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500 câu trắc nghiệm Kinh tế lượng – 11B

Tổng hợp 500 câu trắc nghiệm + tự luận Kinh tế lượng (Elementary Statistics). Tất cả các câu hỏi trắc nghiệm + tự luận đều có đáp án. Nội dung được khái quát trong 13 phần, mỗi phần gồm 3 bài kiểm tra (A, B, C). Các câu hỏi trắc nghiệm + tự luận bám rất sát chương trình kinh tế lượng, đặc biệt là phần thống kê, rất phù hợp cho các bạn củng cố và mở rộng các kiến thức về Kinh tế lượng. Các câu hỏi trắc nghiệm + tự luận của phần 11B bao gồm:

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Provide an appropriate response.
1) Suppose you are to test for equality of four different means, with H0 : \({\mu _A}\) = \({\mu _B}\) = \({\mu _C}\) = \({\mu _D}\). Write the hypotheses for the paired tests. Use methods of probability to explain why the process of ANOVA has a higher degree of confidence than testing each of the pairs separately.

The six paired hypotheses are \({\mu _A}\) = \({\mu _B}\), \({\mu _A}\) = \({\mu _C}\), \({\mu _A}\) = \({\mu _D}\), \({\mu _C}\) = \({\mu _B}\), \({\mu _D}\) = \({\mu _B}\), \({\mu _C}\) = \({\mu _D}\). Suppose we test
each with a 5% significance level (95% confidence level). Then, the degree of confidence for all six would
be 0.95% or 0.735, yielding an excessively high risk of a type I error. ANOVA maintains the 5% significance level while testing equivalence of all four means.

2) Define the term “treatment”. What other term means the same thing? Give an example.

A treatment (also known as a factor) is a property or characteristic that allows us to distinguish the
different populations from one another. Examples will vary.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
3) Fill in the missing entries in the following partially completed one-way ANOVA table.

Source df SS MS=SS/df F
Treatment 3 11.17
Error 13.72 0.686
Total

Source df SS MS=SS/df F
Treatment 3 2.55 7.66 11.17
Error 20 13.72 0.686
Total 23 16.27

Source df SS MS=SS/df F
Treatment 3 0.184 0.061 11.17
Error 20 13.72 0.686
Total 23 13.90

Source df SS MS=SS/df F
Treatment 3 22.97 7.66 11.17
Error 20 13.72 0.68 6
Total 23 36.69

Source df SS MS=SS/df F
Treatment 3 48.80 16.27 11.17
Error 20 13.72 0.686
Total 23 62.52

Given below are the analysis of variance results from a Minitab display. Assume that you want to use a
0.05 significance level in testing the null hypothesis that the different samples come from populations with
the same mean.
4)

Source DF SS MS F p
Factor 3 13.500 4.500 5.17 0.011
Error 16 13.925 0.870
Total 19 27.425

Identify the p-value.
○ 5.17
○ 4.500
○ 0.870
● 0.011

5)

Source DF SS MS F p
Factor 3 30 10.00 1.6 0.264
Error 8 50 6.25
Total 11 80

What can you conclude about the equality of the population means?
○ Reject the null hypothesis since the p-value is greater than the significance level.
○ Accept the null hypothesis since the p-value is less than the significance level.
● Accept the null hypothesis since the p-value is greater than the significance level.
○ Reject the null hypothesis since the p-value is less than the significance level.

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Test the claim that the samples come from populations with the same mean. Assume that the populations are normally distributed with the same variance.
6) A consumer magazine wants to compare the lifetimes of ballpoint pens of three different types. The magazine takes a random sample of pens of each type. Results are shown in the following table.

Brand 1 Brand 2 Brand 3
260 181 238
218 240 257
184 162 241
219 218 213

Do the data indicate that there is a difference in mean lifetime for the three brands of ballpoint pens? Use \(\alpha \) = 0.01.

Test statistic: F = 1.620. Critical value: F = 8.0215.
Fail to reject the claim of equal means. The data do not provide sufficient evidence to conclude that there is a difference in the mean lifetime of the three brands of ballpoint pen.

7) At the 0.025 significance level, test the claim that the four brands have the same mean if the following sample results have been obtained.

Brand A Brand B Brand C Brand D
15 20 21 15
25 17 22 15
21 22 20 14
23 23 19 23
22 18 22
20 28
28

H0: \({\mu _1}\) = \({\mu _2}\) = \({\mu _3}\) = \({\mu _4}\). H1: The means are not all equal.
Test statistic: F = 0.0555. Critical value: F = 3.9539.
Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the four brands have the same mean.

Provide an appropriate response.
8) At the same time each day, a researcher records the temperature in each of three greenhouses.
The table shows the temperatures in degrees Fahrenheit recorded for one week.

Greenhouse #1 Greenhouse #2 Greenhouse #3
73 71 67
72 69 63
73 72 62
66 72 61
68 65 60
71 73 62
72 71 59

i) Use a 0.025 significance level to test the claim that the average temperature is the same in each greenhouse.
ii) How are the analysis of variance results affected if the same constant is added to every one of the original sample values?

i) Reject the claim that the average temperature is the same in each greenhouse since F = 24.3 > F0.025(2, 18) = 4.5597.
ii) The analysis of variance results are not affected