Question

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distributions follow the normal probability distribution and the population standard deviations are equal. The information is summarized below. Statistic Sample mean Sample standard deviation Sample size Men Women 24.51 22.69 4.48 3.86 35 40 difference in the mean number of times men and women order take-out dinners At the 0.01 significance level, is there in a month? a. State the decision rule for 0.01 significance level: Ho: Men=hWomen Hi: Wen # Mwomen. (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.) Answer is not complete. or Reject HO ifts b. Compute the value of the test statistic. (Round you answer to 3 decimal places.) Answer is complete and correct. Value of the test 1.890 statistic c. What is your decision regarding the null hypothesis? Answer is complete and correct. The decision is do not reject the null hypothesis that the means are the same. d. What is the p-value? (Round your answer to 3 decimal places.) p-value

Answer 1

We have to test the below hypotheses

H₀: µ_Men=µ_Women

_­H_A: µ_Men ≠ µ_Women

We have following information        

	Men	Women
Sample mean	24.51	22.69
Sample standard deviation	4.48	3.86
Sample size	35	40

Total degrees of freedom is

df_total =n₁ +n₂ -2

df_total =35+40 -2

df_total =73

Answer(a):

We have the two tailed test, so the critical value for this test will be the table value of t at α/2=0.005 level of significance with 73df.

t_(0.995,73) = 2.645

Hence the decision rule is

Reject H0 if t< -2.645 or t > 2.645

Answer(b):

The test statistic to test the null hypothesis is

X 1 t = - Xal 1 + ny nz Com Sp

Where sp is pooled standard deviation

S P

Se = (n4 - 1)s? +(n2 - 1)sz? Ny tn - 2 P

S (35 – 1)4.482 + (40 – 1)3.862 35 + 40 - 2

1263.47% 2 EL 73

2 Sp 17.308

Sp = 4.16

The test statistic is

24.51 – 22.69 t= 4.16* (460+33)

1.82 t = 4.16 * 0.231455

1.82 t= 0.963

t 1.890

Value of the test statistic = 1.890

Answer(c):

The value of test statistic is within the acceptance region, so we fail to reject the null hypothesis.

The decision is do not reject the null hypothesis that the means are same.

Answer(d):

The p-value of above test is 0.031

p-value = 0.031