500 câu trắc nghiệm Kinh tế lượng – 11A
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 11A bao gồm:
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Provide an appropriate response.
1) Describe the null and alternative hypotheses for one-way ANOVA. Give an example.
The null hypothesis for one-way ANOVA is that three or more means are equal. The alternative hypothesis is that the means are not all equal. Examples will vary.
2) When using statistical software packages, the critical value is typically not given. What method is used to determine whether you reject or fail to reject the null hypothesis?
The decision to reject or fail to reject is based on P-values. If the P-value is less than or equal to the significance level, you reject the null hypothesis. Otherwise you fail to reject.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
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.
3)
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 value of the test statistic.
○ 13.500
● 5.17
○ 4.500
○ 0.011
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 |
What can you conclude about the equality of the population means?
○ 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 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.
5) Given the sample data below, test the claim that the populations have the same mean. Use a significance level of 0.05.
Brand A | Brand B | Brand C |
n = 10 | n = 10 | n = 10 |
\({\bar x}\) = 31.1 | \({\bar x}\) = 31.8 | \({\bar x}\) = 27.3 |
s2 = 4.29 | s2 = 4.84 | s2 = 3.73 |
Test statistic: F = 13.678. Critical value: F = 3.35.
Reject the claim of equal means. The different brands do not appear to have the same mean
6) 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 |
17 | 18 | 21 | 22 |
20 | 18 | 24 | 25 |
21 | 23 | 25 | 27 |
22 | 25 | 26 | 29 |
21 | 26 | 29 | 35 |
29 | 36 | ||
37 |
H0: \({\mu _1}\) = \({\mu _2}\) = \({\mu _3}\) = \({\mu _4}\). H1: The means are not all equal.
Test statistic: F = 6.6983. Critical value: F = 3.9034.
Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the four brands have the same mean.
Provide an appropriate response.
7) Four independent samples of 100 values each are randomly drawn from populations that are normally distributed with equal variances. You wish to test the claim that \({\mu _1}\) = \({\mu _2}\) = \({\mu _3}\) = \({\mu _4}\).
i) If you test the individual claims \({\mu _1}\) = \({\mu _2}\), \({\mu _1}\) = \({\mu _3}\), \({\mu _1}\) = \({\mu _4}\), . . . , \({\mu _3}\) = \({\mu _4}\), how many ways can you pair off the 4 means?
ii) Assume that the tests are independent and that for each test of equality between two means, there is a 0.99 probability of not making a type I error. If all possible pairs of means are tested for equality, what is the probability of making no type I errors?
iii) If you use analysis of variance to test the claim that \({\mu _1}\) = \({\mu _2}\) = \({\mu _3}\) = \({\mu _4}\) at the 0.01 level of significance, what is the probability of not making a type I error?
i) 6
ii) 0.996= 0.9415
iii) 0.99
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 0.05 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 8oC is added to each temperature listed for greenhouse #3?
i) Reject the claim that the average temperature is the same in each greenhouse since F = 24.3 > F0.05(2, 18) = 3.55.
ii) Accept the claim that the average temperature is the same in each greenhouse since F = 0.128 < F0.05(2, 18) = 3.55.