500 câu trắc nghiệm Kinh tế lượng – 10A

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

Number of successes	0	1	2
Frequency	47	98	55

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)

Number of successes	0	1	2
Probability	0.25	0.50	0.25

ii)

Number of successes	0	1/22
Expected frequency	50	100	50

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).

	Drug	Placebo
Headaches	11	7
No headaches	73	91

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.

	Yes	No	Undecided
Employed	30	15	5
Unemployed	20	25	10

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.

	Win	Loss
Suit	22	28
T-shirt	28	22

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.

	Men	Women
Health insurance	39	24
No health insurance	36	22

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.