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Sagot :
Answer:
The 98% confidence interval for the mean of the population is (59, 68.2).
Step-by-step explanation:
Before building the confidence interval, we need to find the sample mean and the sample standard deviation.
Sample mean:
[tex]\overline{x} = \frac{55+64+58+61+69+64+59+69+72+65}{10} = 63.6[/tex]
Sample standard deviation:
[tex]s = \sqrt{\frac{(55-63.6)^2+(64-63.6)^2+(58-63.6)^2+(61-63.6)^2+(69-63.6)^2+(64-63.6)^2+(59-63.6)^2+(69-63.6)^2+(72-63.6)^2+(65-63.6)^2}{10}} = 5.142[/tex]
Confidence interval:
We have the standard deviation for the sample, and thus, we use the t-distribution.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 10 - 1 = 9
98% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.98}{2} = 0.99[/tex]. So we have T = 2.821
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 2.821\frac{5.142}{\sqrt{10}} = 4.6[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 63.6 - 4.6 = 59
The upper end of the interval is the sample mean added to M. So it is 63.6 + 4.6 = 68.2.
The 98% confidence interval for the mean of the population is (59, 68.2).
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