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]

Use the [tex]$t$[/tex]-table to find the appropriate [tex]$t$[/tex]-value.

Answer :

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

1. Identify the Given Values:
- Mean of the differences: 193 points.
- Standard deviation of the differences: 62.73 points.
- Sample size: 8 students (since there are 8 differences given).

2. Determine the Degrees of Freedom:
- Degrees of freedom (df) = sample size - 1 = 8 - 1 = 7.

3. Find the Appropriate [tex]\( t \)[/tex]-value:
- For a 98% confidence level and 7 degrees of freedom, the [tex]\( t \)[/tex]-value is approximately 2.821.

4. Calculate the Standard Error:
- Standard error (SE) is calculated using the formula:
[tex]\[
\text{SE} = \frac{\text{Standard deviation}}{\sqrt{\text{Sample size}}}
\][/tex]
- So the standard error is:
[tex]\[
\text{SE} = \frac{62.73}{\sqrt{8}} \approx 22.18
\][/tex]

5. Calculate the Margin of Error:
- Margin of error (ME) is calculated by multiplying the [tex]\( t \)[/tex]-value by the standard error:
[tex]\[
\text{ME} = t \times \text{SE} = 2.821 \times 22.18 \approx 62.57
\][/tex]

6. Determine the Confidence Interval:
- The confidence interval is calculated by adding and subtracting the margin of error from the mean difference:
- Lower bound = Mean difference - ME = 193 - 62.57 = 130.43
- Upper bound = Mean difference + ME = 193 + 62.57 = 255.57

So, the 98% confidence interval for the mean difference in scores is approximately [tex]\( (130.43, 255.57) \)[/tex].