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Answer:

[tex]50889.33-2.09\frac{16346.34}{\sqrt{21}}=43434.17[/tex]    

[tex]50889.33+2.09\frac{16346.34}{\sqrt{21}}=58344.49[/tex]    

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".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] 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:

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

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

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

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

The mean calculated for this case is [tex]\bar X=50889.33[/tex]

The sample deviation calculated [tex]s=16346.34[/tex]

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

[tex]df=n-1=21-1=20[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]t_{\alpha/2}=2.09[/tex]

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

[tex]50889.33-2.09\frac{16346.34}{\sqrt{21}}=43434.17[/tex]    

[tex]50889.33+2.09\frac{16346.34}{\sqrt{21}}=58344.49[/tex]    

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

The confidence interval for the mean paid attendance of the game will be (41833, 52109).

What is a confidence interval?

A confidence interval simply means the probability that a parameter will fall between the values that are around the mean.

In the complete question, the confidence interval can be depicted at the last column. Here, the number was 30 and the mean was 46971. The confidence interval is (41833, 52109)

Learn more about confidence interval on:

https://brainly.com/question/25779324