Question

Restaurant Bill by Gender In the study described in Exercise 6.224 the diners were also chosen so that half the people at each table were female and half were male. Thus we can also test for a difference in mean meal cost between females \(\left(n_{f}=24, \bar{x}_{f}=44.46, s_{f}=15.48\right)\) and males \(\left(n_{m}=24, \bar{x}_{m}=43.75, s_{m}=14.81\right) .\) Show all details for doing this test.

Short answer

The t-value obtained is 0.101 with degrees of freedom 46. This information can be used to find the p-value which will be used to determine whether there is statistical evidence that the two means are different.

Step-by-step solution

Calculate the Sample Mean Difference

Calculate the difference between the sample means which is simply \( \bar{x}_{f} - \bar{x}_{m} = 44.46 - 43.75 = 0.71 \)

Calculate the Pooled Variance

Compute the pooled variance. You do this by summing the squares of the standard deviations of both samples divided by their respective sample sizes minus one then divided by the total sample size minus two. This gives us: \(S_p^2 = \frac{ ((n_{f}-1) * s_{f}^2 + (n_{m}-1) * s_{m}^2) }{ (n_{m}+n_{f}-2) } = \frac{ ((24-1)*15.48^2 + (24-1)*14.81^2)}{ (24+24-2) } = 232.08 \)

Calculate the Standard Error

Calculate the standard error using the pooled variance and the sample sizes. It represents the average amount that the sample mean 'misses' the population mean. It's given by: \(SE = \sqrt{ S_p^2*(1/n_{m} + 1/n_{f}) } = \sqrt{232.08 * (1/24 + 1/24) } = 7.06 \)

Compute the t-value

Compute the t-value. The t-value measures how many standard errors are between the sample mean and the null hypothesis. It's given by: \(t = \frac{ \bar{x}_{f} - \bar{x}_{m} } {SE} = \frac{0.71}{7.06} = 0.101 \)

Determine the Degrees of Freedom

The degrees of freedom for this two-sample t-test is \(df = n_{m} + n_{f} - 2 = 24 + 24 - 2 = 46 \)