A study found the body temperatures of 121 healthy adults Th

A study found the body temperatures of 121 healthy adults. The sample mean was 98.2 degrees. Suppose the population standard deviation is sigma = 0.62 degrees. Find a 95% confidence interval for p. (Don\'t forget to check the conditions.) A study found the body temperatures of 121 healthy adults. The sample mean was 98.2 degrees and the sample standard deviation was 0.62 degrees. Find a 95% confidence interval for p. (Don?t forget to check the conditions.) We found the Flesch ease of reading scores for 12 different pages randomly selected from J.K. Rowling?s Harry Potter and the Sorcerer?s Stone. The sample mean was 80.75, and the sample standard deviation was 4.68. Assuming the population is approximately normal, find a 90% confidence interval for p. (Don?t forget to check the conditions.) How large a sample would we need to estimate the mean body temperature of healthy adults to within 0.05 degrees with 95% confidence? (Use the sample standard deviation from #2 above as an estimate for a.)

Solution

We need to assume that the population follows Normal distribution.

Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

xbar - Z*s/vn = 98.2 -1.96*0.62/sqrt(121) = 98.08953

So the upper bound is

xbar + Z*s/vn = 98.2 +1.96*0.62/sqrt(121) =98.31047

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The degree of freedom =n-1=12-1=11

Given a=1-0.9=0.1, t(0.05, df=11) = 1.796 (from student t table)

So the lower bound is

xbar - t*s/vn = 80.75- 1.796*4.68/sqrt(12)=78.3236

So the upper bound is

xbar + t*s/vn = 80.75+ 1.796*4.68/sqrt(12)=83.1764

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Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So n=(Z*s/E)^2

=(1.96*4.68/0.05)^2

=33656.1

Take n=33657