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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

A certain television station has been providing live coverage of a particularly sensational criminal trial. The station's program director wishes to know whether more than half the potential viewers prefer a return to regular daytime programming. A survey of randomly selected viewers is conducted. Let \(p\) represent the proportion of all viewers who prefer regular daytime programming. What hypotheses should the program director test to answer the question of interest?

According to a survey of 1000 adult Americans conducted by Opinion Research Corporation, 210 of those surveyed said playing the lottery would be the most practical way for them to accumulate \(\$ 200,000\) in net wealth in their lifetime ("One in Five Believe Path to Riches Is the Lottery," San Luis Obispo Tribune, January 11, 2006 ). Although the article does not describe how the sample was selected, for purposes of this exercise, assume that the sample can be regarded as a random sample of adult Americans. Is there convincing evidence that more than \(20 \%\) of adult Americans believe that playing the lottery is the best strategy for accumulating \(\$ 200,000\) in net wealth?

A credit bureau analysis of undergraduate students credit records found that the average number of credit cards in an undergraduate's wallet was 4.09 ("Undergraduate Students and Credit Cards in 2004," Nellie Mae, May 2005\()\). It was also reported that in a random sample of 132 undergraduates, the sample mean number of credit cards that the students said they carried was 2.6. The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was 1.2 . Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of \(4.09 ?\)

In a survey of 1000 women age 22 to 35 who work full time, 540 indicated that they would be willing to give up some personal time in order to make more money (USA Today. March 4, 2010 ). The sample was selected in a way that was designed to produce a sample that was representative of women in the targeted age group. a. Do the sample data provide convincing evidence that the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money? Test the relevant hypotheses using \(\alpha=.01\). b. Would it be reasonable to generalize the conclusion from Part (a) to all working women? Explain why or why not.

The report "Highest Paying Jobs for \(2009-10\) Bachelor's Degree Graduates" (National Association of Colleges and Employers, February 2010 ) states that the mean yearly salary offer for students graduating with a degree in accounting in 2010 is \(\$ 48,722\). Suppose that a random sample of 50 accounting graduates at a large university who received job offers resulted in a mean offer of \(\$ 49,850\) and a standard deviation of \(\$ 3300 .\) Do the sample data provide strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of \(\$ 48,722 ?\) Test the relevant hypotheses using \(\alpha=.05 .\)

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