Question

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? 0.58  0.68  0.10  0.93  1.37  0.50  0.96

Answer 1

Solution :

Given that,

From the given data,

Point estimate = sample mean = = Xi / n = 5.12 / 7 = 0.7314

sample standard deviation = s = ( Xi - ) / n - 1 = 0.8072 / 6 = 0.40

sample size = n = 7

Degrees of freedom = df = n - 1 = 6

At 99% confidence level the t is ,

t _/2,df = t_0.005,6 = 3.707

Margin of error = E = t_/2,df * (s /n)

= 3.707 * ( 0.40/ 7)

= 0.560

The 99% confidence interval estimate of the population mean is,

- E < < + E

0.731 - 0.560 < < 0.731 + 0.560

0.141 < < 1.291

( 0.141 , 1.291)