The mean of the differences is 193 points, and the standard deviation of the differences is 62.73 points. The conditions for inference are met.

What is the correct [tex]$98 \%$[/tex] confidence interval for the mean difference (after - before) in score?

A. [tex]$193 \pm 2.764\left(\frac{62.73}{\sqrt{9}}\right)$[/tex]

B. [tex]$193 \pm 2.764\left(\frac{62.73}{\sqrt{10}}\right)$[/tex]

C. [tex]$193 \pm 2.821\left(\frac{62.73}{\sqrt{9}}\right)$[/tex]

D. [tex]$193 \pm 2.821\left(\frac{62.73}{\sqrt{10}}\right)$[/tex]

Question

High School

Answer :

imranhasan123 imranhasan123 · Answer 1 · 2024-12-20T02:54:57-05:00

Answer:

Step-by-step explanation:To find the correct 98% confidence interval, we need to:

1. Identify the sample size (n): Not explicitly given, but we have two options: n = 9 or n = 10.

2. Determine the degrees of freedom (df): df = n - 1.

3. Find the critical t-value for the 98% confidence interval using the t-table.

Assuming df = 9 - 1 = 8 (for n = 9) or df = 10 - 1 = 9 (for n = 10), we can look up the critical t-value:

For df = 8 and 98% confidence, the critical t-value is approximately 2.896.

For df = 9 and 98% confidence, the critical t-value is approximately 2.821.

Comparing the options:

A. 193±2.764(962.73): Incorrect critical t-value.

B. 193±2.764(1062.73): Incorrect critical t-value.

C. 193±2.821(962.73): Correct critical t-value for df = 9.

D. 193±2.821(1062.73): Correct critical t-value, but the sample size might be 9.

Given the options and the critical t-value, the most likely correct answer is:

C. 193±2.821(962.73)