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Answer :
We are given the following information from a paired-sample study:
- Sample mean difference:
[tex]$$\bar{d} = 193$$[/tex]
- Sample standard deviation of the differences:
[tex]$$s_d = 62.73$$[/tex]
- Sample size:
[tex]$$n = 9$$[/tex]
Since the conditions for inference are met, we construct the confidence interval for the mean difference using the paired t‐test formula:
[tex]$$
\bar{d} \pm t^* \left(\frac{s_d}{\sqrt{n}}\right)
$$[/tex]
Step 1. Determine the standard error.
The standard error (SE) is computed by:
[tex]$$
\text{SE} = \frac{s_d}{\sqrt{n}} = \frac{62.73}{\sqrt{9}} = \frac{62.73}{3} \approx 20.91
$$[/tex]
Step 2. Identify the appropriate t critical value.
For a [tex]$98\%$[/tex] confidence interval with [tex]$n-1 = 8$[/tex] degrees of freedom, the t critical value is given as:
[tex]$$
t^* \approx 2.821
$$[/tex]
Step 3. Compute the margin of error (ME).
Multiply the t critical value by the standard error:
[tex]$$
\text{ME} = t^* \cdot \text{SE} = 2.821 \times 20.91 \approx 58.99
$$[/tex]
Step 4. Construct the confidence interval.
Using the formula, the confidence interval for the mean difference is:
[tex]$$
\bar{d} \pm \text{ME} = 193 \pm 58.99
$$[/tex]
Thus, the lower and upper bounds are computed as follows:
- Lower bound:
[tex]$$
193 - 58.99 \approx 134.01
$$[/tex]
- Upper bound:
[tex]$$
193 + 58.99 \approx 251.99
$$[/tex]
Step 5. Write the final answer.
The [tex]$98\%$[/tex] confidence interval for the mean difference in score is approximately:
[tex]$$
\boxed{193 \pm 2.821\left(\frac{62.73}{\sqrt{9}}\right) \quad \text{or} \quad (134.01, \, 251.99)}
$$[/tex]
This provides the interval within which we are [tex]$98\%$[/tex] confident the true mean difference lies.
- Sample mean difference:
[tex]$$\bar{d} = 193$$[/tex]
- Sample standard deviation of the differences:
[tex]$$s_d = 62.73$$[/tex]
- Sample size:
[tex]$$n = 9$$[/tex]
Since the conditions for inference are met, we construct the confidence interval for the mean difference using the paired t‐test formula:
[tex]$$
\bar{d} \pm t^* \left(\frac{s_d}{\sqrt{n}}\right)
$$[/tex]
Step 1. Determine the standard error.
The standard error (SE) is computed by:
[tex]$$
\text{SE} = \frac{s_d}{\sqrt{n}} = \frac{62.73}{\sqrt{9}} = \frac{62.73}{3} \approx 20.91
$$[/tex]
Step 2. Identify the appropriate t critical value.
For a [tex]$98\%$[/tex] confidence interval with [tex]$n-1 = 8$[/tex] degrees of freedom, the t critical value is given as:
[tex]$$
t^* \approx 2.821
$$[/tex]
Step 3. Compute the margin of error (ME).
Multiply the t critical value by the standard error:
[tex]$$
\text{ME} = t^* \cdot \text{SE} = 2.821 \times 20.91 \approx 58.99
$$[/tex]
Step 4. Construct the confidence interval.
Using the formula, the confidence interval for the mean difference is:
[tex]$$
\bar{d} \pm \text{ME} = 193 \pm 58.99
$$[/tex]
Thus, the lower and upper bounds are computed as follows:
- Lower bound:
[tex]$$
193 - 58.99 \approx 134.01
$$[/tex]
- Upper bound:
[tex]$$
193 + 58.99 \approx 251.99
$$[/tex]
Step 5. Write the final answer.
The [tex]$98\%$[/tex] confidence interval for the mean difference in score is approximately:
[tex]$$
\boxed{193 \pm 2.821\left(\frac{62.73}{\sqrt{9}}\right) \quad \text{or} \quad (134.01, \, 251.99)}
$$[/tex]
This provides the interval within which we are [tex]$98\%$[/tex] confident the true mean difference lies.
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