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For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Medical Supply Sales According to the U.S. Bureau of Labor Statistics, approximately equal numbers of men and women are engaged in sales and related occupations. Although that may be true for total numbers, perhaps the proportions differ by industry. A random sample of 200 salespersons from the industrial sector indicated that 114 were men, and in the medical supply sector, 80 of 200 were men. At the 0.05 level of significance, can we conclude that the proportion of men in industrial sales differs from the proportion of men in medical supply sales?

Short Answer

Expert verified
Yes, the proportion of men in industrial sales differs from that in medical supply sales.

Step by step solution

01

State the Hypotheses and Identify the Claim

In this problem, we are comparing two proportions: the proportion of men in industrial sales and the proportion of men in medical supply sales. Let \( p_1 \) be the proportion of men in industrial sales and \( p_2 \) be the proportion of men in medical supply sales. Our null hypothesis \( H_0 \) is that the two proportions are equal: \( H_0: p_1 = p_2 \). The alternative hypothesis \( H_a \), which is the claim, is that the two proportions are not equal: \( H_a: p_1 eq p_2 \). This is a two-tailed test.
02

Find the Critical Value(s)

For a two-tailed test at the \( 0.05 \) level of significance, we need to find the critical z-values that correspond to the significance level of \( 0.025 \) in each tail. Using a standard normal distribution table, the critical z-values are approximately \( z = -1.96 \) and \( z = 1.96 \).
03

Compute the Test Value

First, calculate the sample proportions: \( \hat{p_1} = \frac{114}{200} = 0.57 \) and \( \hat{p_2} = \frac{80}{200} = 0.40 \). Next, find the pooled sample proportion \( \hat{p} \) using \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{114 + 80}{200 + 200} = 0.485 \). The test statistic is calculated using the formula: \[ z = \frac{\hat{p_1} - \hat{p_2}}{\sqrt{\hat{p}(1 - \hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \].Plug in the values: \[ z = \frac{0.57 - 0.40}{\sqrt{0.485 \times (1 - 0.485) \times \left(\frac{1}{200} + \frac{1}{200}\right)}} \approx 3.08 \].
04

Make the Decision

The computed z-value of \( 3.08 \) is greater than the critical value of \( 1.96 \) (and also greater than \( -1.96 \) if considering the absolute value). Since the test statistic falls in the rejection region, we reject the null hypothesis \( H_0 \).
05

Summarize the Results

At the \( 0.05 \) level of significance, there is enough evidence to support the claim that the proportion of men in industrial sales differs from the proportion of men in medical supply sales.

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Key Concepts

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

Proportion Comparison
Proportion comparison is a critical technique in hypothesis testing when you want to evaluate if a significant difference exists between two groups. In our exercise, we're comparing the proportion of men in industrial sales to those in medical supply sales. Instead of comparing raw numbers, which may be misleading due to varying sample sizes, proportions standardize the comparison by considering the relative size of each group.
To start, we calculate the sample proportions of men in each sector:
  • The industrial sales sector has a sample proportion (\( \hat{p_1} \)) of \( 0.57 \) (114 out of 200).
  • The medical supply sector has a sample proportion (\( \hat{p_2} \)) of \( 0.40 \) (80 out of 200).
From this, proportions provide a normalized basis for comparison, allowing for more accurate hypothesis testing. This is a foundational step before moving forward with any statistical tests.
Two-Tailed Test
A two-tailed test in hypothesis testing looks for evidence of either direction of effect, meaning it checks if one proportion is simply not equal to another. This differs from a one-tailed test which only seeks evidence in one specific direction (greater or lesser).
In the case of our exercise, the two-tailed test is employed to determine if the proportion of men in industrial sales significantly differs, either way, from those in medical supply sales. This means:
  • If the test statistic is too high or too low beyond critical values, it signifies a significant difference.
  • The test is unbiased towards which side the difference might lie.
Thus, adopting a two-tailed test covers all bases for potential variation or discrepancy between the two groups being compared.
Significance Level
The significance level, often denoted as \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true (Type I error). Common values for \( \alpha \) include 0.05, 0.01, and 0.10, with 0.05 being the most commonly used in many scientific disciplines.
For our exercise, a significance level of 0.05 is chosen. It implies that there is a 5% risk of concluding that a difference exists when there is none. In more layman's terms, the decision to reject the null hypothesis reflects a trade-off between the risk of making a Type I error and the desire to identify a true effect. By setting \( \alpha = 0.05 \), we express a high level of confidence in our test results.
Critical Values
Critical values are threshold values that the test statistic must exceed for the null hypothesis to be rejected. They are determined based on the chosen significance level and the type of test (one-tailed or two-tailed).
In our specific example, for a two-tailed test at a 0.05 significance level, the critical values correspond to the outer edges of the standard normal distribution. They are approximately \( z = -1.96 \) and \( z = 1.96 \). If the computed test statistic falls beyond these values:
  • It's considered significant enough to reject the null hypothesis.
  • An observed effect is deemed statistically meaningful.
Thus, critical values serve a pivotal role in hypothesis testing, offering the basis for decision making about the null hypothesis.

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Most popular questions from this chapter

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Noise Levels in Hospitals In a hospital study, it was found that the standard deviation of the sound levels from 20 randomly selected areas designated as "casualty doors" "was 4.1 dB A and the standard deviation of 24 randomly selected areas designated as operating theaters was 7.5 dB A. At \(\alpha=0.05,\) can you substantiate the claim that there is a difference in the standard deviations?

When a researcher selects all possible pairs of samples from a population in order to find the difference between the means of each pair, what will be the shape of the distribution of the differences when the original distributions are normally distributed? What will be the mean of the distribution? What will be the standard deviation of the distribution?

Teacher Salaries A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers in a large school district. The mean of the salaries of a random sample of 26 elementary school teachers is dollar 48,256, and the sample standard deviation is dollar 3,912.40 . The mean of the salaries of a random sample of 24 secondary school teachers is dollar 45,633 . The sample standard deviation is dollar 5533 . At \(\alpha=0.05,\) can it be concluded that the mean of the salaries of the elementary school teachers is greater than the mean of the salaries of the secondary school teachers? Use the \(P\) -value method.

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Mistakes in a Song A random sample of six music students played a short song, and the number of mistakes in music each student made was recorded. After they practiced the song 5 times, the number of mistakes each student made was recorded. The data are shown. At \(\alpha=0.05,\) can it be concluded that there was a decrease in the mean number of mistakes? $$ \begin{array}{l|cccccc}{\text { Student }} & {\mathrm{A}} & {\mathrm{B}} & {\mathrm{C}} & {\mathrm{D}} & {\mathrm{E}} & {\mathrm{F}} \\ \hline \text { Before } & {10} & {6} & {8} & {8} & {13} & {8} \\ \hline \text { After } & {4} & {2} & {2} & {7} & {8} & {9}\end{array} $$

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Wolf Pack Pups Does the variance in the number of pups per pack differ between Montana and Idaho wolf packs? Random samples of packs were selected for each area, and the numbers of pups per pack were recorded. At the 0.05 level of significance, can a difference in variances be concluded? $$ \begin{array}{l|cccccccc}{\text { Montana }} & {4} & {3} & {5} & {6} & {1} & {2} & {8} & {2} \\ \hline \text { wolf packs } & {3} & {1} & {7} & {6} & {5} & {} & {} \\ \hline \text { Idaho } & {2} & {4} & {5} & {4} & {2} & {4} & {6} & {3} \\ \hline \text { wolf packs } & {1} & {4} & {2} & {1} & {}\end{array} $$

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