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?

0.57 0.68 0.10 0.93 1.29 0.55 0.86
What is the confidence interval estimate of the population mean mu​?
_ppm < mu < _ ppm

Answer 1

hi there friend

Step-by-step explanation:

CI=(0.431,1.0376)CI=(0.431,1.0376)

Step-by-step explanation:

Given that:

The sample size , n = 7

The mean of the observation:

Mean = Sum of observation / Total number of observation

= (0.56+ 0.72+ 0.10 + 0.99 + 1.32 + 0.52 + 0.93) / 7 = 0.7343

The standard deviation:

S.D. = \sqrt {\frac {\sum_{i=1}^{i=7}(x_i-\bar{x})^2}{n-1}}S.D.=

n−1

∑

i=1

i=7

(x

i

−

x

ˉ

)

2

Calculating SD as:

S.D. = \sqrt {\frac {(0.56-0.7343)^2+(0.72-0.7343)^2+.....+(0.52-0.7343)^2+(0.93-0.7343)^2}{7-1}}S.D.=

7−1

(0.56−0.7343)

2

+(0.72−0.7343)

2

+.....+(0.52−0.7343)

2

+(0.93−0.7343)

2

SD = 0.3928

Degree of freedom = n-1 = 6

The critical value for t at 2% level of significance and 6 degree of freedom is 2.043.

So,

90 \% \ confidence\ interval=Mean\pm Z\times \frac {SD}{\sqrt {n}}90% confidence interval=Mean±Z×

n

SD

So, applying values , we get:

CI=0.7343\pm 2.043\times \frac {0.3928}{\sqrt {7}}CI=0.7343±2.043×

7

0.3928

CI=(0.431,1.0376)CI=(0.431,1.0376)