500 câu trắc nghiệm Kinh tế lượng – 7C
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution, or neither.
14) Claim \(\mu \) = 77. Sample data: n = 20, \({\bar x}\) = 110, s = 15.2. The sample data appear to come from a population with a distribution that is very far from normal and \(\sigma \) is unknown.
○ Student t
○ Normal
● Neither
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Assume that a simple random sample has been selected from a normally distributed population. Find the test statistics, p-value, critical value(s), and state the final conclusion.
15) Test the claim that the mean lifetime of car engines of a particular type is greater than 220,000 miles. Sample data are summarized as n = 23, \({\bar x}\) = 226,450 miles, and s = 11,500 miles. Use a significane level of \(\alpha \) = 0.01
\(\alpha \) = 0.05, Test statistics, t = 2.690, 0.05 < p-value < 0.01. Critical value, t = 2.508. Because the test statistics exceeds the critical value, we fail to reject the null hypothesis. The sample data support the claim that \(\mu \) > 220,000 miles
Test the given claim using the traditional method of hypothesis testing. Assume that the sample has been randomly selected from a population with a normal distribution.
16) A manufacturer makes ball bearings that are supposed to have a mean weight of 30 g. A retailer suspects that the mean weight is actually less than 30g. The mean weight for a random sample of 16 ball bearings is 29.5g with a standard deviation of 4.1g. At the 0.05 significane level, test the claim that the mean is less than 30g.
Test statistics, t = -0.488, Critical value, t = -1.753. Fail to reject the null hypothesis. There is not sufficent sample evidence to support the claim that the mean is less than 30g.
17) In test of a coumputer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increased the time between failures. Tests on random sample of 10 modified components resulted in the following times (in hours) between failures.
518 548 561 523 536 499 538 557 528 563
At the 0.05 significane level, test the claim that for the modified components, the mean time between failures is greater than 520 hours.
Test statistics, t = 2.612, Critical value, t = 1.833. Reject H0: \(\mu \) = 520 hours. The sample data support the claim that the mean is greater than 520 hours.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Find the critical value or values of \({\chi ^2}\) based on the given information.
18) H1: \(\sigma \) \( \ne \) 9.3, n = 28, \(\alpha \) = 0.05
● 14.573; 43,194
○ -14.573; 14.573
○ -40.113; 40.113
○ 16.151; 40.113
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Use the traditional method to test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected.
19) A manufacturer uses a new production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standard deviation of 2.4 cm. At the 0.10 significane level, test the claim that the new production method has lengths with a standard deviation different from 3.5cm, which was standard deviation for the old method.
Test statistics: \(\chi _{n – 1}^2 = \frac{{(n – 1){s^2}}}{{{\sigma ^2}}}\) = 7.523; Critical value: \({\chi ^2}\) = 7.962; 26.296. Reject H0. The sample data support the claim that the standard deviation is different from 3.5.
20) Height of men aged 25 to 34 have a standard deviation of 2.9 inches. Use a 0.05 significane level to test the claim that the heights of women aged 25 to 34 have a different standard deviation. The heights (in inches) of 16 randomly selected women aged 25 to 34 are listed below: 62.13 65.09 64.18 66.72 63.09 61.15 67.50 64.65 63.80 64.21 60.17 68.28 66.49 62.10 65.73 64.72
Test statistics: \(\chi _{n – 1}^2 = \frac{{(n – 1){s^2}}}{{{\sigma ^2}}}\) = 9.260; Critical value: \({\chi ^2}\) = 6.262; 27.488. Fail to reject H0. There is not sufficent sample evidence to support the claim that the heights of women aged 25 to 34 have a standard deviation different from 2.9 in.