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 24.85 21.33 Sample standard deviation 5.54 4.93 Sample size 34 36

Answer 1

The two-sample t-test suggests a significant difference in monthly take-out dinner purchases between men (24.85) and women (21.33).

To compare the take-out dinner habits of men and women who live alone, we can conduct a hypothesis test for the difference in means. Let's define the null hypothesis (H0) as the mean take-out dinners for men and women being equal, and the alternative hypothesis (H1) as the mean take-out dinners being different for men and women.

`\[ H_0: \mu_{\text{men}} = \mu_{\text{women}} \]`

`\[ H_1: \mu_{\text{men}} \\eq \mu_{\text{women}} \]`

We can use a two-sample t-test since the sample sizes are relatively small, and the population standard deviations are assumed to be equal.

The test statistic is calculated using the formula:

`\[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{(s_1^2)/(n_1) + (s_2^2)/(n_2)}} \]`

where
`\(\bar{x}_1\), \(\bar{x}_2\)` are the sample means,
`\(s_1\), \(s_2\)` are the sample standard deviations, and
`\(n_1\), \(n_2\)` are the sample sizes for men and women, respectively.

Substituting the given values:

`\[ t = \frac{(24.85 - 21.33)}{\sqrt{(5.54^2)/(34) + (4.93^2)/(36)}} \]`

After calculating the test statistic, we can compare it to the critical value or p-value to determine if there is enough evidence to reject the null hypothesis.

If the p-value is less than the significance level (commonly 0.05), we reject the null hypothesis, suggesting that there is a significant difference in the mean number of take-out dinners between men and women who live alone.