Question

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

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

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, \(H_1: p > 0.33\).

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.

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, \(z = (0.35 - 0.33)/sqrt((0.33*(1-0.33))/2765) = 1.75\).

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.

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.

Key concepts

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.