Question

14.

A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city. Construct a

99​% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna​ sushi?

0.51, 0.70, 0.10, 0.91, 1.24, 0.55, 0.84

What is the confidence interval estimate of the population mean μ​?

15.

Listed below are the amounts of net worth​ (in millions of​ dollars) of the ten wealthiest celebrities in a country. Construct a 90​% confidence interval. What does the result tell us about the population of all​ celebrities? Do the data appear to be from a normally distributed population as​ required?

267, 207, 184, 161, 156, 154, 146, 146, 146, 141

What is the confidence interval estimate of the population mean μ​?

Answer 1

#14.
sample mean, xbar = 0.6929
sample standard deviation, s = 0.3591
sample size, n = 7
degrees of freedom, df = n - 1 = 6

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 3.707

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (0.6929 - 3.707 * 0.3591/sqrt(7) , 0.6929 + 3.707 * 0.3591/sqrt(7))
CI = (0.19 , 1.20)

#15.
sample mean, xbar = 170.8
sample standard deviation, s = 39.5384
sample size, n = 10
degrees of freedom, df = n - 1 = 9

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.833

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (170.8 - 1.833 * 39.5384/sqrt(10) , 170.8 + 1.833 * 39.5384/sqrt(10))
CI = (147.88 , 193.72)