Question

Rating service at five-star hotels. A study published in the Journal of American Academy of Business, Cambridge (March 2002) examined whether the perception of service quality at five-star hotels in Jamaica differed by gender. Hotel guests were randomly selected from the lobby and restaurant areas and asked to rate 10 service-related items (e.g., “The personal attention you received from our employees”). Each item was rated on a 5-point scale (1 = “much worse than I expected,” 5 = “much better than I expected”), and the sum of the items for each guest was determined. A summary of the guest scores is provided in the table.

Gender	Sample size	Mean score	Standard deviation
Males	127	39.08	6.73
Females	114	38.79	6.94

a. Construct a 90% confidence interval for the difference between the population mean service-rating scores given by male and female guests at Jamaican five-star hotels.

Short answer

a. The 90% confidence interval for the two population means service-rating scores is (-1.162, 1.742).

Step-by-step solution

Given Information

The sample size of males, \({n_1} = 127.\)

The sample size of females, \({n_2} = 114.\)

The mean score of the males,\({\bar x_1} = 39.08.\)

The mean score of the females, \({\bar x_2} = 38.79.\)

The standard deviation of the males, \({\sigma _1} = 6.73.\)

The standard deviation of females, \({\sigma _2} = 6.94.\)

State the large, independent samples confidence interval for \(\left( {{\mu _1} - {\mu _2}} \right)\).

If \(\sigma _1^2\) and \(\sigma _2^2\) be known the confidence interval for \(\left( {{\mu _1} - {\mu _2}} \right)\) is,\(C.I = \left( {\left( {{{\bar x}_1} - {{\bar x}_2}} \right) \pm {Z_{\frac{\alpha }{2}}}\sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}} } \right)\)

Where \({\bar x_1}\) and \({\bar x_2}\) are the means of the population, \(\sigma _1^2\) and \(\sigma _2^2\) are variance of the population, \({n_1}\) and \({n_2}\) are the sample size, and \({Z_{\frac{\alpha }{2}}}\) is the tabled value obtained from the standard normal table.

\(\)

Compute the 90% confidence interval for the difference between the two population mean service rating scores.

Let \({\mu _1}\) be the mean score for males and \({\mu _2}\) be the mean score of females respectively.

Let the confidence level be 0.90.

\(\begin{aligned}{c}\left( {1 - \alpha } \right) &= 0.90\\\alpha &= 0.1\\\frac{\alpha }{2} &= 0.05\end{aligned}\)

The \({Z_{\frac{\alpha }{2}}}\) value for the 90% of confidence level obtained from the standard normal table is 1.645.

The confidence interval for \(\left( {{\mu _1} - {\mu _2}} \right)\) is computed as:

\(\begin{aligned}{c}C.I &= \left( {\left( {{{\bar x}_1} - {{\bar x}_2}} \right) \pm {Z_{\frac{\alpha }{2}}}\sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}} } \right)\\ &= \left( {\left( {39.08 - 38.79} \right) \pm 1.645\sqrt {\frac{{{{6.73}^2}}}{{127}} + \frac{{{{6.94}^2}}}{{114}}} } \right)\\ &= \left( {0.29 \pm 1.452} \right)\\ &= \left( { - 1.162,1.742} \right)\end{aligned}\)

Hence, the 90% confidence interval for the two population means service-rating scores is (-1.162, 1.742).