-3-22 16:23
A Type I error: A)
| rejects a false null hypothesis. |
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B)
| fails to reject a false null hypothesis. |
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C)
| rejects a true null hypothesis. |
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A Type I Error is defined as rejecting the null hypothesis when it is actually true. The probability of committing a Type I error is the significance level or alpha risk.
作者: JonnyKay 时间: 2012-3-22 16:23
Which of the following statements regarding Type I and Type II errors is most accurate? A)
| A Type I error is rejecting the null hypothesis when it is actually true. |
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B)
| A Type I error is failing to reject the null hypothesis when it is actually false. |
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C)
| A Type II error is rejecting the alternative hypothesis when it is actually true. |
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A Type I Error is defined as rejecting the null hypothesis when it is actually true. The probability of committing a Type I error is the risk level or alpha risk.
作者: JonnyKay 时间: 2012-3-22 16:24
A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $59,000 per year. What is the test statistic given a sample of 135 newly acquired CFA charterholders with a mean starting salary of $64,000 and a standard deviation of $5,500?
With a large sample size (135) the z-statistic is used. The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (population standard deviation / (sample size)1/2) = (X − µ) / (σ / n1/2) = (64,000 – 59,000) / (5,500 / 1351/2) = (5,000) / (5,500 / 11.62) = 10.56.
作者: JonnyKay 时间: 2012-3-22 16:24
A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $58,500 per year. What is the test statistic given a sample of 175 newly acquired CFA charterholders with a mean starting salary of $67,000 and a standard deviation of $5,200?
With a large sample size (175) the z-statistic is used. The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (population standard deviation / (sample size)1/2 = (X − µ) / (σ / n1/2) = (67,000 – 58,500) / (5,200 / 1751/2) = (8,500) / (5,200 / 13.22) = 21.62.
作者: JonnyKay 时间: 2012-3-22 16:25
A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $54,000 per year. Assuming a normal distribution, what is the test statistic given a sample of 75 newly acquired CFA charterholders with a mean starting salary of $57,000 and a standard deviation of $1,300?
With a large sample size (75) the z-statistic is used. The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (population standard deviation / (sample size)1/2 = (X − µ) / (σ / n1/2) = (57,000 – 54,000) / (1,300 / 751/2) = (3,000) / (1,300 / 8.66) = 19.99.
作者: JonnyKay 时间: 2012-3-22 16:25
Identify the error type associated with the level of significance and the meaning of a 5 percent significance level. | Error type | α = 0.05 means there is a 5 percent probability of |
A)
| | Type I error | failing to reject a true null hypothesis |
|
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B)
| | Type II error | rejecting a true null hypothesis |
|
|
C)
| | Type I error | rejecting a true null hypothesis |
|
|
The significance level is the risk of making a Type 1 error and rejecting the null hypothesis when it is true.
作者: JonnyKay 时间: 2012-3-22 16:25
A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $57,000 per year. Assuming a normal distribution, what is the test statistic given a sample of 115 newly acquired CFA charterholders with a mean starting salary of $65,000 and a standard deviation of $4,500?
With a large sample size (115) the z-statistic is used. The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (population standard deviation / (sample size)1/2 = (X − µ) / (σ / n1/2) = (65,000 – 57,000) / (4,500 / 1151/2) = (8,000) / (4,500 / 10.72) = 19.06.
作者: JonnyKay 时间: 2012-3-22 16:26
If a two-tailed hypothesis test has a 5% probability of rejecting the null hypothesis when the null is true, it is most likely that the: A)
| probability of a Type I error is 2.5%. |
|
B)
| significance level of the test is 5%. |
|
C)
| power of the test is 95%. |
|
Rejecting the null hypothesis when it is true is a Type I error. The probability of a Type I error is the significance level of the test. The power of a test is one minus the probability of a Type II error, which cannot be calculated from the information given.
作者: chunty 时间: 2012-3-22 16:27
Which of the following statements about hypothesis testing is most accurate? A Type II error is the probability of:A)
| rejecting a true alternative hypothesis. |
|
B)
| failing to reject a false null hypothesis. |
|
C)
| rejecting a true null hypothesis. |
|
The Type II error is the error of failing to reject a null hypothesis that is not true.
作者: chunty 时间: 2012-3-22 16:27
If the probability of a Type I error decreases, then the probability of: A)
| incorrectly accepting the null decreases. |
|
B)
| incorrectly rejecting the null increases. |
|
C)
| a Type II error increases. |
|
If P(Type I error) decreases, then P(Type II error) increases. A null hypothesis is never accepted. We can only fail to reject the null.
作者: chunty 时间: 2012-3-22 16:28
Which of the following statements about hypothesis testing is most accurate?A)
| A Type I error is rejecting the null hypothesis when it is true, and a Type II error is accepting the alternative hypothesis when it is false. |
|
B)
| When the critical Z-statistic is greater than the sample Z-statistic in a two-tailed test, reject the null hypothesis and accept the alternative hypothesis. |
|
C)
| A hypothesized mean of 3, a sample mean of 6, and a standard error of the sampling means of 2 give a sample Z-statistic of 1.5. |
|
Z = (6 - 3)/2 = 1.5. A Type II error is wrongly accepting the null hypothesis. The null hypothesis should be rejected when the sample Z-statistic is greater than the critical Z-statistic.
作者: chunty 时间: 2012-3-22 16:28
A researcher is testing whether the average age of employees in a large firm is statistically different from 35 years (either above or below). A sample is drawn of 250 employees and the researcher determines that the appropriate critical value for the test statistic is 1.96. The value of the computed test statistic is 4.35. Given this information, which of the following statements is least accurate? The test:A)
| indicates that the researcher will reject the null hypothesis. |
|
B)
| indicates that the researcher is 95% confident that the average employee age is different than 35 years. |
|
C)
| has a significance level of 95%. |
|
This test has a significance level of 5%. The relationship between confidence and significance is: significance level = 1 − confidence level. We know that the significance level is 5% because the sample size is large and the critical value of the test statistic is 1.96 (2.5% of probability is in both the upper and lower tails).
作者: chunty 时间: 2012-3-22 16:28
Given a mean of 10% and a standard deviation of 14%, what is a 95% confidence interval for the return next year?
10% +/- 14(1.96) = -17.44% to 37.44%.
作者: chunty 时间: 2012-3-22 16:29
An analyst calculates that the mean of a sample of 200 observations is 5. The analyst wants to determine whether the calculated mean, which has a standard error of the sample statistic of 1, is significantly different from 7 at the 5% level of significance. Which of the following statements is least accurate?: A)
| The mean observation is significantly different from 7, because the calculated Z-statistic is less than the critical Z-statistic. |
|
B)
| The alternative hypothesis would be Ha: mean > 7. |
|
C)
| The null hypothesis would be: H0: mean = 7. |
|
The way the question is worded, this is a two tailed test.The alternative hypothesis is not Ha: M > 7 because in a two-tailed test the alternative is =, while < and > indicate one-tailed tests. A test statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (standard error of the sample statistic) = (5 - 7) / (1) = -2. The calculated Z is -2, while the critical value is -1.96. The calculated test statistic of -2 falls to the left of the critical Z-statistic of -1.96, and is in the rejection region. Thus, the null hypothesis is rejected and the conclusion is that the sample mean of 5 is significantly different than 7. What the negative sign shows is that the mean is less than 7; a positive sign would indicate that the mean is more than 7. The way the null hypothesis is written, it makes no difference whether the mean is more or less than 7, just that it is not 7.
作者: chunty 时间: 2012-3-22 16:29
A goal of an “innocent until proven guilty” justice system is to place a higher priority on: A)
| avoiding type II errors. |
|
|
C)
| avoiding type I errors. |
|
In an “innocent until proven guilty” justice system, the null hypothesis is that the accused is innocent. The hypothesis can only be rejected by evidence proving guilt beyond a reasonable doubt, favoring the avoidance of type I errors.
作者: chunty 时间: 2012-3-22 16:29
If the null hypothesis is innocence, then the statement “It is better that the guilty go free, than the innocent are punished” is an example of preferring a: A)
| type II error over a type I error. |
|
B)
| higher level of significance. |
|
C)
| type I error over a type II error. |
|
The statement shows a preference for accepting the null hypothesis when it is false (a type II error), over rejecting it when it is true (a type I error).
作者: chunty 时间: 2012-3-22 16:30
A bottler of iced tea wishes to ensure that an average of 16 ounces of tea is in each bottle. In order to analyze the accuracy of the bottling process, a random sample of 150 bottles is taken. Using a t-distributed test statistic of -1.09 and a 5% level of significance, the bottler should: A)
| not reject the null hypothesis and conclude that bottles contain an average 16 ounces of tea. |
|
B)
| not reject the null hypothesis and conclude that bottles do not contain an average of 16 ounces of tea. |
|
C)
| reject the null hypothesis and conclude that bottles contain an average 16 ounces of tea. |
|
Ho: µ = 16; Ha: µ ≠ 16. Do not reject the null since |t| = 1.09 < 1.96 (critical value).
作者: chunty 时间: 2012-3-22 16:30
The power of the test is: A)
| the probability of rejecting a false null hypothesis. |
|
B)
| the probability of rejecting a true null hypothesis. |
|
C)
| equal to the level of confidence. |
|
This is the definition of the power of the test: the probability of correctly rejecting the null hypothesis (rejecting the null hypothesis when it is false).
作者: chunty 时间: 2012-3-22 16:30
For a t-distributed test statistic with 30 degrees of freedom, a one-tailed test specifying the parameter greater than some value and a 95% confidence level, the critical value is:
This is the critical value for a one-tailed probability of 5% and 30 degrees of freedom.
作者: chunty 时间: 2012-3-22 16:30
Ryan McKeeler and Howard Hu, two junior statisticians, were discussing the relation between confidence intervals and hypothesis tests. During their discussion each of them made the following statement:
McKeeler: A confidence interval for a two-tailed hypothesis test is calculated as adding and subtracting the product of the standard error and the critical value from the sample statistic. For example, for a level of confidence of 68%, there is a 32% probability that the true population parameter is contained in the interval.
Hu: A 99% confidence interval uses a critical value associated with a given distribution at the 1% level of significance. A hypothesis test would compare a calculated test statistic to that critical value. As such, the confidence interval is the range for the test statistic within which a researcher would not reject the null hypothesis for a two-tailed hypothesis test about the value of the population mean of the random variable.
With respect to the statements made by McKeeler and Hu:
McKeeler’s statement is incorrect. Specifically, for a level of confidence of say, 68%, there is a 68% probability that the true population parameter is contained in the interval. Therefore, there is a 32% probability that the true population parameter is not contained in the interval. Hu’s statement is correct.
作者: chunty 时间: 2012-3-22 16:31
Of the following explanations, which is least likely to be a valid explanation for divergence between statistical significance and economic significance?
While data errors would certainly come to bear on the analysis, in their presence we would not be able to assert either statistical or economic significance. In other words, data errors are not a valid explanation. The others are all mitigating factors that can cause statistically significant results to be less than economically significant.
作者: chunty 时间: 2012-3-22 16:31
Which of the following statements about statistical results is most accurate? A)
| A result may be statistically significant, but may not be economically meaningful. |
|
B)
| If a result is statistically significant and economically meaningful, the relationship will continue into the future. |
|
C)
| If a result is statistically significant, it must also be economically meaningful. |
|
It is possible for an investigation to determine that something is both statistically and economically significant. However, statistical significance does not ensure economic significance. Even if a result is both statistically significant and economically meaningful, the analyst needs to examine the reasons why the economic relationship exists to discern whether it is likely to be sustained in the future.
作者: chunty 时间: 2012-3-22 16:31
An analyst is testing the hypothesis that the mean excess return from a trading strategy is less than or equal to zero. The analyst reports that this hypothesis test produces a p-value of 0.034. This result most likely suggests that the: A)
| best estimate of the mean excess return produced by the strategy is 3.4%. |
|
B)
| smallest significance level at which the null hypothesis can be rejected is 6.8%. |
|
C)
| null hypothesis can be rejected at the 5% significance level. |
|
A p-value of 0.035 means the hypothesis can be rejected at a significance level of 3.5% or higher. Thus, the hypothesis can be rejected at the 10% or 5% significance level, but cannot be rejected at the 1% significance level.
作者: chunty 时间: 2012-3-22 16:32
Brandee Shoffield is the public relations manager for Night Train Express, a local sports team. Shoffield is trying to sell advertising spots and wants to know if she can say with 90% confidence that average home game attendance is greater than 3,000. Attendance is approximately normally distributed. A sample of the attendance at 15 home games results in a mean of 3,150 and a standard deviation of 450. Which of the following statements is most accurate? A)
| With an unknown population variance and a small sample size, no statistic is available to test Shoffield's hypothesis. |
|
B)
| The calculated test statistic is 1.291. |
|
C)
| Shoffield should use a two-tailed Z-test. |
|
We will use the process of Hypothesis testing to determine whether Shoffield should reject Ho:
Step 1: State the Hypothesis
Ho: μ ≤ 3,000
Ha: μ > 3,000
Step 2: Select Appropriate Test Statistic
Here, we have a normally distributed population with an unknown variance (we are given only the sample standard deviation) and a small sample size (less than 30.) Thus, we will use the t-statistic.
Step 3: Specify the Level of Significance
Here, the confidence level is 90%, or 0.90, which translates to a 0.10 significance level.
Step 4: State the Decision Rule
This is a one-tailed test. The critical value for this question will be the t-statistic that corresponds to an α of 0.10, and 14 (n-1) degrees of freedom. Using the t-table , we determine that the appropriate critical value = 1.345. Thus, we will reject the null hypothesis if the calculated test statistic is greater than 1.345.
Step 5: Calculate sample (test) statistic
The test statistic = t = (3,150 – 3,000) / (450 / √ 15) = 1.291
Step 6: Make a decision
Fail to reject the null hypothesis because the calculated statistic is less than the critical value. Shoffield cannot state with 90% certainty that the home game attendance exceeds 3,000.
The other statements are false. As shown above, the appropriate test is a t-test, not a Z-test. There is a test statistic for an normally distributed population, an unknown variance and a small sample size – the t-statistic. There is no test for a non-normal population with unknown variance and small sample size.
作者: chunty 时间: 2012-3-22 16:32
In order to test if the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken. The sample value of the computed z-statistic = 3.4. The appropriate decision at a 5% significance level is to: A)
| reject the null hypotheses and conclude that the population mean is greater than 100. |
|
B)
| reject the null hypothesis and conclude that the population mean is not equal to 100. |
|
C)
| reject the null hypothesis and conclude that the population mean is equal to 100. |
|
Ho:µ ≤ 100; Ha: µ > 100. Reject the null since z = 3.4 > 1.65 (critical value).
作者: chunty 时间: 2012-3-22 16:32
Maria Huffman is the Vice President of Human Resources for a large regional car rental company. Last year, she hired Graham Brickley as Manager of Employee Retention. Part of the compensation package was the chance to earn one of the following two bonuses: if Brickley can reduce turnover to less than 30%, he will receive a 25% bonus. If he can reduce turnover to less than 25%, he will receive a 50% bonus (using a significance level of 10%). The population of turnover rates is normally distributed. The population standard deviation of turnover rates is 1.5%. A recent sample of 100 branch offices resulted in an average turnover rate of 24.2%. Which of the following statements is most accurate? A)
| For the 50% bonus level, the critical value is -1.65 and Huffman should give Brickley a 50% bonus. |
|
B)
| Brickley should not receive either bonus. |
|
C)
| For the 50% bonus level, the test statistic is -5.33 and Huffman should give Brickley a 50% bonus. |
|
Using the process of Hypothesis testing:
Step 1: State the Hypothesis. For 25% bonus level - Ho: m ≥ 30% Ha: m < 30%; For 50% bonus level - Ho: m ≥ 25% Ha: m < 25%.
Step 2: Select Appropriate Test Statistic. Here, we have a normally distributed population with a known variance (standard deviation is the square root of the variance) and a large sample size (greater than 30.) Thus, we will use the z-statistic.
Step 3: Specify the Level of Significance. α = 0.10.
Step 4: State the Decision Rule. This is a one-tailed test. The critical value for this question will be the z-statistic that corresponds to an α of 0.10, or an area to the left of the mean of 40% (with 50% to the right of the mean). Using the z-table (normal table), we determine that the appropriate critical value = -1.28 (Remember that we highly recommend that you have the “common” z-statistics memorized!) Thus, we will reject the null hypothesis if the calculated test statistic is less than -1.28.
Step 5: Calculate sample (test) statistics. Z (for 50% bonus) = (24.2 – 25) / (1.5 / √ 100) = −5.333. Z (for 25% bonus) = (24.2 – 30) / (1.5 / √ 100) = −38.67.
Step 6: Make a decision. Reject the null hypothesis for both the 25% and 50% bonus level because the test statistic is less than the critical value. Thus, Huffman should give Soberg a 50% bonus.
The other statements are false. The critical value of –1.28 is based on the significance level, and is thus the same for both the 50% and 25% bonus levels.
作者: chunty 时间: 2012-3-22 16:33
Which of the following statements about test statistics is least accurate?A)
| In a test of the population mean, if the population variance is unknown and the sample is small, we should use a z-distributed test statistic. |
|
B)
| In the case of a test of the difference in means of two independent samples, we use a t-distributed test statistic. |
|
C)
| In a test of the population mean, if the population variance is unknown, we should use a t-distributed test statistic. |
|
If the population sampled has a known variance, the z-test is the correct test to use. In general, a t-test is used to test the mean of a population when the population is unknown. Note that in special cases when the sample is extremely large, the z-test may be used in place of the t-test, but the t-test is considered to be the test of choice when the population variance is unknown. A t-test is also used to test the difference between two population means while an F-test is used to compare differences between the variances of two populations.
作者: chunty 时间: 2012-3-22 16:33
In a test of the mean of a population, if the population variance is:A)
| known, a z-distributed test statistic is appropriate. |
|
B)
| known, a t-distributed test statistic is appropriate. |
|
C)
| unknown, a z-distributed test statistic is appropriate. |
|
If the population sampled has a known variance, the z-test is the correct test to use. In general, a t-test is used to test the mean of a population when the population variance is unknown. Note that in special cases when the sample is extremely large, the z-test may be used in place of the t-test, but the t-test is considered to be the test of choice when the population variance is unknown.
作者: chunty 时间: 2012-3-22 16:33
In order to test whether the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken and the sample value of the computed test statistic, tn-1 = 3.4. If you choose a 5% significance level you should: A)
| reject the null hypothesis and conclude that the population mean is greater that 100. |
|
B)
| fail to reject the null hypothesis and conclude that the population mean is less than or equal to 100. |
|
C)
| fail to reject the null hypothesis and conclude that the population mean is greater than 100. |
|
At a 5% significance level, the critical t-statistic using the Student’s t distribution table for a one-tailed test and 29 degrees of freedom (sample size of 30 less 1) is 1.699 (with a large sample size the critical z-statistic of 1.645 may be used). Because the calculated t-statistic of 3.4 is greater than the critical t-statistic of 1.699, meaning that the calculated t-statistic is in the rejection range, we reject the null hypothesis and we conclude that the population mean is greater than 100.
作者: chunty 时间: 2012-3-22 16:34
In a two-tailed hypothesis test, Jack Olson observes a t-statistic of -1.38 based on a sample of 20 observations where the population mean is zero. If you choose a 5% significance level, you should: A)
| reject the null hypothesis and conclude that the population mean is significantly different from zero. |
|
B)
| fail to reject the null hypothesis that the population mean is not significantly different from zero. |
|
C)
| reject the null hypothesis and conclude that the population mean is not significantly different from zero. |
|
At a 5% significance level, the critical t-statistic using the Student’s t distribution table for a two-tailed test and 19 degrees of freedom (sample size of 20 less 1) is ± 2.093 (with a large sample size the critical z-statistic of 1.960 may be used). Because the critical t-statistic of -2.093 is to the left of the calculated t-statistic of -1.38, meaning that the calculated t-statistic is not in the rejection range, we fail to reject the null hypothesis that the population mean is not significantly different from zero.
作者: chunty 时间: 2012-3-22 16:34
A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $62,500 per year. What is the test statistic given a sample of 125 newly acquired CFA charterholders with a mean starting salary of $65,000 and a standard deviation of $2,600?
With a large sample size (125) and an unknown population variance, either the t-statistic or the z-statistic could be used. Using the z-statistic, it is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. The test statistic = (sample mean – hypothesized mean) / (sample standard deviation / (sample size1/2)) = (X − µ) / (s / n1/2) = (65,000 – 62,500) / (2,600 / 1251/2) = (2,500) / (2,600 / 11.18) = 10.75.
作者: chunty 时间: 2012-3-22 16:34
Ken Wallace is interested in testing whether the average price to earnings (P/E) of firms in the retail industry is 25. Using a t-distributed test statistic and a 5% level of significance, the critical values for a sample of 40 firms is (are):
There are 40 − 1 = 39 degrees of freedom and the test is two-tailed. Therefore, the critical t-values are ± 2.023. The value 2.023 is the critical value for a one-tailed probability of 2.5%.
作者: chunty 时间: 2012-3-22 16:35
Simone Mak is a television network advertising executive. One of her responsibilities is selling commercial spots for a successful weekly sitcom. If the average share of viewers for this season exceeds 8.5%, she can raise the advertising rates by 50% for the next season. The population of viewer shares is normally distributed. A sample of the past 18 episodes results in a mean share of 9.6% with a standard deviation of 10.0%. If Mak is willing to make a Type 1 error with a 5% probability, which of the following statements is most accurate? A)
| Mak cannot charge a higher rate next season for advertising spots based on this sample. |
|
B)
| With an unknown population variance and a small sample size, Mak cannot test a hypothesis based on her sample data. |
|
C)
| The null hypothesis Mak needs to test is that the mean share of viewers is greater than 8.5%. |
|
Mak cannot conclude with 95% confidence that the average share of viewers for the show this season exceeds 8.5 and thus she cannot charge a higher advertising rate next season.
Hypothesis testing process:
Step 1: State the hypothesis. Null hypothesis: mean ≤ 8.5%; Alternative hypothesis: mean > 8.5%
Step 2: Select the appropriate test statistic. Use a t statistic because we have a normally distributed population with an unknown variance (we are given only the sample variance) and a small sample size (less than 30). If the population were not normally distributed, no test would be available to use with a small sample size.
Step 3: Specify the level of significance. The significance level is the probability of a Type I error, or 0.05.
Step 4: State the decision rule. This is a one-tailed test. The critical value for this question will be the t-statistic that corresponds to a significance level of 0.05 and n-1 or 17 degrees of freedom. Using the t-table, we determine that we will reject the null hypothesis if the calculated test statistic is greater than the critical value of 1.74.
Step 5: Calculate the sample (test) statistic. The test statistic = t = (9.6 – 8.5) / (10.0 / √ 18) = 0.479 (Note: Remember to use standard error in the denominator because we are testing a hypothesis about the population mean based on the mean of 18 observations.)
Step 6: Make a decision. The calculated statistic is less than the critical value. Mak cannot conclude with 95% confidence that the mean share of viewers exceeds 8.5% and thus she cannot charge higher rates.
Note: By eliminating the two incorrect choices, you can select the correct response to this question without performing the calculations.
作者: chunty 时间: 2012-3-22 16:35
Roy Fisher, CFA, wants to determine whether there is a significant difference, at the 5% significance level, between the mean monthly return on Stock GHI and the mean monthly return on Stock JKL. Fisher assumes the variances of the two stocks’ returns are equal. Using the last 12 months of returns on each stock, Fisher calculates a t-statistic of 2.0 for a test of equality of means. Based on this result, Fisher’s test: A)
| rejects the null hypothesis, and Fisher can conclude that the means are equal. |
|
B)
| rejects the null hypothesis, and Fisher can conclude that the means are not equal. |
|
C)
| fails to reject the null hypothesis. |
|
The null hypothesis for a test of equality of means is H0: μ1 − μ2 = 0. Assuming the variances are equal, degrees of freedom for this test are (n1 + n2 − 2) = 12 + 12 − 2 = 22. From the table of critical values for Student’s t-distribution, the critical value for a two-tailed test at the 5% significance level for df = 22 is 2.074. Because the calculated t-statistic of 2.0 is less than the critical value, this test fails to reject the null hypothesis that the means are equal.
作者: chunty 时间: 2012-3-22 16:36
A test of a hypothesis that the means of two normally distributed populations are equal based on two independent random samples:A)
| is a paired-comparisons test. |
|
B)
| is based on a Chi Square statistic. |
|
C)
| is done with a t-statistic. |
|
We have two formulas for test statistics for the hypothesis of equal sample means. Which one we use depends on whether or not we assume the samples have equal variances. Either formula generates a test statistic that follows a T-distribution.
作者: chunty 时间: 2012-3-22 16:36
Joe Sutton is evaluating the effects of the 1987 market decline on the volume of trading. Specifically, he wants to test whether the decline affected trading volume. He selected a sample of 500 companies and collected data on the total annual volume for one year prior to the decline and for one year following the decline. What is the set of hypotheses that Sutton is testing?A)
| H0: µd = µd0 versus Ha: µd ≠ µd0. |
|
B)
| H0: µd ≠ µd0 versus Ha: µd = µd0. |
|
C)
| H0: µd = µd0 versus Ha: µd > µd0. |
|
This is a paired comparison because the sample cases are not independent (i.e., there is a before and an after for each stock). Note that the test is two-tailed, t-test.
作者: chunty 时间: 2012-3-22 16:36
An analyst wants to determine whether the monthly returns on two stocks over the last year were the same or not. What test should she use if she is willing to assume that the returns are normally distributed? A)
| A difference in means test only if the variances of monthly returns are equal for the two stocks. |
|
B)
| A difference in means test with pooled variances from the two samples. |
|
C)
| A paired comparisons test because the samples are not independent. |
|
A paired comparisons test must be used. The difference in means test requires that the samples be independent. Portfolio theory teaches us that returns on two stocks over the same time period are unlikely to be independent since both have some systematic risk.
作者: chunty 时间: 2012-3-22 16:36
The use of the F-distributed test statistic, F = s12 / s22, to compare the variances of two populations does NOT require which of the following?A)
| populations are normally distributed. |
|
B)
| two samples are of the same size. |
|
C)
| samples are independent of one another. |
|
The F-statistic can be computed using samples of different sizes. That is, n1 need not be equal to n2.
作者: torontoanalyst 时间: 2012-3-22 16:38
The variance of 100 daily stock returns for Stock A is 0.0078. The variance of 90 daily stock returns for Stock B is 0.0083. Using a 5% level of significance, the critical value for this test is 1.61. The most appropriate conclusion regarding whether the variance of Stock A is different from the variance of Stock B is that the: |
B)
| variance of Stock B is significantly greater than the variance of Stock A. |
|
C)
| variances are not equal. |
|
A test of the equality of variances requires an F-statistic. The calculated F-statistic is 0.0083/0.0078 = 1.064. Since the calculated F value of 1.064 is less than the critical F value of 1.61, we cannot reject the null hypothesis that the variances of the 2 stocks are equal.
作者: torontoanalyst 时间: 2012-3-22 16:38
Which of the following statements about the variance of a normally distributed population is least accurate?A)
| The Chi-squared distribution is a symmetric distribution. |
|
B)
| The test of whether the population variance equals σ02 requires the use of a Chi-squared distributed test statistic, [(n − 1)s2] / σ02. |
|
C)
| A test of whether the variance of a normally distributed population is equal to some value σ02, the hypotheses are: H0: σ2 = σ02, versus Ha: σ2 ≠ σ02. |
|
The Chi-squared distribution is not symmetrical, which means that the critical values will not be numerically equidistant from the center of the distribution, though the probability on either side of the critical values will be equal (that is, if there is a 5% level of significance and a two-sided test, 2.5% will lie outside each of the two critical values).
作者: torontoanalyst 时间: 2012-3-22 16:39
A test of the population variance is equal to a hypothesized value requires the use of a test statistic that is: |
|
C)
| Chi-squared distributed. |
|
In tests of whether the variance of a population equals a particular value, the chi-squared test statistic is appropriate.
作者: torontoanalyst 时间: 2012-3-22 16:39
The test of the equality of the variances of two normally distributed populations requires the use of a test statistic that is: |
B)
| Chi-squared distributed. |
|
|
The F-distributed test statistic, F = s12 / s22, is used to compare the variances of two populations.
作者: torontoanalyst 时间: 2012-3-22 16:39
Abby Ness is an analyst for a firm that specializes in evaluating firms involved in mineral extraction. Ness believes that the earnings of copper extracting firms are more volatile than those of bauxite extraction firms. In order to test this, Ness examines the volatility of returns for 31 copper firms and 25 bauxite firms. The standard deviation of earnings for copper firms was $2.69, while the standard deviation of earnings for bauxite firms was $2.92. Ness’s Null Hypothesis is σ12 = σ22. Based on the samples, can we reject the null hypothesis at a 95% confidence level using an F-statistic and why? Null is: A)
| rejected. The F-value exceeds the critical value by 0.849. |
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B)
| not rejected. The critical value exceeds the F-value by 0.71. |
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C)
| rejected. The F-value exceeds the critical value by 0.71. |
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F = s12 / s22 = $2.922 / $2.692 = 1.18
From an F table, the critical value with numerator df = 24 and denominator df = 30 is 1.89.
作者: torontoanalyst 时间: 2012-3-22 16:40
In order to test if Stock A is more volatile than Stock B, prices of both stocks are observed to construct the sample variance of the two stocks. The appropriate test statistics to carry out the test is the:
The F test is used to test the differences of variance between two samples.
作者: torontoanalyst 时间: 2012-3-22 16:40
Which of the following statements about parametric and nonparametric tests is least accurate?A)
| The test of the difference in means is used when you are comparing means from two independent samples. |
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B)
| Nonparametric tests rely on population parameters. |
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C)
| The test of the mean of the differences is used when performing a paired comparison. |
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Nonparametric tests are not concerned with parameters; they make minimal assumptions about the population from which a sample comes. It is important to distinguish between the test of the difference in the means and the test of the mean of the differences. Also, it is important to understand that parametric tests rely on distributional assumptions, whereas nonparametric tests are not as strict regarding distributional properties.
作者: torontoanalyst 时间: 2012-3-22 16:40
Which of the following statements about parametric and nonparametric tests is least accurate?A)
| Parametric tests are most appropriate when a population is heavily skewed. |
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B)
| Nonparametric tests have fewer assumptions than parametric tests. |
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C)
| Nonparametric tests are often used in conjunction with parametric tests. |
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For a distribution that is non-normally distributed, a nonparametric test may be most appropriate. A nonparametric test tends to make minimal assumptions about the population, while parametric tests rely on assumptions regarding the distribution of the population. Both kinds of tests are often used in conjunction with one another.
作者: terpsichorefan 时间: 2013-3-20 02:09
The hypothesis testing is the trickiest part to me..... thanks a lot!
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