Use the given information to construct a 99% confidence interval estimate of the mean of the population. n=39, standard deviation= 3.8

Question

Use the given information to construct a 99% confidence interval estimate of the mean of the population. n=39, standard deviation= 3.86, sample mean = 41.7

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Iris 3 weeks 2021-11-15T12:21:13+00:00 1 Answer 0 views 0

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    2021-11-15T12:22:46+00:00

    Answer:

    41.7-2.71\frac{3.86}{\sqrt{39}}=40.02    

    41.7+2.71\frac{3.86}{\sqrt{39}}=43.38    

    So on this case the 99% confidence interval would be given by (40.02;43.38)    

    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=41.7 represent the sample mean for the sample  

    \mu population mean (variable of interest)

    s=3.86 represent the sample standard deviation

    n=39 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=39-1=38

    Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: “=-T.INV(0.005,38)”.And we see that t_{\alpha/2}=2.71

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

    41.7-2.71\frac{3.86}{\sqrt{39}}=40.02    

    41.7+2.71\frac{3.86}{\sqrt{39}}=43.38    

    So on this case the 99% confidence interval would be given by (40.02;43.38)    

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