Question

A food service safety guideline is that the mercury in fish should be below 1 part per million. Listed below are the amounts of mercury found in tuna sampled at different stores in a major city. Construct a 98% confidence interval level estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi? 0.51 0.80 0.10 0.91 1.28 0.55 0.96

Answer 1

98% confidence interval level estimate of the mean amount of mercury in the population is [0.28 , 1.18].

Explanation:

We are given that for the 69 second-year students in the study at the university, the sample mean procrastination score was 41.00 and the sample standard deviation was 6.89.

Firstly, the pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. =
`(\bar X -\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample mean amount =
`(\sum X)/(n)` = 0.73

s = sample standard deviation =
`\sqrt{(\sum (X-\bar X)^(2) )/(n-1) }` = 0.381

n = sample size = 7

`\mu` = population mean amount of mercury

Here for constructing 98% confidence interval we have used One-sample t test statistics because we don't know about population standard deviation.

So, 98% confidence interval for the population mean,
`\mu` is ;

P(-3.143 <
`t_6` < 3.143) = 0.98 {As the critical value of t at 6 degree

of freedom are -3.143 & 3.143 with P = 1%}

P(-3.143 <
`(\bar X -\mu)/((s)/(√(n) ) )` < 3.143) = 0.98

P(
`-3.143 * {(s)/(√(n) ) }` <
`{\bar X -\mu}` <
`3.143 * {(s)/(√(n) ) }` ) = 0.98

P(
`\bar X-3.143 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+3.143 * {(s)/(√(n) ) }` ) = 0.98

98% confidence interval for
`\mu` = [
`\bar X-3.143 * {(s)/(√(n) ) }` ,
`\bar X+3.143 * {(s)/(√(n) ) }` ]

= [
`0.73-3.143 * {(0.381)/(√(7) ) }` ,
`0.73+3.143 * {(0.381)/(√(7) ) }` ]

= [0.28 , 1.18]

Therefore, 98% confidence interval level estimate of the mean amount of mercury in the population is [0.28 , 1.18].

The interpretation of the above confidence interval is that we are 98% confident that the mean amount of mercury in the population will lie between 0.28 and 1.18.