Mathematics, 11.02.2020 03:25 IsoSaysHi

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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Answer from: liz3022

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

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

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

Step-by-step explanation:

Assuming the following dataset:

47,59668,75145,83869,83128,84353,10731,391 48,829 50,70692,89255,10563,97456,67438,362 51,54931,938 31,85156,08834,90638,35972,086

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)

Answer from: Quest

henry ordered 5 large pizzas

step-by-step explanation:

sorry no step by step

Answer from: Quest

40%=2/550=1/230=3/1056=14/25

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