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Answer :
To find the correct [tex]\(98\%\)[/tex] confidence interval for the mean difference in score, we need to follow these steps:
1. Identify the Mean and Standard Deviation:
- Mean of differences = 193 points
- Standard deviation of differences = 62.73 points
2. Determine the Sample Size:
- The sample size = 8 students
3. Identify the Confidence Level and Corresponding [tex]\(t\)[/tex]-value:
- We are given a confidence level of [tex]\(98\%\)[/tex]. For this level, the appropriate [tex]\(t\)[/tex]-value from a [tex]\(t\)[/tex]-distribution table for [tex]\(n - 1 = 8 - 1 = 7\)[/tex] degrees of freedom is 2.821.
4. Calculate the Standard Error (SE):
[tex]\[
SE = \frac{\text{standard deviation}}{\sqrt{\text{sample size}}} = \frac{62.73}{\sqrt{8}} \approx 22.18
\][/tex]
5. Calculate the Margin of Error:
[tex]\[
\text{Margin of Error} = t \times SE = 2.821 \times 22.18 \approx 62.57
\][/tex]
6. Calculate the Confidence Interval:
[tex]\[
\text{Lower Bound} = \text{Mean} - \text{Margin of Error} = 193 - 62.57 \approx 130.43
\][/tex]
[tex]\[
\text{Upper Bound} = \text{Mean} + \text{Margin of Error} = 193 + 62.57 \approx 255.57
\][/tex]
Thus, the [tex]\(98\%\)[/tex] confidence interval for the mean difference in score is approximately [tex]\( (130.43, 255.57) \)[/tex].
Based on these calculations and given options, the fourth option [tex]\(193 \pm 2.821\left(\frac{62.73}{\sqrt{10}}\right)\)[/tex] uses a slightly incorrect sample size for the square root. However, with our process using 8, the interval was found as above, highlighting that proper sample recognition should yield [tex]\(193 \pm 2.821\left(\frac{62.73}{\sqrt{8}}\right)\)[/tex] to match calculated results if sample size were reflected accurately in adjustment terms involving the table.
1. Identify the Mean and Standard Deviation:
- Mean of differences = 193 points
- Standard deviation of differences = 62.73 points
2. Determine the Sample Size:
- The sample size = 8 students
3. Identify the Confidence Level and Corresponding [tex]\(t\)[/tex]-value:
- We are given a confidence level of [tex]\(98\%\)[/tex]. For this level, the appropriate [tex]\(t\)[/tex]-value from a [tex]\(t\)[/tex]-distribution table for [tex]\(n - 1 = 8 - 1 = 7\)[/tex] degrees of freedom is 2.821.
4. Calculate the Standard Error (SE):
[tex]\[
SE = \frac{\text{standard deviation}}{\sqrt{\text{sample size}}} = \frac{62.73}{\sqrt{8}} \approx 22.18
\][/tex]
5. Calculate the Margin of Error:
[tex]\[
\text{Margin of Error} = t \times SE = 2.821 \times 22.18 \approx 62.57
\][/tex]
6. Calculate the Confidence Interval:
[tex]\[
\text{Lower Bound} = \text{Mean} - \text{Margin of Error} = 193 - 62.57 \approx 130.43
\][/tex]
[tex]\[
\text{Upper Bound} = \text{Mean} + \text{Margin of Error} = 193 + 62.57 \approx 255.57
\][/tex]
Thus, the [tex]\(98\%\)[/tex] confidence interval for the mean difference in score is approximately [tex]\( (130.43, 255.57) \)[/tex].
Based on these calculations and given options, the fourth option [tex]\(193 \pm 2.821\left(\frac{62.73}{\sqrt{10}}\right)\)[/tex] uses a slightly incorrect sample size for the square root. However, with our process using 8, the interval was found as above, highlighting that proper sample recognition should yield [tex]\(193 \pm 2.821\left(\frac{62.73}{\sqrt{8}}\right)\)[/tex] to match calculated results if sample size were reflected accurately in adjustment terms involving the table.
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