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	Men	Women
Sample mean	24.71	21.94
Sample standard deviation	5.53	4.71
Sample size	36	41

At the 0.01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month?

1. State the decision rule for 0.01 significance level: H₀: μ_Men= μ_Women   H₁: μ_Men ≠ μ_Women. (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.)

2. Compute the value of the test statistic. (Round your answer to 3 decimal places.)

3. What is your decision regarding the null hypothesis?

4. What is the p-value? (Round your answer to 3 decimal places.)

Answer 1

a)

degree of freedom   v=		(n1+n2-2)=	75.000
for 0.01 level ,two tail test &75 df, critical t=			2.643
Decision rule :                   reject Ho if absolute value of test statistic |t|>2.643

b)

Pooled Variance Sp2=((n1-1)s21+(n2-1)*s22)/(n1+n2-2)=

26.1026

standard error se =Sp*√(1/n1+1/n2)=			1.1669
test stat t =(x1-x2-Δo)/Se=		2.374

c)

fail to reject Ho

d_)

p value=0.020