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Answer :
Answer:
The 90% confidence interval of the mean high temperature of towns is (97.40, 99.17)°F.
Explanation:
To find the 90% confidence interval (CI) of the mean high temperature of towns, we can use the t-distribution since the sample size is small (n = 8) and the population standard deviation is unknown.
First, let's calculate the sample mean ([tex]\( \bar{x} \)[/tex]) and the sample standard deviation ([tex]\( s \)[/tex]) from the given data:
[tex]\[ \text{Sample Mean (\( \bar{x} \))} = \frac{99.6 + 97.3 + 97.7 + 98.4 + 99.2 + 99.7 + 96.6 + 97.8}{8} \][/tex]
[tex]\[ \text{Sample Mean (\( \bar{x} \))} = \frac{786.3}{8} \][/tex]
[tex]\[ \text{Sample Mean (\( \bar{x} \))} = 98.2875 \][/tex]
Next, we calculate the sample standard deviation:
[tex]\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \][/tex]
Substituting the given values:
[tex]\[ s = \sqrt{\frac{(99.6 - 98.2875)^2 + (97.3 - 98.2875)^2 + \ldots + (97.8 - 98.2875)^2}{8 - 1}} \][/tex]
[tex]\[ s = \sqrt{\frac{4.27625 + 0.459375 + \ldots + 0.409375}{7}} \][/tex]
[tex]\[ s = \sqrt{\frac{12.17625}{7}} \][/tex]
[tex]\[ s \approx \sqrt{1.73946429} \][/tex]
[tex]\[ s \approx 1.31822 \][/tex]
Now, we need to find the critical value (t*) from the t-distribution for a 90% confidence level with [tex]\( n - 1 = 8 - 1 = 7 \)[/tex] degrees of freedom.
Using a t-table or statistical software, we find that [tex]\( t^* \)[/tex] for a 90% confidence level and 7 degrees of freedom is approximately 1.8946.
Finally, we can calculate the margin of error (ME):
[tex]\[ \text{ME} = t^* \times \frac{s}{\sqrt{n}} \][/tex]
Substituting the values:
[tex]\[ \text{ME} = 1.8946 \times \frac{1.31822}{\sqrt{8}} \][/tex]
[tex]\[ \text{ME} \approx 1.8946 \times \frac{1.31822}{2.8284} \][/tex]
[tex]\[ \text{ME} \approx 1.8946 \times 0.4660 \][/tex]
[tex]\[ \text{ME} \approx 0.8829 \][/tex]
Now, we can construct the confidence interval:
[tex]\[ \text{CI} = \left( \bar{x} - \text{ME}, \bar{x} + \text{ME} \right) \][/tex]
[tex]\[ \text{CI} = \left( 98.2875 - 0.8829, 98.2875 + 0.8829 \right) \][/tex]
[tex]\[ \text{CI} = \left( 97.4046, 99.1704 \right) \][/tex]
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Rewritten by : Jeany
The 95% confidence interval that estimates the population mean is approximately (97.53, 99.05)
What is the confidence interval?
The sample mean of the given data is:
Mean(x') = (99.6 + 97.3 + 97.7 + 98.4 + 99.2 + 99.7 + 96.6 + 97.8)/8
x' = 98.2875
Using standard deviation calculator, we have:
Sample standard deviation: s = 1.132
Because n is not greater than 30, and because we don't know the population standard deviation (sigma), this means we must use a T distribution (instead of a Z distribution)
At 95% confidence and with degrees of freedom = n - 1 = 8 - 1 = 7, the critical t value is roughly t = 1.8946
The lower bound (L) of the confidence interval is
L = x' - t*s/√(n)
L = 98.2875 - (1.8946 * 1.132)/√8
L = 97.53
The upper bound (U) of the confidence interval is:
U = x' + t*s/√(n)
U = 98.2875 + (1.8946 * 1.132)/√8
U = 99.05
The 95% confidence interval that estimates the population mean is approximately (97.53, 99.05)
