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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Melanie 2 days 2021-10-13T07:23:33+00:00 1 Answer 0

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    2021-10-13T07:25:06+00:00

    Answer:

    2.89-2.06\frac{0.65}{\sqrt{25}}=2.62    

    2.89+2.06\frac{0.65}{\sqrt{25}}=3.16    

    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”.

    \bar X= 2.89 represent the sample mean for the sample  

    \mu 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:

    \bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

    In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

    df=n-1=25-1=24

    Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, 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 t_{\alpha/2}=2.06

    Now we have everything in order to replace into formula (1):

    2.89-2.06\frac{0.65}{\sqrt{25}}=2.62    

    2.89+2.06\frac{0.65}{\sqrt{25}}=3.16    

    So on this case the 95% confidence interval would be given by (2.62;3.16)    

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