## 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

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## Answers ( )

Answer: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 intervalis “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 erroris 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 represent the sample standard deviation

n=21 represent the sample size

Solution to the problemThe confidence interval for the mean is given by the following formula:

(1)

In order to calculate the mean and the sample deviation we can use the following formulas:

(2)

(3)

The mean calculated for this case is

The sample deviation calculated

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(0.025,20)”.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 (43434.17;58344.49)