The 90% confidence interval that estimates the proportion of 8th graders that are involved in an after-school activity is (0.80853,0.93147).
The confidence interval for a population proportion for a given sample is given by the formula:
[tex](p-Z\sqrt{\frac{p(1-p}{n} } , p+Z\sqrt{\frac{p(1-p}{n} })[/tex],
where p is the population proportion, Z is the Z-score value for the confidence interval, and n is the sample size.
In the question, we are given a random sample of 81 8th graders at a large suburban middle school. This implies that the sample size, n = 81. Also, we are told that they indicated 87% of them were involved with some type of after-school activity. This implies that the population proportion, p = 87% = 0.87.
We are asked to find the 90% confidence interval that estimates the proportion of them that are involved in an after-school activity.
Z-score (Z) corresponding to a 90% confidence interval is 1.645.
Thus, we can find the confidence interval as follows:
(0.87 - 1.645√{(0.87(1 - 0.87))/81},0.87 + 1.645√{(0.87(1 - 0.87))/81})
= (0.87 - 0.061469,0.87 + 0.061469)
= (0.80853,0.93147).
Thus, the 90% confidence interval that estimates the proportion of 8th graders that are involved in an after-school activity is (0.80853,0.93147).
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