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A computer manufacturer offers a help line that purchasers can call for help 24 hours a day 7 days a week. Clearing these calls for help in a timely fashion is important to the company's image. After telling the caller that resolution of the problem is important the caller is asked whether the issue is "software" or "hardware" related. The mean time it takes a technician to resolve a software issue is 18 minutes with a standard deviation of 4.2 minutes. This information was obtained from a sample of 35 monitored calls. For a study of 45 hardware issues, the mean time for the technician to resolve the problem was 15.5 minutes with a standard deviation of 3.9 minutes. This information was also obtained from monitored calls. At the . 05 significance level is it reasonable to conclude that it takes longer to resolve software issues? What is the \(p\) -value?

Step by step solution

01

State the Hypotheses

To determine if software issues take longer to resolve than hardware issues, we will formulate the null and alternative hypotheses. The null hypothesis (H0) is that the mean time to resolve software issues is less than or equal to the mean time to resolve hardware issues: H0: \( \mu_1 \leq \mu_2 \).The alternative hypothesis (H1) is that the mean time to resolve software issues is greater than the mean time to resolve hardware issues: H1: \( \mu_1 > \mu_2 \).

02

Calculate the Test Statistic

To compare the means of two independent samples, we use a two-sample t-test. The formula for the test statistic is:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \(\bar{x}_1 = 18\), \(s_1 = 4.2\), \(n_1 = 35\), \(\bar{x}_2 = 15.5\), \(s_2 = 3.9\), and \(n_2 = 45\).Substitute the values: \[ t = \frac{18 - 15.5}{\sqrt{\frac{4.2^2}{35} + \frac{3.9^2}{45}}} \].

03

Calculate and Interpret the p-value

First, calculate the test statistic using the values:\[ t = \frac{2.5}{\sqrt{\frac{17.64}{35} + \frac{15.21}{45}}} = \frac{2.5}{\sqrt{0.504 + 0.338}} = \frac{2.5}{\sqrt{0.842}} = \frac{2.5}{0.917} \approx 2.73 \].To find the p-value, use a t-distribution with degrees of freedom (df) calculated using:\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \] approximately 76.82.Using df ≈ 76, the p-value for t = 2.73 can be found in t-distribution tables or statistical software. It is ascertained that p < 0.05.

04

Decision Making

With a p-value < 0.05, we reject the null hypothesis at the 5% significance level. This suggests there is enough statistical evidence to conclude that the mean time to resolve software issues is significantly greater than for hardware issues.

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Key Concepts

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

Hypothesis Testing

In hypothesis testing, we compare two groups to determine if there is a significant difference between them. Here, the goal is to see whether it takes longer for technicians to resolve software issues than hardware issues. We begin by setting up two hypotheses:

• Null Hypothesis (H0): The mean resolution time for software issues is less than or equal to that for hardware issues. Mathematically, this is expressed as: \( \mu_1 \leq \mu_2 \).
• Alternative Hypothesis (H1): The mean resolution time for software issues is greater than for hardware issues: \( \mu_1 > \mu_2 \).

Hypothesis testing helps us make data-informed conclusions by providing a structured way to assess the evidence against the null hypothesis. In this exercise, the two-sample t-test is used to compare mean resolution times. This test is appropriate because we're comparing means from two independent samples: one for software and one for hardware issues.

Significance Level

The significance level, often denoted by \( \alpha \), defines the threshold for deciding whether to reject the null hypothesis. In this exercise, a significance level of 0.05 is used, which is a common choice in many statistical tests.

• A 5% significance level means you would expect to reject the null hypothesis only 5 times out of 100 if it were true.
• This level indicates a willingness to accept a 5% risk of concluding there is an effect when there isn't (a Type I error).

When the p-value calculated in the test is less than 0.05, it suggests that the observed effect is statistically significant. This signals that there is strong evidence against the null hypothesis, prompting its rejection. In this case, the significance level aids us in determining whether the observed time difference for resolving software and hardware issues is due to random variation or a real difference.

P-Value

The p-value in the context of hypothesis testing indicates the probability of observing results as extreme as those measured, under the null hypothesis. In simpler terms, it tells us how extreme our observed data is relative to the null hypothesis being true.

• A small p-value (typically \( < \alpha = 0.05 \)) indicates strong evidence against the null hypothesis, suggesting that the difference in average resolution times is statistically significant.
• In this exercise, a t-test yielded a p-value that was less than 0.05.

This means the observed mean resolution time for software issues being greater than that for hardware issues is unlikely to have arisen by chance alone. Thus, with a p-value under the significance threshold, we reject the null hypothesis, concluding there is indeed a greater mean resolution time for software.

Degrees of Freedom

Degrees of freedom (df) are crucial in determining the distribution of the test statistic, which in turn affects the precision of the p-value estimation. The formula used for calculating degrees of freedom when performing a two-sample t-test is usually a bit complex when the variances are unequal, as it is case-dependent.

• In this exercise, the degrees of freedom were calculated using a specific formula :
• \[df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]
• This results in approximately 76.82 degrees of freedom.

With these degrees of freedom, we then reference the t-distribution to find the critical value or p-value for our test statistic. The choice of using degrees of freedom in this way accounts for variability and assures the accuracy of the statistical results, strengthening the validity of our conclusions.