Answer:
The 95% confidence interval the average maximum power is (596.0 to 644.0)
Step-by-step explanation:
Average maximum of the sample = x = 620 HP
Standard Deviation = s = 45 HP
Sample size = n = 16
We have to calculate the 95% confidence interval. The value of Population standard deviation is unknown, and value of sample standard deviation is known. Therefore, we will use one sample t-test to build the confidence interval.
Degrees of freedom = df = n - 1 = 15
Critical t-value associated with 95% confidence interval and 15 degrees of freedom, as seen from t-table = [tex]t_{\frac{\alpha}{2}}[/tex] = 2.131
The formula to calculate the confidence interval is:
[tex](x-t_{\frac{\alpha}{2} } \times \frac{s}{\sqrt{n} }, x+t_{\frac{\alpha}{2} } \times \frac{s}{\sqrt{n} })[/tex]
We have all the required values. Substituting them in the above expression, we get:
[tex](620-2.131 \times \frac{45}{\sqrt{16} }, 620+2.131 \times \frac{45}{\sqrt{16} })\\\\ =(596.0 , 644.0)[/tex]
Thus, the 95% confidence interval the average maximum power is (596.0 to 644.0)
A 95% confidence interval for average maximum HP will be "596.0 to 644.0". To understand the calculation, check below.
According to the question,
Sample's average max, x = 620 HP
Standard deviation, s = 45 HP
Sample size, n = 16
Degree of freedom, df = n - 1
= 15
Critical t-value:
[tex]t_{\frac{\alpha}{2} }[/tex] = 2.131
We know the formula,
Confidence level, CI = (x - [tex]t_{\frac{\alpha}{2} }[/tex] × [tex]\frac{s}{\sqrt{n} }[/tex], x + [tex]t_{\frac{\alpha}{2} }[/tex] × [tex]\frac{s}{\sqrt{n} }[/tex])
By substituting the values,
= (620 - 2.131 × [tex]\frac{45}{\sqrt{16} }[/tex], 620 + 2.131 × [tex]\frac{45}{\sqrt{16} }[/tex])
= (596.0, 644.0)
Thus the approach above is correct.
Find out more information about standard deviation here:
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