Q5. 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) fail to reject the null hypothesis and conclude that the population mean is less than or equal to 100.
B) fail to reject the null hypothesis and conclude that the population mean is greater than 100.
C) reject the null hypothesis and conclude that the population mean is greater that 100.
Correct answer is C)
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.
Q6. 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) fail to reject the null hypothesis that the population mean is not significantly different from zero.
B) reject the null hypothesis and conclude that the population mean is significantly different from zero.
C) reject the null hypothesis and conclude that the population mean is not significantly different from zero.
Correct answer is A)
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.
Q7. The table below is for five samples drawn from five separate populations. The far left columns give information on the population distribution, population variance, and sample size. The right-hand columns give three choices for the appropriate tests: z = z-statistic, and t = t-statistic. “None” means that a test statistic is not available.
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Sampling From
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Test Statistic Choices
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Distribution
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Variance
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n
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One
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Two
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Three
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|
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Non-normal
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0.75
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100
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z
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z
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z
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|
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Normal
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5.60
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75
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z
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z
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z
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|
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Non-normal
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n/a
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15
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t
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t
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none
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|
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Normal
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n/a
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18
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t
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t
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t
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|
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Non-normal
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14.3
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15
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z
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t
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none
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Which set of test statistic choices (One, Two, or Three) matches the correct test statistic to the sample for all five samples?
A) Three.
B) One.
C) Two.
Correct answer is A)
For the exam: COMMIT THE FOLLOWING TABLE TO MEMORY!
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When you are sampling from a:
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and the sample size is small, use a:
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and the sample size is large, use a:
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Normal distribution with a known variance
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z-statistic
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z-statistic
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Normal distribution with an unknown variance
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t-statistic
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t-statistic
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Nonnormal distribution with a known variance
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not available
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z-statistic
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Nonnormal distribution with an unknown variance
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not available
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t-statistic
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Q8. 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) 0.96.
C) 10.75.
Correct answer is C)
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.
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发表于 2009-5-30 21:31
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