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Answer :
We are given the following sample statistics:
- Sample size: [tex]$$n = 50$$[/tex]
- Sample mean: [tex]$$\bar{x} = 38.8$$[/tex]
- Sample standard deviation: [tex]$$s = 9.9$$[/tex]
Since the population standard deviation is unknown and the sample size is moderate, we use the [tex]$$t$$[/tex]-distribution.
1. Determine the degrees of freedom (df):
[tex]$$\text{df} = n - 1 = 50 - 1 = 49.$$[/tex]
2. For a 95% confidence interval, the level of significance is [tex]$$\alpha = 0.05$$[/tex]. In a two-tailed test, the critical value is found with probability [tex]$$1 - \frac{\alpha}{2} = 0.975.$$[/tex]
The critical value is:
[tex]$$t_{0.975, \, 49} \approx 2.01.$$[/tex]
3. Calculate the standard error of the mean (SE):
[tex]$$\text{SE} = \frac{s}{\sqrt{n}} = \frac{9.9}{\sqrt{50}} \approx 1.40.$$[/tex]
4. Compute the margin of error (ME) by multiplying the critical value by the standard error:
[tex]$$\text{ME} = t_{0.975, \, 49} \times \text{SE} \approx 2.01 \times 1.40 \approx 2.81.$$[/tex]
5. Construct the confidence interval by subtracting and adding the margin of error from the sample mean:
[tex]\[
\begin{aligned}
\text{Lower bound} &= \bar{x} - \text{ME} \approx 38.8 - 2.81 \approx 35.99,\\[1mm]
\text{Upper bound} &= \bar{x} + \text{ME} \approx 38.8 + 2.81 \approx 41.61.
\end{aligned}
\][/tex]
Thus, the 95% confidence interval for the population mean [tex]$$\mu$$[/tex] is
[tex]$$[35.99,\ 41.61].$$[/tex]
- Sample size: [tex]$$n = 50$$[/tex]
- Sample mean: [tex]$$\bar{x} = 38.8$$[/tex]
- Sample standard deviation: [tex]$$s = 9.9$$[/tex]
Since the population standard deviation is unknown and the sample size is moderate, we use the [tex]$$t$$[/tex]-distribution.
1. Determine the degrees of freedom (df):
[tex]$$\text{df} = n - 1 = 50 - 1 = 49.$$[/tex]
2. For a 95% confidence interval, the level of significance is [tex]$$\alpha = 0.05$$[/tex]. In a two-tailed test, the critical value is found with probability [tex]$$1 - \frac{\alpha}{2} = 0.975.$$[/tex]
The critical value is:
[tex]$$t_{0.975, \, 49} \approx 2.01.$$[/tex]
3. Calculate the standard error of the mean (SE):
[tex]$$\text{SE} = \frac{s}{\sqrt{n}} = \frac{9.9}{\sqrt{50}} \approx 1.40.$$[/tex]
4. Compute the margin of error (ME) by multiplying the critical value by the standard error:
[tex]$$\text{ME} = t_{0.975, \, 49} \times \text{SE} \approx 2.01 \times 1.40 \approx 2.81.$$[/tex]
5. Construct the confidence interval by subtracting and adding the margin of error from the sample mean:
[tex]\[
\begin{aligned}
\text{Lower bound} &= \bar{x} - \text{ME} \approx 38.8 - 2.81 \approx 35.99,\\[1mm]
\text{Upper bound} &= \bar{x} + \text{ME} \approx 38.8 + 2.81 \approx 41.61.
\end{aligned}
\][/tex]
Thus, the 95% confidence interval for the population mean [tex]$$\mu$$[/tex] is
[tex]$$[35.99,\ 41.61].$$[/tex]
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