/*! 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 seeing the sample evidence? Use \(\alpha=.05\) in making the statistical test. What is the \(p\) -value?

Short Answer

Expert verified
The sample evidence does not contradict the radio station's claim.

Step by step solution

01

Define the Hypotheses

We need to set up the null and alternative hypotheses to test the radio station's claim. - Null Hypothesis (H0): The proportion of hotels and motels that are full is at least 90%, or \( p \geq 0.90 \).- Alternative Hypothesis (H1): The proportion of hotels and motels that are full is less than 90%, or \( p < 0.90 \).
02

Compute Sample Proportion

Calculate the sample proportion of hotels and motels that are full. The sample size is 58 and the number of full hotels is 49.\[ \hat{p} = \frac{49}{58} \approx 0.8448 \]
03

Standardize the Sample Proportion

Calculate the test statistic using the standard normal distribution since the sample size is large enough. The formula for the test statistic is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]Here, \( p_0 = 0.90 \), \( n = 58 \), and \( \hat{p} \approx 0.8448 \).\[ z = \frac{0.8448 - 0.90}{\sqrt{\frac{0.90(1-0.90)}{58}}} \approx \frac{-0.0552}{0.0394} \approx -1.40 \]
04

Determine Critical Value and Decision Rule

Find the critical value for \( z \) at \( \alpha = 0.05 \) in a one-tailed test. The critical value is about \( -1.645 \). Decision Rule: If the test statistic \( z < -1.645 \), reject the null hypothesis. Otherwise, do not reject it.
05

Compute the P-value

The p-value is the probability that the test statistic \( z \) is less than or equal to the observed \( z \)-value. From standard z-tables or using a calculator, \( P(z < -1.40) \approx 0.0808 \).
06

Conclusion Based on P-value and Decision Rule

Since the p-value \( 0.0808 \) is greater than \( \alpha = 0.05 \) and the test statistic \( -1.40 \) is not less than the critical value \( -1.645 \), we do not reject the null hypothesis. Therefore, there is not enough evidence to refute the radio station's claim that at least 90% of the hotels and motels would be full.

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Key Concepts

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

Statistical Inference
Statistical inference is a process that allows us to make conclusions about a population based on sample data. Imagine you have information on a small group and need to generalize it to a larger group. That's what inference helps with. In the context of hypothesis testing, it involves determining whether you can support a hypothesis about a population parameter, like a proportion or mean, based on sample data.

The steps in statistical inference often involve the following:
  • Forming hypotheses: You develop a null and an alternative hypothesis to test.
  • Gathering data: A sample is collected to be examined.
  • Calculating statistics: Important metrics like sample proportion or mean are calculated.
  • Making decisions: Based on calculated probabilities and test statistics, you decide whether to support or reject the hypothesis.
Each part ensures that you can make educated reasoning about the bigger picture beyond your sample.
Null Hypothesis
The null hypothesis, often denoted as \( H_0 \), is a statement suggesting there is no effect or no difference, and it acts as a starting point for statistical testing. For our hotel example, \( H_0 \) posits that at least 90% of the hotels are full over the weekend. This presumption is made so that we can test it against the alternative hypothesis.

The null hypothesis is crucial because it establishes a default position that the evidence must contradict for the alternative hypothesis to be accepted. The goal of a hypothesis test is usually to challenge the null hypothesis by showing enough statistical evidence against it.
  • Accepted until evidence suggests otherwise.
  • Provides a baseline to compare against your data.
  • Helps in measuring the degree of unusualness of your findings with respect to the null hypothesis.
Rejection of a null hypothesis suggests that the observed data significantly differ from what's expected under this initial assumption.
P-value
A P-value is a vital part of hypothesis testing and represents the probability of obtaining results as extreme as, or more extreme than, the observed results under the assumption that the null hypothesis is true. In essence, it indicates how likely your sample results are under the null hypothesis.

For instance, in our example, a P-value of approximately 0.0808 tells us the likelihood of seeing a sample proportion of hotels full as low as we observed—or even lower—if indeed 90% of hotels are supposed to be full. When the P-value is:
  • Less than or equal to the significance level \( \alpha \): Strong evidence against the null, leading to its rejection.
  • Greater than \( \alpha \): Not enough evidence to reject the null.
P-values offer a robust way to quantify the evidence against the null hypothesis, making them central to the decision-making process in hypothesis testing.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to determine the difference between your observed data and what was expected under the null hypothesis, using a mathematical formula.

For our scenario, the test statistic is calculated using the standard normal distribution since the sample is sufficiently large. Specifically, the statistic examines how far the sample proportion (0.8448) deviates from the hypothesized proportion (0.9) when adjusted for variations expected in random samples. This is derived through:
  • \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \)
  • Where \( \hat{p} \) is the sample proportion, \( p_0 \) the hypothesized proportion, and \( n \) the sample size.
The calculated test statistic, in this case, \( -1.40 \), helps compare the observed outcome directly against expected outcomes under the distribution theory, assisting in decisions about the null hypothesis.

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Most popular questions from this chapter

Nielsen reported that young men in the United States watch 56.2 minutes of prime-time TV daily (The Wall Street Journal Europe, November 18,2003 ). A researcher believes that young men in Germany spend more time watching prime- time TV. A sample of German young men will be selected by the researcher and the time they spend watching TV in one day will be recorded. The sample results will be used to test the following null and alternative hypotheses. \\[ \begin{array}{l} H_{0}: \mu \leq 56.2 \\ H_{\mathrm{a}}: \mu>56.2 \end{array} \\] a. What is the Type I error in this situation? What are the consequences of making this error? b. What is the Type II error in this situation? What are the consequences of making this error?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 80 \\ H_{\mathrm{a}}: \mu<80 \end{array} \\] A sample of 100 is used and the population standard deviation is \(12 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\quad \bar{x}=78.5\) b. \(\bar{x}=77\) c. \(\quad \bar{x}=75.5\) d. \(\bar{x}=81\)

A study by Consumer Reports showed that \(64 \%\) of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, the manufacturer of a national name-brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national brand ketchup. a. Formulate the hypotheses that could be used to determine whether the percentage of supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup differed from \(64 \%\) b. If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good as the national brand, what is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. Should the national brand ketchup manufacturer be pleased with this conclusion? Explain.

The chamber of commerce of a Florida Gulf Coast community advertises that area residential property is available at a mean cost of \(\$ 125,000\) or less per lot. Suppose a sample of 32 properties provided a sample mean of \(\$ 130,000\) per lot and a sample standard deviation of \(\$ 12,500 .\) Use a .05 level of significance to test the validity of the advertising claim.

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \leq 25 \\ H_{\mathrm{a}}: \mu>25 \end{array} \\] A sample of 40 provided a sample mean of \(26.4 .\) The population standard deviation is \(6 .\) a. Compute the value of the test statistic. b. What is the \(p\) -value? c. At \(\alpha=.01,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

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