500 câu trắc nghiệm Kinh tế lượng – 10A

Provide an appropriate response.
6) An observed frequency distribution is as follows:

i) Assuming a binomial distribution with n = 2 and p = 1/2, use the binomial formula to find the probability corresponding to each category of the table.
ii) Using the probabilities found in part (i), find the expected frequency for each category.
iii) Use a 0.05 level of significance to test the claim that the observed frequencies fit a binomial distribution for which n = 2 and p = 1/2.

Use a \({\chi ^2}\) test to test the claim that in the given contingency table, the row variable and the column variable are independent.

i)

ii)

iii) H0: p0= 0.25, p1= 0.50, p3= 0.25; H1: At least one of the above probabilities is different from the
claimed value. The value of the test statistic is \({\chi ^2}\) = 0.72, which is less than the critical value of
\({\chi ^2}\) = 5.991. We fail to reject the null hypothesis.

7) Tests for adverse reactions to a new drug yielded the results given in the table. At the 0.05 significance level, test the claim that the treatment (drug or placebo) is independent of the reaction (whether or not headaches were experienced).

H0: Treatment and reaction are independent.
H1: Treatment and reaction are dependent.
Test statistic: \({\chi ^2}\) = 1.798. Critical value: \({\chi ^2}\) = 3.841.
Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that
treatment and reaction are independent

8) Responses to a survey question are broken down according to employment status and the sample results are given below. At the 0.10 significance level, test the claim that response and employment status are independent.

H0: Employment status and response are independent.
H1: Employment status and response are dependent.
Test statistic: \({\chi ^2}\) = 5.942. Critical value: \({\chi ^2}\) = 4.605.
Reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that response and employment status are independent.

Solve the problem.
9) At a high school debate tournament, half of the teams were asked to wear suits and ties and the rest were asked to wear jeans and t-shirts. The results are given in the table below. Test the hypothesis at the 0.05 level that the proportion of wins is the same for teams wearing suits as for teams wearing jeans and t-shirts.

H0: The proportion of wins is the same for teams wearing suits as for teams wearing jeans and t-shirts.
H1: The proportions are different.
Test statistic: \({\chi ^2}\) = 1.440. Critical value: \({\chi ^2}\) = 3.841.
Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the proportion of wins is the same for teams wearing suits as for teams wearing jeans and t-shirts.

Provide an appropriate response.
10) A survey conducted in a small business yielded the results shown in the table.

i) Test the claim that health care coverage is independent of gender. Use a 0.05 significance
level.

ii) Using Yates’ correction, replace \(\sum {\frac{{{{\left( {O – E} \right)}^2}}}{E}} \) with \(\sum {\frac{{{{\left( {\left| {O – E} \right| – 0.5} \right)}^2}}}{E}} \) and repeat the test. What effect does Yates’ correction have on the value of the test statistic?

i) H0: Gender is independent of the health care coverage.
H1: Gender and health care coverage are dependent.
The value of the test statistic is \({\chi ^2}\) = 0.000346, which is less than the critical value of 3.841. We fail to reject the null hypothesis.
ii) The value of the test statistic is \({\chi ^2}\) = 0.028510, which is still less than the critical value of 3.841. We still fail to reject the null hypothesis. Yates’ correction typically decreases the value of the test statistic. However, due to the close fit of the O and E values in this problem, the Yates’ correction increased the
calculated test statistic.