## I took a sample of the grade point averages for students in my class. For the 25 students, the standard deviation of grade points was 0.65 a

Question

I took a sample of the grade point averages for students in my class. For the 25 students, the standard deviation of grade points was 0.65 and the mean was 2.89. A 95% confidence interval for the average grade point average for all students in my class is: ( 2.62 , 3.16 ) ( 2.53 , 3.25 ) ( 2.64 , 3.14 )

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2 days 2021-10-13T07:23:33+00:00 1 Answer 0  So on this case the 95% confidence interval would be given by (2.62;3.16)

Step-by-step explanation:

Previous concepts

A confidence interval is “a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval”.

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”. represent the sample mean for the sample population mean (variable of interest)

s=0.65 represent the sample standard deviation

n=25 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula: (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by: Since the Confidence is 0.95 or 95%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: “=-T.INV(0.025,24)”.And we see that Now we have everything in order to replace into formula (1):  So on this case the 95% confidence interval would be given by (2.62;3.16)