In a random sample of students who took the SAT Reasoning college entrance exam twice, it was found that 427 of the respondents had paid for coaching courses and that the remaining 2733 had not.

What is the upper bound on the 95% confidence interval for the proportion of coaching among the students who retake the SAT?
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1 Answer

This is a binary situation. We use np to find the mean and np(1-p) to find the variance, where n is the dataset size and p the probability of a randomly selected student who had paid for coaching. We can then apply a normal distribution to determine the confidence interval (CI).

n=427+2733=3160, p=427/3160=0.1351 approx. Mean, μ=427, σ2=np(1-p)=369.3, standard deviation, σ=19.22 approx. The critical value for 95% CI is 5% (0.025) two-tailed significance. The Z-score for 95% is 1.96. To find the upper bound X, we use (X-μ)/σ=Z, X=μ+σZ=427+19.22×1.96=464.67 approx.

Upper bound is 464 (the largest integer lower than 464.67).

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