https://file.aiquickdraw.com/mathbot/file/1729389723519tdlw22qe.jpg

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.

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 13 processing times from computer 1 showed a mean of 53 seconds with a standard deviation of 17 seconds, while a random sample of 10 processing times from computer 2 showed a mean of 67 seconds with a standard deviation of 15 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 μ1 − μ2 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.

Calculate the 95% confidence interval for the difference between the mean processing times of two computers. The problem involves sample means, pooled variance, and t-distribution for computing the interval. Includes detailed steps for deriving the lower and upper limits of the interval.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation