Question

Listed below are the amounts of mercury​ (in parts per​ million, or​ ppm) found in tuna sushi sampled at different stores. The sample mean is 0.980 ppm and the sample standard deviation is 0.412 ppm. Use technology to construct a 90​% confidence interval estimate of the mean amount of mercury in the population. 0.52 0.53 1.49 1.32 0.61 1.17 1.22 What is the confidence interval estimate of the mean amount of mercury in the​ population?

Answer 1

t critical value at 0.10 level with 6 df = 1.943

90% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

0.980 - 1.943 * 0.412 / sqrt(7) < < 0.980 + 1.943 * 0.412 / sqrt(7)

0.6774 < < 1.2826

90% CI is ( 0.6774 , 1.2826 )