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Chapter 11: Problem 22

A computer manufacturer offers technical support that is available 24 hours a day, 7 days a week. Timely resolution of these calls is important to the company's image. For 35 calls that were related to software, technicians resolved the issues in a mean time of 18 minutes with a standard deviation of 4.2 minutes. For 45 calls related to hardware, technicians resolved the problems in a mean time of 15.5 minutes with a standard deviation of 3.9 minutes. At the .05 significance level, does it take longer to resolve software issues? What is the \(p\) -value?

Short Answer

Expert verified
The p-value is calculated to assess if software issues take longer to resolve. Compare p-value to \( \alpha = 0.05 \) to conclude.

Step by step solution

01

Formulate the Hypotheses

First, we need to set up our null and alternative hypotheses. For this test, the null hypothesis \( H_0 \) is that the mean time to resolve software issues is less than or equal to the mean time to resolve hardware issues, \( \mu_{software} \leq \mu_{hardware} \). The alternative hypothesis \( H_a \) is that the mean time to resolve software issues is greater, \( \mu_{software} > \mu_{hardware} \).
02

Identify the Test and Inputs

Given the nature of the problem, we'll use a two-sample t-test for the means of independent samples. We know that:\[ n_1 = 35, \bar{x}_1 = 18, s_1 = 4.2 \] for software issues and\[ n_2 = 45, \bar{x}_2 = 15.5, s_2 = 3.9 \] for hardware issues.
03

Calculate the Test Statistic

The test statistic \( t \) for a two-sample t-test is calculated as:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute in the known values:\[ t = \frac{18 - 15.5}{\sqrt{\frac{4.2^2}{35} + \frac{3.9^2}{45}}} \]
04

Calculate the Critical Value and Determine Significance

Using a t-distribution with degrees of freedom approximated by\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]Substitute the values to find the degrees of freedom, then check the t-distribution table for significance at \( \alpha = 0.05 \) for a one-tailed test.
05

Find the p-value

Using the calculated test statistic and the degrees of freedom, find the p-value using statistical software or a t-distribution table. The p-value indicates the probability of observing the data if the null hypothesis is true.
06

Conclusion

Compare the p-value with the significance level \( \alpha = 0.05 \). If the p-value is less than \( \alpha \), reject the null hypothesis, indicating that resolving software issues takes longer than hardware ones.

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Key Concepts

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

Null Hypothesis
When conducting a statistical test, we begin by formulating two types of hypotheses. The first one is the null hypothesis, represented as \( H_0 \). The null hypothesis is a statement that there is no effect or no difference, and it serves as the starting point for any statistical test. In the context of the computer manufacturer's support issue, the null hypothesis posits that the mean time to resolve software issues is less than or equal to the mean time for hardware issues, which is mathematically expressed as \( \mu_{software} \leq \mu_{hardware} \).
  • The null hypothesis assumes that any observed differences are due to random chance.
  • It is the hypothesis that the test seeks to reject in order to support the alternative hypothesis.
Rejection of the null hypothesis suggests that the sample data provides sufficient evidence that the mean time to resolve software issues is indeed greater than for hardware issues.
Alternative Hypothesis
In contrast to the null hypothesis, the alternative hypothesis asserts a specific effect or difference. It is denoted by \( H_a \). The alternative hypothesis represents the claim that is subject to verification. For the computer manufacturer's technical support question, the alternative hypothesis states that the mean time to resolve software issues is greater than the mean time for hardware issues, symbolized as \( \mu_{software} > \mu_{hardware} \).
  • The alternative hypothesis is what researchers aim to prove through their data analysis.
  • Acceptance of the alternative hypothesis indicates that there is enough statistical evidence to suggest a significant difference or effect.
In testing terms, if the statistical analysis shows that it is unlikely for the null hypothesis to hold true given the sample data, the alternative hypothesis is supported.
P-Value
The p-value is a statistical measure that helps determine the strength of the evidence against the null hypothesis. It indicates the probability of obtaining an observed sample result, or something more extreme, if the null hypothesis is true. A lower p-value suggests stronger evidence to reject the null hypothesis. In practical terms:
  • A very small p-value (typically \(< 0.05\)) indicates strong evidence against the null hypothesis.
  • The p-value helps guide the decision on whether to reject the null hypothesis in favor of the alternative hypothesis.
In this exercise, after calculating the test statistic and its corresponding p-value, you compare it to a significance level (commonly set at 0.05). If the p-value is below this threshold, it suggests that the data supports the conclusion that resolving software issues takes longer than resolving hardware issues.
Degrees of Freedom
Degrees of freedom are a critical concept in statistics, particularly in hypothesis testing such as the t-test. They refer to the number of independent values or quantities that can vary in the statistical analysis. For a two-sample t-test, like the one used to compare resolution times for software versus hardware issues, degrees of freedom are determined using a specific formula. This formula considers the sample sizes and variances from each group involved.
  • Degrees of freedom affect the shape of the t-distribution, which is central to calculating the test statistic's critical value and p-value.
  • More degrees of freedom typically lead to a distribution that more closely resembles a normal distribution.
In practice, once the degrees of freedom have been calculated using the formula, you consult a t-distribution table to find the critical t-value for a given significance level, aiding in the hypothesis test's interpretation.

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Most popular questions from this chapter

A sample of 40 observations is selected from one population with a population standard deviation of \(5 .\) The sample mean is \(102 .\) A sample of 50 observations is selected from a second population with a population standard deviation of \(6 .\) The sample mean is 99. Conduct the following test of hypothesis using the .04 significance level. $$ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \\ H_{1}: \mu_{1} \neq \mu_{2} \end{array} $$ a. Is this a one-tailed or a two-tailed test? b. State the decision rule. c. Compute the value of the test statistic. d. What is your decision regarding \(H_{0} ?\) e. What is the \(p\) -value?

Mary Jo Fitzpatrick is the vice president for Nursing Services at St. Luke's Memorial Hospital. Recently she noticed in the job postings for nurses that those that are unionized seem to offer higher wages. She decided to investigate and gathered the following information. $$ \begin{array}{|lccc|} \hline & & \text { Sample } & \text { Population } \\ \text { Group } & \text { Sample Size } & \text { Mean Wage } & \text { Standard Deviation } \\ \hline \text { Union } & 40 & \$ 20.75 & \$ 2.25 \\ \text { Nonunion } & 45 & \$ 19.80 & \$ 1.90 \\ \hline \end{array} $$ Would it be reasonable for her to conclude that union nurses earn more? Use the .02 significance level. What is the \(p\) -value?

A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 40 subscribers to Plan \(A\) is \(\$ 57,000\) with a standard deviation of \(\$ 9,200 .\) For a sample of 30 subscribers to Plan \(B\), the mean income is \(\$ 61,000\) with a standard deviation of \(\$ 7,100 .\) At the .05 significance level, is it reasonable to conclude the mean income of those selecting Plan \(\mathrm{B}\) is larger? What is the \(p\) -value?

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.

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distributions follow the normal probability distribution and the population standard deviations are equal. The information is summarized below. $$ \begin{array}{|lcc|} \hline \text { Statistic } & \text { Men } & \text { Women } \\ \hline \text { Sample mean } & 24.51 & 22.69 \\ \text { Sample standard deviation } & 4.48 & 3.86 \\ \text { Sample size } & 35 & 40 \\ \hline \end{array} $$ At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? What is the \(p\) -value?
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