Four consecutive positive prime numbers have a sum that is divisible by three. What is the smallest possible value of this sum?
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Using the t-distribution, we have that the 95% confidence interval for the true mean number of pushups that can be done is (9, 21).
For this problem, we have the <u>standard deviation for the sample</u>, thus, the t-distribution is used.
- The sample mean is of 15, thus
.
- The sample standard deviation is of 9, thus
.
- The sample size is of 10, thus
.
First, we find the number of degrees of freedom, which is the one less than the sample size, thus df = 9.
Then, looking at the t-table or using a calculator, we find the critical value for a 95% confidence interval, with 9 df, thus t = 2.2622.
The margin of error is of:
Then:
The confidence interval is:
Then
The confidence interval is (9, 21).
A similar problem is given at brainly.com/question/25157574