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500 câu trắc nghiệm Kinh tế lượng – 10C

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 10C bao gồm:

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Provide an appropriate response.
1) In the chi-square test of independence, the formula used is \({\chi ^2} = \sum {\frac{{{{\left( {O – E} \right)}^2}}}{E}} \). Discuss the meaning of O and E and explain the circumstances under which the c2 values will be smaller or larger. What is the relationship between a significant c2 value and the values of O and E?

The O represents the observed frequencies. The E represents the expected frequencies based on the assumption of independence. The c2value will be smaller when the difference between observed and expected frequencies is small and will be larger when the difference between observed and expected values is large. The c2value will be significant when there is a significant difference between the observed and expected values.

2) Explain the computation of expected values for contingency tables in terms of probabilities. Refer to the assumptions of the null hypothesis as part of your explanation. You might give a brief example to illustrate.

Suppose A and B are two categories in a contingency table. In probability computations, the P(A and B) would be computed as P(A)œP(B), provided A and B are independent. The assumption of the null hypothesis is that A and B are in fact independent, so we use the formula P(A and B) = P(A).P(B)

Since P(A) = (# occurrences A)/(total occurrences)
and P(B) = (# occurrences B) /(total occurrences), then
P(A and B) = [(# occurrences A)/(total occurrences)].[(# occurrences B)/(total occurrences)]

Then the expected number of outcomes for A and B would be
P(A and B).(total occurrences) = [(# occurrences A)/(total occurrences)].[(# occurrences B)/(total occurrences0].[total occurrences] or P(A and B).(total occurrences) = [(# occurrences A).(# occurrences B)]/(total occurrences)
This is also the formula for the expected frequency for each cell in a contingency table,
E = [(# occurrences A).(# occurrences B)]/(total occurrences)

So the computation of the expected values is based on the assumption of independence. Examples may be given and will vary.

3) Draw an example of a representative chi-square distribution and discuss three characteristics of a chi-square distribution. Show an example of the special case of the chi-square distribution for only 1 or 2 degrees of freedom.

Any drawing such as that shown in Figure 10-1 will suffice. The chi-square distribution is not symmetric. The values of a chi-square distribution can be 0 or positive, but they cannot be negative. The chi-square distribution is different for each number of degrees of freedom. Examples will vary. Possibilities are a 2 x 2 contingency table for one degree of freedom and a goodness-of-fit test of three categories for two degrees of freedom.

Perform the indicated goodness-of-fit test.
4) Among the four northwestern states, Washington has 51% of the total population, Oregon has 30%, Idaho has 11%, and Montana has 8%. A market researcher selects a sample of 1000 subjects, with 450 in Washington, 340 in Oregon, 150 in Idaho, and 60 in Montana. At the 0.05 significance level, test the claim that the sample of 1000 subjects has a distribution that agrees with the distribution of state populations.

H0: The distribution of the sample agrees with the population distribution.
H1: It does not agree.
Test statistic: \({\chi ^2}\) = 31.938. Critical value: \({\chi ^2}\) = 7.815. Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the distribution of the sample agrees with the distribution of the state populations.

Provide an appropriate response.
5) An observed frequency distribution of exam scores is as follows:

Exam Score Under 60 60 – 69 70 – 79 80 – 89 90 – 100
Frequency 30 30 140 60 40

i) Assuming a normal distribution with \(\mu \) = 75 and \(\sigma \) = 15, find the probability of a randomly selected subject belonging to each class. (Use boundaries of 59.5, 69.5, 79.5, 89.5.)
ii) Using the probabilities found in part (i), find the expected frequency for each category.

iii) Use a 0.05 significance level to test the claim that the exam scores were randomly selected from a normally distributed population with \(\mu \) = 75 = 75 and \(\sigma \) = 15.

i)

Exam Score Under 60 60 – 69 70 – 79 80 – 89 90 – 100
Probability 0.1515 0.2042 0.2622 0.2161 0.1185

ii)

Exam Score Under 60 60 – 69 70 – 79 80 – 89 90 – 100
Expected Frequency 45.45 61.26 78.66 64.83 35.55

iii) The value of the test statistic is \({\chi ^2}\) = 69.954, which is greater than the critical value of \({\chi ^2}\) = 9.488. We reject the null hypothesis that the exam scores are from a normally distributed population.