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As part of a study of corporate employees, the Director of Human 91Ó°ÊÓ for PNC, Inc. wants to compare the distance traveled to work by employees at their office in downtown Cincinnati with the distance for those in downtown Pittsburgh. A sample of 35 Cincinnati employees showed they travel a mean of 370 miles per month, with a standard deviation of 30 miles per month. A sample of 40 Pittsburgh employees showed they travel a mean of 380 miles per month, with a standard deviation of 26 miles per month. At the .05 significance level, is there a difference in the mean number of miles traveled per month between Cincinnati and Pittsburgh employees? Use the five-step hypothesis- testing procedure.

Short Answer

Expert verified
No, there is no significant difference in the mean miles traveled per month between Cincinnati and Pittsburgh employees.

Step by step solution

01

Understanding the Hypotheses

First, we need to establish our null and alternative hypotheses. The null hypothesis (\(H_0\)) states that there is no difference in the mean number of miles traveled to work per month between Cincinnati and Pittsburgh employees: \( \mu_{Cincinnati} = \mu_{Pittsburgh} \). The alternative hypothesis (\(H_1\)) states that there is a difference: \( \mu_{Cincinnati} eq \mu_{Pittsburgh} \).
02

Setting the Significance Level

The significance level for this test is given as \( \alpha = 0.05 \). This is the probability of rejecting the null hypothesis if it is actually true, and it provides the critical value for determining statistical significance.
03

Selecting the Test and Calculating the Test Statistic

We will use a two-sample t-test for comparing the means of two independent samples. The formula for the t-statistic is: \[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where \( \bar{x}_1 = 370 \), \( \bar{x}_2 = 380 \), \( s_1 = 30 \), \( s_2 = 26 \), \( n_1 = 35 \), and \( n_2 = 40 \). Inputting these values, we calculate the t-statistic as follows:\[t = \frac{370 - 380}{\sqrt{\frac{30^2}{35} + \frac{26^2}{40}}} = \frac{-10}{\sqrt{25.71}} = \frac{-10}{5.07} \approx -1.97 \]Thus, the calculated t-statistic is approximately -1.97.
04

Finding the Critical t-Value

Using a t-distribution table, we need to find the critical t-value for a two-tailed test with \(\alpha = 0.05\) and degrees of freedom based on the formula for Welch's approximation. The degrees of freedom \( df \) can be calculated using: \[df \approx \left(\frac{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}{\frac{\frac{s_1^2}{n_1}}{n_1-1} + \frac{\frac{s_2^2}{n_2}}{n_2-1}}\right)\]Plugging in the given values, the degrees of freedom \( df \) can be estimated as approximately 67. Using a t-table, the critical t-value for \( df = 67 \) and \( \alpha = 0.05 \) (two-tailed) is approximately ±2.00.
05

Making the Decision

The calculated t-statistic is -1.97. Comparing this to the critical t-values of ±2.00, we see that -1.97 does not fall outside the range defined by -2.00 and +2.00. Therefore, we fail to reject the null hypothesis.
06

Conclusion

We conclude that at the 0.05 significance level, there is not enough evidence to support the claim that there is a significant difference in the mean number of miles traveled per month between employees in Cincinnati and Pittsburgh.

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Key Concepts

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

Two-sample t-test
A two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. In this exercise, we apply it to compare the average distance traveled by employees from Cincinnati and Pittsburgh.

The main goal is to see if the difference in their travel distances is due to random variation or if it is statistically significant. To perform a two-sample t-test, we compute a test statistic, called the t-statistic, which helps in determining the likelihood that any observed difference is due to chance.

This type of test is appropriate when the data is approximately normally distributed and the sample sizes are relatively similar, although they do not have to be exactly equal. By calculating the t-statistic and comparing it to a critical value, we can make conclusions about our hypotheses.
Significance Level
The significance level is a key concept in hypothesis testing. It represents the probability of making a Type I error, which occurs if we incorrectly reject the null hypothesis. In simpler terms, it is the threshold that tells us how confident we need to be to declare a result as statistically significant.

In our exercise, this level is set at 0.05 (or 5%). This means we are willing to accept a 5% chance of being wrong when concluding that there is a difference in means.
  • **Why 0.05?**: This is a commonly used standard in many scientific fields; it's a balance between being too lenient (which might cause us to incorrectly claim significance) and being too strict (which might mean we miss detecting a true effect).
  • **Practical implication**: A result is considered statistically significant if the calculated probability (p-value) is less than the significance level, indicating that the observed effect is unlikely to be due to chance alone.
Null and Alternative Hypotheses
In hypothesis testing, we start by establishing two competing hypotheses: the null hypothesis and the alternative hypothesis.

**Null Hypothesis ( (H_0 ))**: This is a default statement suggesting no effect or difference exists. In our context, it means there is no difference in the mean number of miles traveled by Cincinnati and Pittsburgh employees ( μ_{Cincinnati} = μ_{Pittsburgh} ).

**Alternative Hypothesis ( (H_1 ))**: This proposes a contrasting statement to the null. It suggests that a difference does exist ( μ_{Cincinnati} eq μ_{Pittsburgh} ).
  • The null hypothesis is assumed true until evidence suggests otherwise.
  • The outcome of the test influences which hypothesis is accepted. If the null hypothesis is not rejected, we do not have sufficient evidence to conclude a difference exists. If rejected, it implies evidence for the alternative hypothesis.
Critical t-value
The critical t-value is a threshold value obtained from the t-distribution table. It is used to determine whether the calculated t-statistic falls within a range indicating statistical significance.

For the two-sample t-test, with a given degrees of freedom and a significance level, we find the critical t-value corresponding to a two-tailed test. These are the boundaries for deciding if we reject the null hypothesis.

In this exercise, our degrees of freedom were calculated around 67, providing critical t-values of approximately ±2.00.
  • If our calculated t-statistic falls outside this range, we reject the null hypothesis since it indicates a low probability of observing such a difference by chance.
  • If the t-statistic falls within the range, as it did here, there is not enough evidence to conclude a significant difference, and we fail to reject the null hypothesis.

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Most popular questions from this chapter

A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.

The null and alternate hypotheses are: $$\begin{array}{l}H_{0}: \mu_{1}=\mu_{2} \\\H_{1}: \mu_{1} \neq \mu_{2}\end{array}$$ A random sample of 10 observations from one population revealed a sample mean of 23 and a sample deviation of \(4 .\) A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of \(5 .\) At the .05 significance level, is there a difference between the population means?

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the afternoon shift than on the day shift. A sample of 54 day-shift workers showed that the mean number of units produced was \(345,\) with a standard deviation of 21 . A sample of 60 afternoon-shift workers showed that the mean number of units produced was \(351,\) with a standard deviation of 28 units. At the .05 significance level, is the number of units produced on the afternoon shift larger?

The Gibbs Baby Food Company wishes to compare the weight gain of infants using their brand versus their competitor's. A sample of 40 babies using the Gibbs products revealed a mean weight gain of 7.6 pounds in the first three months after birth. The standard deviation of the sample was 2.3 pounds. A sample of 55 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds, with a standard deviation of 2.9 pounds. At the .05 significance level, can we conclude that babies using the Gibbs brand gained less weight? Compute the \(p\) -value and interpret it.

The Willow Run Outlet Mall has two Gap Outlet Stores, one located on Peach Street and the other on Plum Street. The two stores are laid out differently, but both store managers claim their layout maximizes the amounts customers will purchase on impulse. A sample of 10 customers at the Peach Street store revealed they spent the following amounts more than planned: \(\$ 17.58, \$ 19.73, \$ 12.61, \$ 17.79, \$ 16.22, \$ 15.82, \$ 15.40, \$ 15.86, \$ 11.82,\) and \(\$ 15.85 .\) A sample of 14 customers at the Plum Street store revealed they spent the following amounts more than they planned: \(\$ 18.19, \$ 20.22, \$ 17.38, \$ 17.96, \$ 23.92, \$ 15.87\), \(\$ 16.47, \$ 15.96, \$ 16.79, \$ 16.74, \$ 21.40, \$ 20.57, \$ 19.79,\) and \(\$ 14.83 .\) At the .01 significance level, is there a difference in the mean amounts purchased on impulse at the two stores?

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