500 câu trắc nghiệm Kinh tế lượng – 8A

Find sd.
14) The differences between two sets of dependent data are -3 -21 -12 -3 -27. Round to the nearest tenth.
○ 21.4
○ 13.9
● 10.7
○ 8.6

Assume that you want to test the claim that the paired sample data come from a population for which the
mean difference is \({\mu _d}\) = 0. Compute the value of the t test statistic.
15)

○ t = 2.890
● t =1.292
○ t = 0.415
○ t = 0.578

Determine the decision criterion for rejecting the null hypothesis in the given hypothesis test; i.e., describe the values of the test statistic that would result in rejection of the null hypothesis.
16) Suppose you wish to test the claim that \({\mu _d}\), the mean value of the differences d for a population of paired data, is different from 0. Given a sample of n = 23 and a significance level of \(\alpha \) = 0.05, what criterion would be used for rejecting the null hypothesis?
● Reject null hypothesis if test statistic > 2.074 or < -2.074. ○ Reject null hypothesis if test statistic > 1.717 or < -1.717. ○ Reject null hypothesis if test statistic > 2.069 or < -2.069. ○ Reject null hypothesis if test statistic > 1.717.

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Use the traditional method of hypothesis testing to test the given claim about the means of two populations. Assume that two dependent samples have been randomly selected from normally distributed populations.
17) Five students took a math test before and after tutoring. Their scores were as follows.

Using a 0.01 level of significance, test the claim that the tutoring has an effect on the math scores.

Test statistic: t = -2.134. Critical value: t = -3.747.
Fail to reject H0: \({\mu _d}\) = 0. There is not sufficient evidence to support the claim that the tutoring has an
effect.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Construct a confidence interval for \({\mu _d}\), the mean of the differences d for the population of paired data. Assume that the population of paired differences is normally distributed.
18) A test of writing ability is given to a random sample of students before and after they completed a formal writing course. The results are given below. Construct a 99% confidence interval for the mean difference between the before and after scores.

○ -0.2 < \({\mu _d}\) < 4.2
○ 1.2 < \({\mu _d}\) < 2.8
● -0.5 < \({\mu _d}\) < 4.5
○ -0.1 < \({\mu _d}\) < 4.1 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Test the indicated claim about the variances or standard deviations of two populations. Assume that the populations are normally distributed. Assume that the two samples are independent and that they have been randomly selected. 19) A random sample of 16 women resulted in blood pressure levels with a standard deviation of 22.8 mm Hg. A random sample of 17 men resulted in blood pressure levels with a standard deviation of 19.9 mm Hg. Use a 0.025 significance level to test the claim that blood pressure levels for women have a larger variance than those for men. H0: \(\sigma _1^2\) = \(\sigma _2^2\). H1: \(\sigma _1^2\) > \(\sigma _2^2\).
Test statistic: F = 1.3127. Critical value: F = 2.7875.
Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that blood pressure levels for women have a larger variance than those for men.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
20) When performing a hypothesis test for the ratio of two population variances, the upper critical F value is denoted FR. The lower critical F value, FL, can be found as follows: interchange the degrees of freedom, and then take the reciprocal of the resulting F value found in table A-5. FR can be denoted \({F_{\alpha /2}}\) and FL can be denoted \({F_{1- \alpha /2}}\) .
Find the critical values FL and FR for a two-tailed hypothesis test based on the following
values:

n1 = 10, n2 = 16, \(\alpha \) = 0.05
○ 0.3202, 3.1227
○ 3.1227, 3.7743
○ 0.2653, 3.7743
● 0.2653, 3.1227