Suppose the 90% confidence interval for the mean SAT scores of applicants at a b

Question

Suppose the 90% confidence interval for the mean SAT scores of applicants at a business college is given by [1,693, 1,822]. This confidence interval uses the sample mean and the sample standard deviation based on 25 observations. Use Table 2.

What are the sample mean and the sample standard deviation used when computing the interval? (Round "t" value to 3 decimal places and “Sample mean” and "Sample standard deviation" to 2 decimal places.)

Suppose the 90% confidence interval for the mean SAT scores of applicants at a business college is given by [1,693, 1,822]. This confidence interval uses the sample mean and the sample standard deviation based on 25 observations. Use Table 2.

What are the sample mean and the sample standard deviation used when computing the interval? (Round "t" value to 3 decimal places and “Sample mean” and "Sample standard deviation" to 2 decimal places.)

Explanation / Answer

Sample mean =(1693+1822)/2 =1757.5

here margin of error =(1822-1693)/2 =64.5

for (25-1=24 ) degree of freedom and 90% CI ; t =1.711

hence sample std deviation =margin of error*(n)1/2 /t =64.5*(25)1/2/1.711 =188.49

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