/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 A radio station in Myrtle Beach ... [FREE SOLUTION] | 91Ó°ÊÓ

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A radio station in Myrtle Beach announced that at least \(90 \%\) of the hotels and motels would be full for the Memorial Day weekend. The station advised listeners to make reservations in advance if they planned to be in the resort over the weekend. On Saturday night a sample of 58 hotels and motels showed 49 with a no-vacancy sign and 9 with vacancies. What is your reaction to the radio station's claim after secing the sample evidence? Use \(a=.05\) in making the statistical test. What is the \(p\) -value?

Short Answer

Expert verified
Reject the claim; the p-value is 0.0378.

Step by step solution

01

State the Hypotheses

To test the radio station's claim, we set up the null hypothesis, \( H_0: p \geq 0.9 \), where \( p \) is the proportion of hotels filled. The alternative hypothesis is \( H_a: p < 0.9 \). This alternative suggests that the proportion is actually less than 90%.
02

Find the Sample Proportion

Calculate the sample proportion of hotels with no vacancies. The sample size, \( n = 58 \), and the number of hotels full, \( X = 49 \). The sample proportion \( \hat{p} \) is \( \hat{p} = \frac{49}{58} \approx 0.8448 \).
03

Calculate the Test Statistic

For the test statistic, use the formula for a proportion: \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \( p_0 = 0.9 \). Substitute the values: \( z = \frac{0.8448 - 0.9}{\sqrt{\frac{0.9 \times 0.1}{58}}} \approx -1.7769 \).
04

Determine the p-Value

The \( z \)-test statistic is approximately \( -1.7769 \). Using standard normal distribution tables or a calculator, the p-value for this \( z \)-test statistic is approximately 0.0378.
05

Make a Decision

Since the p-value \( 0.0378 \) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis. There is sufficient evidence to conclude that less than 90% of the hotels were full.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
Understanding the concept of sample proportion is crucial when dealing with data from a sample instead of a whole population. In the context of hypothesis testing, the sample proportion helps estimate the true proportion of a particular characteristic in the population. To calculate the sample proportion, we take the number of successes (or what we're measuring) and divide it by the total number of observations in our sample. For example, if we're looking at the proportion of hotels filled to capacity, we use: \[ \hat{p} = \frac{X}{n} \] Where:- \( \hat{p} \) is the sample proportion.- \( X \) is the number of successes (hotels with no vacancies).- \( n \) is the total sample size (number of hotels surveyed).In this scenario, \( X = 49 \) and \( n = 58 \), so the calculation is:\[ \hat{p} = \frac{49}{58} \approx 0.8448 \].This means about 84.48% of the sampled hotels showed no vacancies. By comparing this sample proportion to a claimed population proportion, we can begin to see if the claim is statistically valid.
Null Hypothesis
The null hypothesis is a fundamental element of hypothesis testing. It's a statement that there is no effect or no difference, and it represents a starting point for statistical testing. When testing a claim, like the radio station's assertion that at least 90% of hotels are full, the null hypothesis is set as \( H_0: p \geq 0.9 \). It assumes that the proportion of all hotels that are full is at least 90%. We also have an alternative hypothesis, which in this case is \( H_a: p < 0.9 \). This is what we test against the null. If our sample data provides enough evidence against the null hypothesis, we might reject it in favor of the alternative.The null hypothesis is meant to provide a "default" or "neutral" stance, where any deviation must be proven by the data collected and analyzed.
p-value Calculation
A p-value is a key concept in hypothesis testing that helps determine the significance of your results. It measures the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. In simpler terms, it tells us how likely our sample data would be if the null hypothesis were accurate.To calculate the p-value, we first need the test statistic, calculated here using the formula for the z-test for proportions:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]Where:- \( \hat{p} \) is the sample proportion (0.8448).- \( p_0 \) is the hypothesized population proportion (0.9).- \( n \) is the number of observations (58).Substituting these values, we find:\[ z \approx -1.7769 \]This z-score correlates to a p-value, which can be found using standard normal distribution tables or a calculator, and is approximately 0.0378 in this context.Since this p-value is less than the significance level \( \alpha = 0.05 \), it suggests that the observed data is statistically significant, leading to the rejection of the null hypothesis.

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Most popular questions from this chapter

Because of high production-changeover time and costs, a director of manufacturing must convince management that a proposed manufacturing method reduces costs before the new method can be implemented. The current production method operates with a mean cost of \(\$ 220\) per hour. A research study will measure the cost of the new method over a sample production period. a. Develop the null and alternative hypotheses most appropriate for this study. b. Comment on the conclusion when \(H_{0}\) cannot be rejected. c. Comment on the conclusion when \(H_{0}\) can be rejected.

According to the University of Nevada Center for Logistics Management, \(6 \%\) of all merchandise sold in the United States is returned ( Business Week , January 15,2007 ). A Houston department store sampled 80 items sold in January and found that 12 of the items were returned. a. Construct a point estimate of the proportion of items returned for the population of sales transactions at the Houston store. b. Construct a \(95 \%\) confidence interval for the porportion of returns at the Houston store. c. Is the proportion of returns at the Houston store significantly different from the returns for the nation as a whole? Provide statistical support for your answer.

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation \(\sigma=.8\) ounces is assumed. A quality control inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is \(\alpha=.05\) a. State the hypothesis test for this quality control application. b. If a sample mean of \(\bar{x}=16.32\) ounces were found, what is the \(p\) -value? What action would you recommend? c. If a sample mean of \(\bar{x}=15.82\) ounces were found, what is the \(p\) -value? What action would you recommend? d. Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure? Repeat parts (b) and (c). Do you reach the same conclusion?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \leq 12 \\ H_{\mathrm{a}}: \mu>12 \end{array} \\] A sample of 25 provided a sample mean \(\bar{x}=14\) and a sample standard deviation \(s=4.32\) a. Compute the value of the test statistic. b. Use the \(t\) distribution table (Table 2 in Appendix \(B\) ) to compute a range for the \(p\) -value. c. \(\quad\) At \(a=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

The manager of an automobile dealership is considering a new bonus plan designed to increase sales volume. Currently, the mean sales volume is 14 automobiles per month. The manager wants to conduct a research study to see whether the new bonus plan increases sales volume. To collect data on the plan, a sample of sales personnel will be allowed to sell under the new bonus plan for a one-month period. a. Develop the null and alternative hypotheses most appropriate for this situation. b. Comment on the conclusion when \(H_{0}\) cannot be rejected. c. Comment on the conclusion when \(H_{0}\) can be rejected.

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