Find the 95 confidence interval for the mean paid attendance at the major league all-star games. a random sample of the paid attendances

Question

Find the 95 confidence interval for the mean paid attendance at the major league all-star games. a random sample of the paid attendances

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Amara 2 weeks 2022-01-12T01:42:55+00:00 1 Answer 0 views 0

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    2022-01-12T01:44:40+00:00

    Answer:

    50889.33-2.09\frac{16346.34}{\sqrt{21}}=43434.17    

    50889.33+2.09\frac{16346.34}{\sqrt{21}}=58344.49    

    So on this case the 95% confidence interval would be given by (43434.17;58344.49)    

    Step-by-step explanation:

    Assuming the following dataset:

    47,596 68,751 45,838 69,831 28,843 53,107 31,391       48,829  50,706 92,892 55,105 63,974 56,674 38,362  51,549 31,938  31,851 56,088 34,906 38,359 72,086

    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 represent the sample mean for the sample  

    \mu population mean (variable of interest)

    s represent the sample standard deviation

    n=21  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 mean and the sample deviation we can use the following formulas:  

    \bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

    s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

    The mean calculated for this case is \bar X=50889.33

    The sample deviation calculated s=16346.34

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

    df=n-1=21-1=20

    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(0.025,20)”.And we see that t_{\alpha/2}=2.09

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

    50889.33-2.09\frac{16346.34}{\sqrt{21}}=43434.17    

    50889.33+2.09\frac{16346.34}{\sqrt{21}}=58344.49    

    So on this case the 95% confidence interval would be given by (43434.17;58344.49)    

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