Chapter 10.4, Problem 35E

Hotel Occupancy Rates. Tourism is extremely important to the economy of Florida. Hotel occupancy is an often-reported measure of visitor volume and visitor activity (Orlando Sentinel, https://www.orlandosentinel.com/business/tourism/os-bz-hotel-occupancy-dip-20180522-story.html). Hotel occupancy data for February in two consecutive years are as follows.

1. a. Formulate the hypothesis test that can be used to determine if there has been an increase in the proportion of rooms occupied over the one-year period.
2. b. What is the estimated proportion of hotel rooms occupied each year?
3. c. Using a .05 level of significance, what is your hypothesis test conclusion? What is the p-value?
4. d. What is the 95% confidence interval estimate of the change in occupancy for the one-year period? Do you think area officials would be pleased with the results?

To determine

State the null and alternative hypotheses.

Answer to Problem 35E

Null hypothesis:$H_0:p_1-p_2\le 0$

Alternative hypothesis:$H_a:p_1-p_2>0$

Explanation of Solution

Calculation:

The data about the hotel room occupancy in the month of February in two consecutive years is as follows:

	Current year	Previous year
Occupied Rooms	1,470	1,458
Total Rooms	1,750	1,800

Here, $p_1$ represents the population proportion of rooms occupied in the current year and $p_2$ represents the population proportion of rooms occupied in the previous year.

State the hypothesis:

Null hypothesis:

$H_0:p_1-p_2\le 0$

That is, the proportion of rooms occupied in the current year is less than or equal to the proportion of rooms occupied in the previous year.

Alternative hypothesis:

$H_a:p_1-p_2>0$

That is, the proportion of rooms occupied in the current year is greater than the proportion of rooms occupied in the previous year.

To determine

Find the estimated proportion of hotel rooms occupied in each year.

Answer to Problem 35E

The estimated proportion of hotel rooms occupied in the current year is 0.84.

The estimated proportion of hotel rooms occupied in the previous year is 0.81.

Explanation of Solution

Calculation:

For the current year:

The estimated proportion of hotel rooms occupied in the current year is as follows:

$\begin{array}{c}{\overline{p}}_1=\frac{x_1}{n_1}\\ =\frac{1,470}{1,750}\\ =0.84\end{array}$

Thus, the estimated proportion of hotel rooms occupied in the current year is 0.84.

For the previous year:

The estimated proportion of hotel rooms occupied in the previous year is as follows:

$\begin{array}{c}{\overline{p}}_2=\frac{x_2}{n_2}\\ =\frac{1,458}{1,800}\\ =0.81\end{array}$

Thus, the estimated proportion of hotel rooms occupied in the previous year is 0.81.

To determine

Find the p-value and provide a conclusion at $\alpha =0.05$ from the results.

Answer to Problem 35E

The p-value is 0.009.

There is sufficient evidence to conclude that the proportion of rooms occupied in the current year is greater than the previous year.

Explanation of Solution

Calculation:

The test statistic for hypothesis tests about $p_1-p_2$ is as follows:

$z=\frac{\left({\overline{p}}_1-{\overline{p}}_2\right)}{\sqrt{\overline{p}\left(1-\overline{p}\right)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$

Where ${\overline{p}}_1$ is the sample proportion for a simple random sample from population 1, ${\overline{p}}_2$ is the sample proportion for a simple random sample from population 2, $\overline{p}$ is the pooled estimator of p, $n_1$ is the sample size for population 1, $n_2$ is the sample size for population 2, and $z_{\frac{\alpha }{2}}$ is critical value.

Pooled estimator:

$\begin{array}{c}\overline{p}=\frac{n_1{\overline{p}}_1+n_2{\overline{p}}_2}{n_1+n_2}\\ =\frac{1,750\left(0.84\right)+1,800\left(0.81\right)}{1,750+1,800}\\ =\frac{\text{2,928}}{\text{3,550}}\\ =0.8248\end{array}$

Substitute ${\overline{p}}_1$ as 0.84, ${\overline{p}}_2$ as 0.81, $n_1$ as 1,750, $n_2$ as 1,800, and $\overline{p}$ as 0.8248 in z formula.

$\begin{array}{c}z=\frac{\left(0.84-0.81\right)}{\sqrt{0.8248\left(1-0.8248\right)\left(\frac{1}{1,750}+\frac{1}{1,800}\right)}}\\ =\frac{\text{0}\text{.03}}{\sqrt{\left(\text{0}\text{.14450496}\right)\left(\text{0}\text{.00113}\right)}}\\ =\frac{\text{0}\text{.03}}{\text{0}\text{.01278}}\\ =2.35\end{array}$

Thus, the test statistic z-value is 2.35.

Software Procedure:

Step-by-step procedure to obtain the probability value using Excel:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =NORM.S.DIST(2.35,TRUE) in the cell A1.
• Press Enter.

Output obtained using EXCEL software is given below:

From the output, the p-value for the left tail is 0.990613.

For the upper-tail test, the p-value is $\left(1-\text{ area to the left of 2}\text{.35}\right)$.

From the output, the result is as calculated:

$\begin{array}{c}p\text{-value}=1-0.990613\\ =\text{0}\text{.009}\end{array}$

Thus, the p-value is 0.009.

Rejection rule:

If the $p\text{-value}\le \alpha$, reject the null hypothesis.

If the $p\text{-value}>\alpha$, do not reject the null hypothesis.

Conclusion:

Here, the p-value is less than the level of significance.

That is, the $p\text{-value}\left(=0.009\right)<\alpha \left(=0.05\right)$.

From the rejection rule, the null hypothesis is rejected.

Therefore, there is sufficient evidence to conclude that the proportion of rooms occupied in the current year is greater than the previous year.

To determine

Obtain the 95% confidence interval estimate of the change in occupancy for the one-year period.

Explain whether the area officials are pleased with the results or not.

Answer to Problem 35E

The 95% confidence interval is $\left(0.005,0.055\right)$.

Yes, the area officials are pleased with the results.

Explanation of Solution

Calculation:

The formula for the confidence interval estimate of the difference between two populations is as follows:

${\overline{p}}_1-{\overline{p}}_2\pm z_{\frac{\alpha }{2}}\sqrt{\frac{{\overline{p}}_1\left(1-{\overline{p}}_1\right)}{n_1}+\frac{{\overline{p}}_2\left(1-{\overline{p}}_2\right)}{n_2}}$.

Where ${\overline{p}}_1$ is the sample proportion for a simple random sample from population 1, ${\overline{p}}_2$ is the sample proportion for a simple random sample from population 2, $n_1$ is the sample size for population 1, $n_2$ is the sample size for population 2, and $z_{\frac{\alpha }{2}}$ is critical value.

Critical value:

From Table 1: Cumulative probabilities for the standard normal distribution are given below:

• Locate the value 0.025 in the table.
• Locate the row and column that are corresponding to the value 0.025.
• The intersecting row and column is 1.96.

The 95% confidence interval estimate of the change in occupancy for the one-year period is obtained below:

$\begin{array}{c}{\overline{p}}_1-{\overline{p}}_2\pm z_{\frac{\alpha }{2}}\sqrt{\frac{{\overline{p}}_1\left(1-{\overline{p}}_1\right)}{n_1}+\frac{{\overline{p}}_2\left(1-{\overline{p}}_2\right)}{n_2}}=\left(0.84-0.81\right)\pm 1.96\sqrt{\frac{0.84\left(1-0.84\right)}{1,750}+\frac{0.81\left(1-0.81\right)}{1,800}}\\ =0.03\pm 1.96\sqrt{0.0000768+\text{0}\text{.0000855}}\\ =0.03\pm 1.96\left(0.0127\right)\\ =0.03\pm 0.025\end{array}$

                                                            $\begin{array}{c}=\left(0.03-0.025,0.03+0.025\right)\\ =\left(\text{0}\text{.005},\text{0}\text{.055}\right)\end{array}$

Thus, the 95% confidence interval is $\underset{\_}{\left(\text{0}\text{.005},\text{0}\text{.055}\right)}$.

Interpretation:

There is 95% chance that the proportion of increase in hotel room occupancy over one-year period lies in between 0.005 and 0.055.

Since there is an increase in the hotel occupancy rate; the officials are expected to be pleased with the occupancy rate.