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 1.164 ppm and the sample standard deviation is 0.304 ppm. Use technology to construct a 90​% confidence interval estimate of the mean amount of mercury in the population.
1.13
1.38
1.11
0.74
1.49
1.48
0.82

What is the confidence interval estimate of the mean amount of mercury in the​ population?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 1.164

sample standard deviation = s = 0.304

sample size = n = 7

Degrees of freedom = df = n - 1 = 6

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,6 = 1.943

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

= 1.943 * (0.304 / 7)

= 0.223

The 90% confidence interval estimate of the population mean is,

- E < < + E

1.164 - 0.223 < < 1.164 + 0.223

0.914 < < 1.138

The confidence interval estimate of the mean amount of mercury in the​ population is (0.914 , 1.138)