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 98% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna? sushi?0.51 ,0.69,0.10,0.93,1.31,0.50,0.89What is the confidence interval estimate of the population mean \mu?ppm< \mu< $ ppmmuless than?$nothingppm

Answer 1

98% Confidence interval: (0.25,1.16)

Explanation:

We are given the following data set:

0.51 ,0.69,0.10,0.93,1.31,0.50,0.89

Sample size, n = 7

Formula:

`\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}`

where
`x_i` are data points,
`\bar{x}` is the mean and n is the number of observations.

`Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}`

`Mean =\displaystyle(4.93)/(7) = 0.704`

Sum of squares of differences = 0.897

`S.D = \sqrt{(0.897)/(6)} = 0.386`

98% Confidence interval:

`\bar{x} \pm t_(critical)\displaystyle(s)/(√(n))`

Putting the values, we get,

`t_(critical)\text{ at degree of freedom 6 and}~\alpha_(0.02) = \pm 3.142`

`0.704 \pm 3.142((0.386)/(√(7)) ) = 0.704 \pm 0.458 = (0.25,1.16)`

No, it does not appear that there is too much mercury in the tuna.