the gpa of 15 randomy slected students are shown below on this table.
2.87, 3.35, 3.61, 4.00, 3.73, 2.95, 3.02, 3.46, 3.21, 3.50, 3.89, 3.20, 3.62, 2.77, 3.92
construct a 95% confedence interval using a t-distribution for the mean gpa?
would agiven mean gpa of 3.72 be likly given the calculated confednce interval?
Solution
You have a sample of n=15, with a mean of 3.41 and a standard deviation of 0.395. The formula
for a confidence interval is:
Mean ± t/2 * st. dev. / n
where t/2 is a critical value of the t-distribution with n-1=15-1=14 degrees of freedom. In this
case, that number is 2.14. So the confidence interval is:
3.41 ± 2.14 * 0.395 / 15
3.41 ± 0.22
So from 3.19 to 3.63.
Given that 3.72 is outside this confidence interval, it is unlikely to be the mean..
the gpa of 15 randomy slected students are shown below on this table.pdf
1. the gpa of 15 randomy slected students are shown below on this table.
2.87, 3.35, 3.61, 4.00, 3.73, 2.95, 3.02, 3.46, 3.21, 3.50, 3.89, 3.20, 3.62, 2.77, 3.92
construct a 95% confedence interval using a t-distribution for the mean gpa?
would agiven mean gpa of 3.72 be likly given the calculated confednce interval?
Solution
You have a sample of n=15, with a mean of 3.41 and a standard deviation of 0.395. The formula
for a confidence interval is:
Mean ± t/2 * st. dev. / n
where t/2 is a critical value of the t-distribution with n-1=15-1=14 degrees of freedom. In this
case, that number is 2.14. So the confidence interval is:
3.41 ± 2.14 * 0.395 / 15
3.41 ± 0.22
So from 3.19 to 3.63.
Given that 3.72 is outside this confidence interval, it is unlikely to be the mean.
