High School

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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]

Answer :

To determine the 98% confidence interval for the mean difference in scores (after - before), follow these steps:

1. Calculate the Mean Difference:
The mean difference given is 193 points.

2. Find the Standard Deviation:
The standard deviation of the differences is 62.73 points.

3. Determine the Sample Size:
The sample size is 8 students.

4. Degrees of Freedom:
For the t-distribution, the degrees of freedom is the sample size minus one. Thus, degrees of freedom = 8 - 1 = 7.

5. Locate the t Value:
For a 98% confidence level and 7 degrees of freedom, you use a t-distribution table to find the t value. The correct t* value provided is 2.821.

6. Calculate the Standard Error:
The standard error (SE) is found using the formula:
[tex]\[
SE = \frac{\text{Standard Deviation}}{\sqrt{\text{Sample Size}}}
\][/tex]
So:
[tex]\[
SE = \frac{62.73}{\sqrt{8}} \approx 22.18
\][/tex]

7. Compute the Margin of Error:
The margin of error (ME) is calculated using:
[tex]\[
ME = t^* \times SE
\][/tex]
So:
[tex]\[
ME = 2.821 \times 22.18 \approx 62.57
\][/tex]

8. Determine the Confidence Interval:
The 98% confidence interval is the mean difference plus or minus the margin of error:
- Lower Bound = 193 - 62.57 ≈ 130.43
- Upper Bound = 193 + 62.57 ≈ 255.57

Therefore, the 98% confidence interval for the mean difference in scores is approximately (130.43, 255.57).

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