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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..

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- 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.

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