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Topic 6 Homework (Nonadaptive)
Question 3 of 11 (1 point)|Question Attempt: 1 of Unlimited

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Question 3
The university data center has two main computers. The center wants to examine whether computer 1 is receiving tasks that require processing times comparable to those of computer 2. A random sample of 10 processing times from computer 1 showed a mean of 59 seconds with a standard deviation of 16 seconds, while a random sample of 13 processing times from computer 2 (chosen independently of those for computer 1) showed a mean of 70 seconds with a standard deviation of 20 seconds. Assume that the populations of processing times are normally distributed for each of the two computers and that the variances are equal. Construct a 95% confidence interval for the difference −μ1μ2 between the mean processing time of computer 1, μ1, and the mean processing time of computer 2, μ2. Then find the lower limit and upper limit of the 95% confidence interval.
Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. (If necessary, consult a list of formulas.)

Lower limit:
Upper limit:

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To construct a 95% confidence interval for the difference between the mean processing times of the two computers (\(\mu_1 - \mu_2\)), we can use the formula for a confidence interval when the variances are assumed equal. Here's the process:

### Given Data
- Mean of computer 1, \(\bar{x}_1 = 59\)
- Standard deviation of computer 1, \(s_1 = 16\)
- Sample size of computer 1, \(n_1 = 10\)
- Mean of computer 2, \(\bar{x}_2 = 70\)
- Standard deviation of computer 2, \(s_2 = 20\)
- Sample size of computer 2, \(n_2 = 13\)

### Step 1: Calculate the Pooled Standard Deviation (\(s_p\))
The pooled standard deviation is calculated as:
\[
s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}
\]
Plugging in the values:
\[
s_p = \sqrt{\frac{(10 - 1)(16^2) + (13 - 1)(20^2)}{10 + 13 - 2}}
\]
\[
s_p = \sqrt{\frac{9 \cdot 256 + 12 \cdot 400}{21}}
\]
\[
s_p = \sqrt{\frac{2304 + 4800}{21}}
\]
\[
s_p = \sqrt{\frac{7104}{21}}
\]
\[
s_p \approx \sqrt{338.29} \approx 18.39
\]

### Step 2: Calculate the Standard Error (SE) of the Difference in Means
\[
SE = s_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}
\]
\[
SE = 18.39 \cdot \sqrt{\frac{1}{10} + \frac{1}{13}}
\]
\[
SE = 18.39 \cdot \sqrt{0.1 + 0.0769}
\]
\[
SE = 18.39 \cdot \sqrt{0.1769}
\]
\[
SE \approx 18.39 \cdot 0.4205 \approx 7.74
\]

### Step 3: Find the Critical Value (\(t\)) for a 95% Confidence Level
Since \(n_1 + n_2 - 2 = 21\), we'll use a \(t\)-distribution with 21 degrees of freedom. The critical value \(t_{0.025, 21}\) for a two-tailed test at a 95% confidence level is approximately 2.080.

### Step 4: Construct the Confidence Interval
\[
(\bar{x}_1 - \bar{x}_2) \pm t \cdot SE
\]
\[
(59 - 70) \pm 2.080 \cdot 7.74
\]
\[
-11 \pm 16.0992
\]

So, the confidence interval is:
\[
(-27.10, 5.10)
\]

### Final Answers
- **Lower limit**: \(-27.10\)
- **Upper limit**: \(5.10\)

This interval suggests that the difference in mean processing times between the two computers could range from \(-27.10\) seconds to \(5.10\) seconds.

Would you like further details on any step or clarification?

### Related Questions
1. How would the confidence interval change if the variances were not assumed equal?
2. What happens to the confidence interval if the sample sizes are increased for both computers?
3. What is the interpretation of the confidence interval in the context of hypothesis testing?
4. How would you calculate the confidence interval if only the sample size of computer 1 changed?
5. How does the critical value change if the confidence level is adjusted to 99%?

**Tip:** Always verify the degrees of freedom when using the \(t\)-distribution, as it directly influences the critical value and, thus, the confidence interval.

The university data center has two main computers. The center wants to examine whether computer 1 is receiving tasks that require processing times comparable to those of computer 2. A random sample of 10 processing times from computer 1 showed a mean of 59 seconds with a standard deviation of 16 seconds, while a random sample of 13 processing times from computer 2 (chosen independently of those for computer 1) showed a mean of 70 seconds with a standard deviation of 20 seconds. Assume that the populations of processing times are normally distributed for each of the two computers and that the variances are equal. Construct a 95% confidence interval for the difference −μ1μ2 between the mean processing time of computer 1, μ1, and the mean processing time of computer 2, μ2. Then find the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places.

Learn how to calculate a 95% confidence interval for the difference in mean processing times between two computers. This detailed solution walks through statistical concepts such as pooled standard deviation, standard error, and the t-distribution, with a practical application in comparing the performance of computer systems.

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