A recent study focused on the number of times men and women

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 sample standard deviations are not equal. The information is summarized below. Statistic Men Women Sample mean 24.53 21.89 Sample standard deviation 5.69 4.66 Sample size 35 39 At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? 1. Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic 2.122 2. What is your decision regarding on null hypothesis? The decision is the null hypothesis that the means are the same. 3. What is the p-value? (Round your answer to 4 decimal places.) p-value .034

Solution

Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=24.53
Standard Deviation(s.d1)=5.69 ; Number(n1)=35
Y(Mean)=21.89
Standard Deviation(s.d2)=4.66; Number(n2)=39
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =24.53-21.89/Sqrt((32.3761/35)+(21.7156/39))
to =2.1687
| to | =2.1687
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 34 d.f is 2.728
We got |to| = 2.16872 & | t | = 2.728
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 2.1687 ) = 0.037
Hence Value of P0.01 < 0.037,Here We Do not Reject Ho