To solve this problem, we need to compute the 95% confidence interval for the difference in mean processing times between computer 1 ($\mu_1$) and computer 2 ($\mu_2$). The sample data provided is:

• For Computer 1:

  • Sample size ($n_1$) = 13
  • Sample mean ($\bar{x}_1$) = 53 seconds
  • Sample standard deviation ($s_1$) = 17 seconds

• For Computer 2:

  • Sample size ($n_2$) = 10
  • Sample mean ($\bar{x}_2$) = 67 seconds
  • Sample standard deviation ($s_2$) = 15 seconds

Step 1: Find the difference in sample means

$\bar{x}_1 - \bar{x}_2 = 53 - 67 = -14$

Step 2: Pooled standard deviation formula

Since the problem assumes that the population variances are equal, we use the pooled variance to estimate the standard error of the difference between the two means. The formula for the pooled standard deviation ($S_p$) is:

$S_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}$

Substituting the values:

$S_p = \sqrt{\frac{(13-1)(17^2) + (10-1)(15^2)}{13 + 10 - 2}} = \sqrt{\frac{12(289) + 9(225)}{21}} = \sqrt{\frac{3468 + 2025}{21}} = \sqrt{261.857} \approx 16.18$

Step 3: Standard error of the difference in means

The standard error for the difference in means is given by:

$SE = S_p \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$

Substituting the values:

$SE = 16.18 \times \sqrt{\frac{1}{13} + \frac{1}{10}} = 16.18 \times \sqrt{0.07692 + 0.1} = 16.18 \times \sqrt{0.17692} \approx 16.18 \times 0.4207 \approx 6.81$

Step 4: Find the critical value

For a 95% confidence interval, the critical value for a two-tailed t-distribution with $n_1 + n_2 - 2 = 21$ degrees of freedom can be found using a t-table or calculator. The critical value ($t_{\alpha/2}$) is approximately:

$t_{\alpha/2} \approx 2.080$

Step 5: Compute the confidence interval

The confidence interval is given by:

$(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \times SE$

Substituting the values:

$-14 \pm 2.080 \times 6.81$

$-14 \pm 14.16$

Thus, the confidence interval is:

$(-14 - 14.16, -14 + 14.16) = (-28.16, 0.16)$

Final Answer:

• Lower limit: $-28.16$
• Upper limit: $0.16$

Would you like further details or have any questions?

Here are 5 related questions to expand this:

1. What if the variances were not equal? How would the calculation change?
2. How would the confidence interval change if we wanted a 99% confidence level?
3. How do we interpret the confidence interval in terms of performance between the two computers?
4. Why is it important to assume normal distribution in this calculation?
5. What if the sample sizes were the same for both computers?

Tip: When calculating pooled variance, always ensure the degrees of freedom (n1 + n2 - 2) are correctly applied for precise estimates.