We need to determine if the mean work hours per week for female and male servers at Swank Bar are statistically different. We are given two samples: 70 female servers work an average of 33 hours per week with a standard deviation of 3, and 70 male servers work an average of 22 hours per week with a standard deviation of 2.
1.) Conduct a hypothesis test, such as a two-sample t-test if we assume the samples are independent and normally distributed.
The null hypothesis (H0) would state that the mean hours worked by female servers are equal to the mean hours worked by male servers, while the alternative hypothesis (H1) would state that the means are not equal.
H0: µ1 = µ2
H1: µ1 ≠ µ2
2.) Calculate the point estimate for the difference in means.
µ1 - µ2 = 34 - 20 = 14
3.) For a 88% confidence level, α = 1 - 0.88 = 0.12, and α/2 = 0.06. The critical value zα/2 for a standard normal distribution is approximately 1.18.
4.) Calculate the standard error for the difference in means.
SE = √[(s12/n1) + (s22/n2)] = √[(52/70) + (32/70)] = √[(25/70) + (9/70)] = √(34/70) = √(17/35) ≈ 0.697
5.) Calculate the margin of error.
ME = zα/2 × SE = 1.18 * 0.697 ≈ 0.82
6.) Calculate the confidence interval.
(µ1 - µ2) ± ME = 14 ± 0.82 ⇒ (14 - 0.82, 14 + 0.82) ⇒ (13.18, 14.82)
The answer is 13.18 < μ1 - μ2 < 14.82 .
Since both bounds are positive, µ1 - µ2 > 0 ⇒ µ1 > µ2
This indicates that female servers typically work more hours than male servers at Swank Bar.
There is a significant difference between the two sample means (33 hours for females vs. 22 hours for males). Given the large sample size of each group (n = 70), the Central Limit Theorem tells us that the distribution of the sample means will be approximately normally distributed, even if the population distribution is not normally distributed. This significant difference in the sample means suggests that female and male servers do not work the same number of hours.
