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

0.56  0.72  0.10  0.99  1.32  0.52  0.93
What is the confidence interval estimate of the population mean mu​?

Answer 1

`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}}`

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}}`

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* \frac {SD}{\sqrt {n}}`

So, applying values , we get:

`CI=0.7343\pm 2.043* \frac {0.3928}{\sqrt {7}}`

`CI=(0.431,1.0376)`