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In a survey conducted by CareerBuilder.com, employers were asked if they had ever sent an employee home because they were dressed inappropriately (June 17 . 2008 , www.careerbuilder.com). A total of 2765 employers responded to the survey, with 968 saying that they had sent an employee home for inappropriate attire. In a press release, CareerBuilder makes the claim that more than one- third of employers have sent an employee home to change clothes. Do the sample data provide convincing evidence in support of this claim? Test the relevant hypotheses using \(\alpha=.05 .\) For purposes of this exercise, assume that it is reasonable to regard the sample as representative of employers in the United States.

Short Answer

Expert verified
Yes, the sample data provides convincing evidence to support the CareerBuilder's claim that more than one-third of employers have sent an employee home due to inappropriate attire. The P-value of 0.0401 is less than the significance level of 0.05, leading to the rejection of the null hypothesis in favor of the alternative one.

Step by step solution

01

Define Null and Alternative Hypotheses

The null hypothesis \(H_0\) assumes that the claim is not true. Therefore, it would be \(H_0: p \leq 0.33\), where p is the population proportion of employers who have sent an employee home for inappropriate clothing.\nThe alternative Hypothesis \(H_1\) represents the claim we are testing for. So, \(H_1: p > 0.33\).
02

Calculate the Sample Proportion

Calculate the sample proportion \(p̂\) using the formula: \(p̂ = x/n\), where x represents the number of successes (in this case, the number of employers who have sent an employee home for inappropriate attire) and n the sample size (total number of employers surveyed). Here, \(p̂ = 968/2765 = 0.35\). This is the observed sample proportion.
03

Compute Test Statistic

The test statistic for a one-sample proportion hypothesis test is a z-score (z). A z-score tells us how many standard deviations an element is from the mean. In this case, calculate it as follows: \(z = (p̂ - p)/sqrt((p*(1-p))/n)\). Here, \(z = (0.35 - 0.33)/sqrt((0.33*(1-0.33))/2765) = 1.75\).
04

Determine the P-Value

To get the P-value using a Z-table for a z-value of 1.75, we get the probability of 0.9599, but since this is a one-tailed test, we consider the area in the tail which will be 1 - 0.9599 = 0.0401.
05

Make a Decision and Interpret the Result

Given that the significance level (\(\alpha\)) is 0.05, and our P-value is 0.0401, we can compare them to decide whether or not to reject the null hypothesis. The decision rule is: if P-value ≤ \(α\), then reject the null hypothesis. Here, since 0.0401 < 0.05, the null hypothesis is rejected. Thus, there is convincing evidence at the 0.05 level of significance to support the claim that more than one-third of employers have sent an employee home to change clothes.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis represents the baseline assumption or the default belief. It's a statement that there is no effect or no difference in the situation being analyzed. For this exercise, we start by assuming that the claim made by CareerBuilder.com is not true.

Here's how the null hypothesis is formulated:
  • The null hypothesis, denoted as \(H_0\), posits that the proportion of employers who have sent employees home for inappropriate clothing is less than or equal to one-third (i.e., 33%).
  • Mathematically, this is expressed as \(H_0: p \leq 0.33\), where \(p\) symbolizes the actual proportion of the employer population.
By assuming the null hypothesis, we can conduct tests to evaluate if there is strong enough statistical evidence to support a change in this initial belief.
Alternative Hypothesis
The alternative hypothesis suggests the opposite of the null, and it is what we aim to provide evidence for. It reflects the claim that is being tested and verified.

In this exercise, the alternative hypothesis stands as:
  • Symbolized as \(H_1\), the alternative hypothesis indicates that the proportion of employers who have sent employees home due to inappropriate attire is greater than one-third.
  • In mathematical terms, this is: \(H_1: p > 0.33\).
The alternative hypothesis attempts to showcase a significant change or effect, thus justifying the need to reject the null hypothesis if sufficient evidence is present. It's essentially the goal or the claim we are examining in the context of statistical testing.
Proportion Hypothesis Test
A proportion hypothesis test applies when we're interested in analyzing the proportion of a certain characteristic within a population, based on sample data. This type of test is suitable for this scenario, where we examine the percentage of employers who enforce dress standards strictly enough to send employees home.

Key steps involved include:
  • Calculating the observed sample proportion \(\hat{p}\), which in this exercise is determined by \(\hat{p} = \frac{968}{2765}\).
  • Comparing \(\hat{p}\) to a hypothesized population proportion \(p\), which is 0.33, through a standard statistical method (such as the z-score).
  • Checking the result against a predefined significance level (\(\alpha\)), which commonly sits at 0.05 in such analyses.
The test aims to discern if the difference between the sample proportion and the hypothesized proportion is statistically significant, paving the way to validate or refute the claim under examination.
Z-Score
The z-score is an integral part of hypothesis testing, acting as the test statistic in this assessment. It measures how far the sample proportion is from the population proportion under the null hypothesis, standardized by the population's standard deviation.

Computation involves:
  • Using the formula: \(z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\), where \(\hat{p}\) is the sample proportion, \(p\) is the purported proportion, and \(n\) is the sample size.
  • For this scenario, inserting the relevant values gives: \(z = \frac{0.35 - 0.33}{\sqrt{\frac{0.33 \times (1-0.33)}{2765}}} \approx 1.75\).
The z-score maps to a point on the standard normal distribution, providing the means to assess how likely it is to observe a sample proportion this extreme if the null hypothesis were true. This foundation allows us to calculate the p-value, aiding in the decision-making process to reject or not reject the null hypothesis.

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Most popular questions from this chapter

The article referenced in the previous exercise also reported that 470 of 1000 randomly selected adult Americans thought that the quality of movies being produced was getting worse. a. Is there convincing evidence that fewer than half of adult Americans believe that movie quality is getting worse? Use a significance level of .05 . b. Suppose that the sample size had been 100 instead of 1000 , and that 47 thought that the movie quality was getting worse (so that the sample proportion is still .47). Based on this sample of 100 , is there convincing evidence that fewer than half of adult Americans believe that movie quality is getting worse? Use a significance level of .05 . c. Write a few sentences explaining why different conclusions were reached in the hypothesis tests of Parts (a) and (b).

10.48 A study of fast-food intake is described in the paper "What People Buy From Fast-Food Restaurants" (Obesity [2009]: \(1369-1374\) ). Adult customers at three hamburger chains (McDonald's, Burger King, and Wendy's) at lunchtime in New York City were approached as they entered the restaurant and asked to provide their receipt when exiting. The receipts were then used to determine what was purchased and the number of calories consumed was determined. In all, 3857 people participated in the study. The sample mean number of calories consumed was 857 and the sample standard deviation was 677 . a. The sample standard deviation is quite large. What does this tell you about number of calories consumed in a hamburger-chain lunchtime fast-food purchase in New York City? b. Given the values of the sample mean and standard deviation and the fact that the number of calories consumed can't be negative, explain why it is not reasonable to assume that the distribution of calories consumed is normal. c. Based on a recommended daily intake of 2000 calories, the online Healthy Dining Finder (www .healthydiningfinder.com) recommends a target of 750 calories for lunch. Assuming that it is reasonable to regard the sample of 3857 fast-food purchases as representative of all hamburger-chain lunchtime purchases in New York City, carry out a hypothesis test to determine if the sample provides convincing evidence that the mean number of calories in a New York City hamburger-chain lunchtime purchase is greater than the lunch recommendation of 750 calories. Use \(\alpha=.01\). d. Would it be reasonable to generalize the conclusion of the test in Part (c) to the lunchtime fast-food purchases of all adult Americans? Explain why or why not. e. Explain why it is better to use the customer receipt to determine what was ordered rather than just asking a customer leaving the restaurant what he or she purchased. f. Do you think that asking a customer to provide his or her receipt before they ordered could have introduced a potential bias? Explain.

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose that the specifications state that the mean strength of welds should exceed \(100 \mathrm{lb} / \mathrm{in}^{2}\). The inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100 .\) Explain why this alternative hypothesis was chosen rather than \(\mu<100\).

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