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Sagot :
Answer:
The margin of error for this estimate is of 14.79 yards per game.
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
T interval
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 20 - 1 = 19
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.093
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}}[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
You randomly select 20 games and see that the average yards per game is 273.7 with a standard deviation of 31.64 yards.
This means that [tex]n = 20, s = 31.61[/tex]
What is the margin of error for this estimate?
[tex]M = T\frac{s}{\sqrt{n}}[/tex]
[tex]M = 2.093\frac{31.61}{\sqrt{20}}[/tex]
[tex]M = 14.79[/tex]
The margin of error for this estimate is of 14.79 yards per game.
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