Question 1196934: 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?
Giva Farmula for calculating when you have two sample test of mean and SD is unknown?
State what will be null hypothesis and alternate hypotheses at05 significance level. Dose it take longer to resolve software issues
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! We will conduct a two-sample t-test to determine whether it takes longer to resolve software issues compared to hardware issues. Here's the step-by-step process:
---
### Formula for Two-Sample t-Test (Unequal Variances)
\[
t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]
Where:
- \(\bar{X}_1, \bar{X}_2\): Sample means
- \(s_1, s_2\): Sample standard deviations
- \(n_1, n_2\): Sample sizes
Degrees of freedom (\(df\)) are approximated as:
\[
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}
\]
---
### Hypotheses
- **Null Hypothesis (\(H_0\))**: There is no difference in the time taken to resolve software and hardware issues (\(\mu_1 = \mu_2\)).
- **Alternative Hypothesis (\(H_a\))**: It takes longer to resolve software issues (\(\mu_1 > \mu_2\)).
This is a **one-tailed test** at the 0.05 significance level.
---
### Given Data
- Software calls:
- \(n_1 = 35\), \(\bar{X}_1 = 18\), \(s_1 = 4.2\)
- Hardware calls:
- \(n_2 = 45\), \(\bar{X}_2 = 15.5\), \(s_2 = 3.9\)
---
### Step 1: Calculate the Test Statistic
Substitute the values into the t-test formula:
\[
t = \frac{18 - 15.5}{\sqrt{\frac{4.2^2}{35} + \frac{3.9^2}{45}}}
\]
\[
t = \frac{2.5}{\sqrt{\frac{17.64}{35} + \frac{15.21}{45}}}
\]
\[
t = \frac{2.5}{\sqrt{0.504 + 0.338}}
\]
\[
t = \frac{2.5}{\sqrt{0.842}} = \frac{2.5}{0.9177} \approx 2.72
\]
---
### Step 2: Degrees of Freedom
Calculate \(df\):
\[
df = \frac{\left(0.504 + 0.338\right)^2}{\frac{0.504^2}{34} + \frac{0.338^2}{44}}
\]
\[
df = \frac{0.842^2}{\frac{0.254}{34} + \frac{0.114}{44}} = \frac{0.708}{0.00747 + 0.00259} = \frac{0.708}{0.01006} \approx 70.34
\]
Round \(df\) to 70 for simplicity.
---
### Step 3: Compare to Critical Value and Find p-Value
For a one-tailed test at \(\alpha = 0.05\) and \(df = 70\), the critical \(t\)-value is approximately **1.667**.
Since \(t = 2.72 > 1.667\), we reject \(H_0\).
To find the p-value, use a t-distribution table or software. The p-value for \(t = 2.72\) with \(df = 70\) is approximately **0.004**.
---
### Conclusion
- **At the 0.05 significance level**, we conclude that it takes longer to resolve software issues.
- **p-value**: \(0.004\)
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