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Answer :
We are given a sample of [tex]\( n = 8 \)[/tex] high temperature measurements:
[tex]\[
98.6,\; 96.8,\; 97.7,\; 97.4,\; 96.7,\; 99,\; 98.2,\; 99.6
\][/tex]
and we wish to construct a [tex]\( 99\% \)[/tex] confidence interval for the population mean assuming that the high temperatures are normally distributed.
Step 1. Compute the Sample Mean
The sample mean ([tex]\( \bar{x} \)[/tex]) is given by
[tex]\[
\bar{x} = \frac{98.6 + 96.8 + 97.7 + 97.4 + 96.7 + 99 + 98.2 + 99.6}{8} = 98.0.
\][/tex]
Step 2. Compute the Sample Standard Deviation
The sample standard deviation ([tex]\( s \)[/tex]) is calculated (using [tex]\( n-1 \)[/tex] in the denominator) and we obtain
[tex]\[
s \approx 1.04 \; (\text{to 3 decimal places}).
\][/tex]
Step 3. Determine the Standard Error of the Mean
The standard error (SE) of the mean is
[tex]\[
\text{SE} = \frac{s}{\sqrt{n}} = \frac{1.04}{\sqrt{8}} \approx \frac{1.04}{2.83} \approx 0.37.
\][/tex]
Step 4. Find the Appropriate t-Critical Value
Because the sample size is small and the population standard deviation is unknown, we use the [tex]\( t \)[/tex]-distribution with degrees of freedom [tex]\( \text{df} = n - 1 = 7 \)[/tex]. For a [tex]\( 99\% \)[/tex] confidence level, the two-tailed critical probability is [tex]\( 1 - \alpha/2 = 0.995 \)[/tex]. The corresponding critical value is
[tex]\[
t_{0.005,7} \approx 3.50.
\][/tex]
Step 5. Calculate the Margin of Error
The margin of error (ME) is given by
[tex]\[
\text{ME} = t \times \text{SE} \approx 3.50 \times 0.37 \approx 1.28.
\][/tex]
Step 6. Construct the Confidence Interval
The [tex]\( 99\% \)[/tex] confidence interval for the population mean is
[tex]\[
\left( \bar{x} - \text{ME},\; \bar{x} + \text{ME} \right) = (98.0 - 1.28,\; 98.0 + 1.28) = (96.72,\; 99.28).
\][/tex]
Thus, the [tex]\( 99\% \)[/tex] confidence interval for the mean high temperature of the small towns is:
[tex]\[
\boxed{(96.72,\; 99.28)}.
\][/tex]
[tex]\[
98.6,\; 96.8,\; 97.7,\; 97.4,\; 96.7,\; 99,\; 98.2,\; 99.6
\][/tex]
and we wish to construct a [tex]\( 99\% \)[/tex] confidence interval for the population mean assuming that the high temperatures are normally distributed.
Step 1. Compute the Sample Mean
The sample mean ([tex]\( \bar{x} \)[/tex]) is given by
[tex]\[
\bar{x} = \frac{98.6 + 96.8 + 97.7 + 97.4 + 96.7 + 99 + 98.2 + 99.6}{8} = 98.0.
\][/tex]
Step 2. Compute the Sample Standard Deviation
The sample standard deviation ([tex]\( s \)[/tex]) is calculated (using [tex]\( n-1 \)[/tex] in the denominator) and we obtain
[tex]\[
s \approx 1.04 \; (\text{to 3 decimal places}).
\][/tex]
Step 3. Determine the Standard Error of the Mean
The standard error (SE) of the mean is
[tex]\[
\text{SE} = \frac{s}{\sqrt{n}} = \frac{1.04}{\sqrt{8}} \approx \frac{1.04}{2.83} \approx 0.37.
\][/tex]
Step 4. Find the Appropriate t-Critical Value
Because the sample size is small and the population standard deviation is unknown, we use the [tex]\( t \)[/tex]-distribution with degrees of freedom [tex]\( \text{df} = n - 1 = 7 \)[/tex]. For a [tex]\( 99\% \)[/tex] confidence level, the two-tailed critical probability is [tex]\( 1 - \alpha/2 = 0.995 \)[/tex]. The corresponding critical value is
[tex]\[
t_{0.005,7} \approx 3.50.
\][/tex]
Step 5. Calculate the Margin of Error
The margin of error (ME) is given by
[tex]\[
\text{ME} = t \times \text{SE} \approx 3.50 \times 0.37 \approx 1.28.
\][/tex]
Step 6. Construct the Confidence Interval
The [tex]\( 99\% \)[/tex] confidence interval for the population mean is
[tex]\[
\left( \bar{x} - \text{ME},\; \bar{x} + \text{ME} \right) = (98.0 - 1.28,\; 98.0 + 1.28) = (96.72,\; 99.28).
\][/tex]
Thus, the [tex]\( 99\% \)[/tex] confidence interval for the mean high temperature of the small towns is:
[tex]\[
\boxed{(96.72,\; 99.28)}.
\][/tex]
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