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?
State the decision rule for 0.01 significance level: H_{0}: μ_{Men}= μ_{Women} H_{1}: μ_{Men} ≠ μ_{Women.} (Negative values should be indicated by a minus sign. Round your answers to 3 decimal places.)
Compute the value of the test statistic. (Round your answer to 3 decimal places.)
What is your decision regarding the null hypothesis?
What is the p-value? (Round your answer to 3 decimal places.)
a)
degree of freedom v= | (n_{1}+n_{2}-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 Sp^{2}=((n_{1}-1)s^{2}_{1}+(n_{2}-1)*s^{2}_{2})/(n_{1}+n_{2}-2)= | 26.1026 |
standard error se =S_{p}*√(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
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