Chapter 8.6, Problem 8.51E

Hotel Costs Refer to Exercise 8.20. The means and standard deviations for 50 billing statements from each of the computer databases of each of the three hotel chains are given in the table

a. Find a 95% confidence interval for the difference in the average room rates for the Marriott and the Westin hotel chains.

b. Find a 99% confidence interval for the difference in the average room rates for the Westin and the Doubletree hotel chains.

c. Do the intervals in parts a and b contain the value Why is this of interest to the researcher?

d. Do the data indicate a difference in the average room rates between the Marriott and the Westin chains? Between the Westin and the Doubletree chains?

To determine

To find: the 95% confidence interval for the difference between two population means.

Answer to Problem 8.51E

The confidence interval is $(-22.8502,-7.1498)$ .

Explanation of Solution

Given:

The sample size of Region 1 ( $n_1$ ) is 50.

The sample size of Region 2 ( $n_2$ ) is 50.

The level of significance ( $\alpha$ ) is (1-0.95) i.e. 0.05.

The sample mean of region 1 is 150.

The sample mean of region 2 is 165.

The sample variance of region 1 is 17.2.

The sample variance of region 2 is 22.5.

Formulae used:

Consider two sample sizes $n_1$ and $n_2$ , then for a given ( $\alpha$ ) level of significance, the $100(1-\alpha )\%$ Confidence interval for the difference between two population means ( ${\mu }_1-{\mu }_2$ ) is given by:-

  $\left(({\overline{x}}_1-{\overline{x}}_2)-\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}{Z}_{\frac{\alpha }{2}},({\overline{x}}_1-{\overline{x}}_2)+\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}{Z}_{\frac{\alpha }{2}}\right)$

Calculations:

Consider, ${\mu }_1$ and ${\mu }_2$ be two population means of Region 1 and Region 2 respectively and ${\overline{x}}_1$ and ${\overline{x}}_2$ be the unbiased estimators of ${\mu }_1$ and ${\mu }_2$

Also, Consider ${\sigma }_1$ and ${\sigma }_2$ be two population variance of Region 1 and Region 2 respectively and $s_1$ and $s_2$ be the unbiased estimators of ${\sigma }_1$ and ${\sigma }_2$

The Z Critical value at 95% level of significance is 1.96 calculated from the standard normal table.

The $100(1-\alpha )\%$ Confidence interval for the difference between two population means ( ${\mu }_1-{\mu }_2$ ) is calculated as:-

  $\begin{array}{c}\left(({\overline{x}}_1-{\overline{x}}_2)-\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}{Z}_{\frac{\alpha }{2}},({\overline{x}}_1-{\overline{x}}_2)+\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}{Z}_{\frac{\alpha }{2}}\right)\\ =\left((150-165)-1.96\left(\sqrt{\frac{{17.2}^2}{50}+\frac{{22.5}^2}{50}}\right),(150-165)+1.96\left(\sqrt{\frac{{17.2}^2}{50}+\frac{{22.5}^2}{50}}\right)\right)\\ =(-22.8502,-7.1498)\end{array}$

Therefore, the confidence interval for difference between two population means ( ${\mu }_1-{\mu }_2$ ) is $(-22.8502,-7.1498)$ .

To determine

To find: the 99% confidence interval for the difference between two population means.

Answer to Problem 8.51E

The confidence interval is $(30.5550,49.4450)$ .

Explanation of Solution

Given:

The sample size of Region 1 ( $n_1$ ) is 50.

The sample size of Region 2 ( $n_2$ ) is 50.

The level of significance ( $\alpha$ ) is (1-0.95) i.e. 0.05.

The sample mean of region 1 is 165.

The sample mean of region 2 is 125.

The sample variance of region 1 is 22.5.

The sample variance of region 2 is 12.8.

Calculations:

The Z Critical value at 99% level of significance is 2.58 calculated from the standard normal table.

The $100(1-\alpha )\%$ Confidence interval for the difference between two population means ( ${\mu }_1-{\mu }_2$ ) is calculated as:-

  $\begin{array}{c}\left(({\overline{x}}_1-{\overline{x}}_2)-\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}{Z}_{\frac{\alpha }{2}},({\overline{x}}_1-{\overline{x}}_2)+\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}{Z}_{\frac{\alpha }{2}}\right)\\ =\left((165-125)-2.58\left(\sqrt{\frac{{22.5}^2}{50}+\frac{{12.8}^2}{50}}\right),(165-125)+2.58\left(\sqrt{\frac{{22.5}^2}{50}+\frac{{12.8}^2}{50}}\right)\right)\\ =(30.5550,49.4450)\end{array}$

Therefore, the confidence interval for difference between two population means ( ${\mu }_1-{\mu }_2$ ) is $(30.5550,49.4450)$ .

To determine

To check: if there exists difference between two population means or not..

Answer to Problem 8.51E

Yes, there is a difference between two means.

Explanation of Solution

The confidence interval for difference between two population means ( ${\mu }_1-{\mu }_2$ ) is $(-22.8502,-7.1498)$ for part (a).

The confidence interval for difference between two population means ( ${\mu }_1-{\mu }_2$ ) is $(30.5550,49.4450)$ for part (b).

Since, 0 is not contained in either of the intervals.

Therefore, there is a significant difference in both the population means.

To determine

To check: if there exists difference between two population means or not.

Answer to Problem 8.51E

Yes, there is a difference between two means.

Explanation of Solution

The confidence interval for difference between two population means ( ${\mu }_1-{\mu }_2$ ) is $(-22.8502,-7.1498)$ for part (a).

The confidence interval for difference between two population means ( ${\mu }_1-{\mu }_2$ ) is $(30.5550,49.4450)$ for part (b).

Since, 0 is not contained in either of the intervals.

Therefore, there is a significant difference in both the population means.

This shows that there does exist significant rates difference.