chunty

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.

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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.

chunty

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.

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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.

chunty

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.

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Ho:µ ≤ 100; Ha: µ > 100. Reject the null since z = 3.4 > 1.65 (critical value).

chunty

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.

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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.

chunty

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.

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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.

chunty

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.

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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.

chunty

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.

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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 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.

chunty

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.

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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 -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.

chunty

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?

A) 10.75.

B) -10.75.

C) 0.96.

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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.

chunty

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):

A) -2.023 and 2.023.

B) -1.96 and 1.96.

C) -1.685 and 1.685.

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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%.