Please help asap 20 pts
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Answer:
The 99% confidence interval for the scores of men is (64.406,75.594).
The 99% confidence interval for the scores of women is (73.609,90.391).
Step-by-step explanation:
We have the standard deviations for the sample, which means that the t-distribution will be used to solve this question.
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 = 25 - 1 = 24
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.797.
For men:
Standard deviation of 10.
The margin of error is:
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 70 - 5.594 = 64.406
The upper end of the interval is the sample mean added to M. So it is 70 + 5.594 = 75.594
The 99% confidence interval for the scores of men is (64.406,75.594).
For women:
Standard deviation of 15.
The margin of error is:
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 82 - 8.391 = 73.609
The upper end of the interval is the sample mean added to M. So it is 82 + 8.391 = 90.391
The 99% confidence interval for the scores of women is (73.609,90.391).