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In analyzing data from over 700 games in the National Football League, economist David Romer identified 1068 fourth-down situations in which, based on his analysis, the right call would have been to go for it and not to punt. Nonetheless, in 959 of those situations, the teams punted. Find a 95% confidence interval for the proportion of times NFL teams punt on a fourth down when, statistically speaking, they shouldn’t be punting.1 Assume the sample is reasonably representative of all such fourth down situations in the NFL.

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temdan2001

Answer: between 0.88 and 0.916

Explanation: parameters given are:

n = 1068

CI = 95 % = 1.96 

p = 959

Let us Divide p by n.

959 \ 1068 = 0.898

Then,

0.898 -\+ 1.96 (√0.898 (1 - 0.898) \ 1068)

Find standard error

√0.898 (1 - 0.898) \ 1068 = 0.009

Also let us Find Margin of error.

(1.96)(0.009) = 0.018

Then calculate confidence interval.

0.898 -\+ 0.018 = 0.880 to 0.916

The 95% confidence, that the proportion of times NFL teams punt on a fourth down when they shouldn't be punting is between 0.880 and 0.916

abiolataiwo2015

The 95% confidence interval for the proportion of times NFL teams punt on a fourth down is: 0.880; 0.916.

Confidence interval

n = 1068

z-score for 95% CI = 1.96

p = 959

Hence:

p=959 \ 1068

p= 0.898

Standard error=0.898 -\+ 1.96 (√0.898 (1 - 0.898) \ 1068)

Standard error=√0.898 (1 - 0.898) \ 1068

Standard error=√0.898 (0.102) \ 1068

Standard error = 0.009

Margin of error=(1.96)(0.009)

Margin of error= 0.018

Confidence interval=0.898 -\+ 0.018

Confidence interval =(0.898-0.018); (0.898+0.018)

Confidence interval = 0.880; 0.916

Inconclusion the 95% confidence interval for the proportion of times NFL teams punt on a fourth down is: 0.880; 0.916.

Learn more about confidence interval here:https://brainly.com/question/15712887

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