Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Sheila Smith, the manager of a large resort\'s main hotel, hasbeen receiving com

ID: 3339602 • Letter: S

Question

Sheila Smith, the manager of a large resort's main hotel, hasbeen receiving complaints from some guests that they are notbeing provided with prompt service upon approaching the front desk.In particular, she is concerned that desk staff might be providingfemale guests with less prompt service that their malecounterparts. In observing a sample of 34 male guests, she finds ittakes an average of 15.2 seconds, with a standard deviation of 5.9seconds, from the to greeted after their arrival at the front desk.For a sample of 39 female guests, the mean and standard deviationare 17.4 seconds and 6.4 seconds, respectively. Assuming the population standard deviations to be equal, use the .05 level of significance in examining wheter the population mean time forservicing female guests might actually be no greater than that forservicing male guests

Explanation / Answer

We need to use the two-sample t-test for unpaired data, assuming population standard deviations are equal.

Assuming that the population Mean Service Times for Males and Females are M1 amd M2 respectively.

Then the null hypothesis is H0: M2 = M1 and alternate hypothesis is H1: M2 > M1

Hence we need to do the one tailed t-test at 0.05 level of significance.

Let the sample sizes be n1 and n2, means be m1 and m2 and standard seviations be s1 and s2.

Then the formula for the Test Statistic,T is: T = (m2 - m1) / {sp.sqrt(1/n1+1/n2)}, with (n1+n2-2) degrees of freedom

where sp = sqrt[{(n1-1)s12 + (n2-1)s22)}/(n1+n2-2)].

Here n1= 34, m1 = 15.2, s1 = 5.9, n2 = 39, m2 = 17.4, s2 = 6.4

Hence sp = sqrt{(1148.73 + 1556.48)/71} = 6.1726 and

T = (17.4-15.2) / {6.1726.sqrt(1/34+1/39)} = 2.2 / (6.1726 x 0.2346) = 1.5192 with 71 df.

Using the TDIST function in Excel, probability of getting a value of T >= 1.5192 with 71 df for a one-tailed test is 0.066576, which is > the specified significance level of 0.05

Hence we accept the null Hypothesis H0 and conclude that the population mean time for servicing female guests is not greater than the population mean time for servicing male guests.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote