Math Problem Statement

Tourism is extremely important to the economy of Florida. Hotel occupancy is an often-reported measure of visitor volume and visitor activity (Orlando Sentinel). Hotel occupancy data for February in two consecutive years are as follows.

Current Year: Occupied Rooms = 1,440 Total Rooms = 1,800

Previous year : Occupied Rooms = 1,463 Total rooms = 1,900. Conduct a hypothesis test, what is the p value to 4 decimals

Solution

We are conducting a hypothesis test to determine whether there is a significant difference in the hotel occupancy rates between the current year and the previous year.

The hotel occupancy rate can be calculated as:

Occupancy Rate=Occupied RoomsTotal Rooms\text{Occupancy Rate} = \frac{\text{Occupied Rooms}}{\text{Total Rooms}}

The sample proportions for the current and previous years are:

p^1=14401800=0.8\hat{p}_1 = \frac{1440}{1800} = 0.8 p^2=14631900=0.769\hat{p}_2 = \frac{1463}{1900} = 0.769

Step 1: Hypotheses

  • Null Hypothesis (H0H_0): There is no difference in occupancy rates between the current and previous year. p^1=p^2\hat{p}_1 = \hat{p}_2
  • Alternative Hypothesis (HaH_a): There is a difference in occupancy rates between the two years. p^1p^2\hat{p}_1 \neq \hat{p}_2

Step 2: Calculate the test statistic

We can use a two-proportion z-test for this hypothesis test.

The pooled proportion is given by:

p^=Occupied Rooms in both yearsTotal Rooms in both years=1440+14631800+1900\hat{p} = \frac{\text{Occupied Rooms in both years}}{\text{Total Rooms in both years}} = \frac{1440 + 1463}{1800 + 1900}

The standard error (SE) for the difference in proportions is calculated as:

SE=p^(1p^)(1n1+1n2)SE = \sqrt{\hat{p}(1-\hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}

The test statistic (z-value) is:

z=p^1p^2SEz = \frac{\hat{p}_1 - \hat{p}_2}{SE}

Step 3: Calculate p-value

We will calculate this and find the p-value using a two-tailed test. Let me compute these values.The p-value for this hypothesis test is approximately 0.0265.

Since this p-value is less than the common significance level of 0.05, we can reject the null hypothesis, meaning there is a statistically significant difference in the hotel occupancy rates between the current and previous year.

Would you like further details or explanations on this?

Here are some related questions to explore further:

  1. How is a two-proportion z-test used in other contexts?
  2. What is the effect size and how could it be calculated in this scenario?
  3. What other factors might influence hotel occupancy besides the year difference?
  4. How do confidence intervals for proportions compare to hypothesis testing?
  5. What assumptions are made in the z-test for proportions?

Tip: In hypothesis testing, the p-value helps you determine whether the observed data significantly deviates from the null hypothesis, but always consider the context and possible real-world implications before drawing conclusions.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Two-Proportion Z-Test
P-Value Calculation

Formulas

Occupancy Rate = (Occupied Rooms) / (Total Rooms)
Pooled Proportion = (Occupied Rooms in both years) / (Total Rooms in both years)
Standard Error (SE) = sqrt(p(1-p) * (1/n1 + 1/n2))
Test Statistic (z-value) = (p1 - p2) / SE

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate Level Statistics