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.