A study conducted by pre-med students to determine the normal body temperature f

Question

A study conducted by pre-med students to determine the normal body temperature found that a random sample of 200 healthy persons living in this city had a sample average of 98.3degree F with a sample standard deviation of s = 0.75degree F. Construct a 90% confidence interval for the average body temperature of people in this city. a.(96.3, 100.3) b. (98.21, 98.39) c. (98.12, 98.48) d.97.55, 99.05) The daily salaries, in dollars, for workers in a large company are shown. Find the 90% confidence interval of the true mean for the salaries 69, 69,52,70, 81, 60, 65, 67 a ($61.76 $71.49) b.($59.65,$73.62) c ($60.83, $72.42) d.($61.02, $72.23) Find the 95% confidence interval for the standard deviation of the lengths of pipes if a sample of 16 pipes has a standard deviation of 12 inches 14.0

Explanation / Answer

Question 7)

sample size i.e n= 200

sample average = 98.3

sample standard deviation = 0.75

standard deviation of avearge body temperature = sample standard deviation/n

= 0.75/200 = 0.053

90% confidence interval is (mean-1.645*sd, mean+1.645*sd)

90% confidence interval of avearge body temperature is (98.3-1.645*0.053, 98.3+1.645*0.053) i.e (98.21,98.39)

so answer is (B)

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