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?(Round to three decimal places as? needed.)

Answer 1

The 98% confidence interval would be given by (0.245;1.163)

Since the upper value for the confidence interval is higher than 1 we can't conclude that the specification required is meeting at 2% of significance.

Explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

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

2) Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)`

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}`

In Excel the formula is:

=AVERAGE(0.51 ,0.69,0.10,0.93,1.31,0.50,0.89)

On this case the average is
`\bar X= 0.704`

In excel the formula is:

=STDEV.S(0.51 ,0.69,0.10,0.93,1.31,0.50,0.89)

The sample standard deviation obtained was s=0.3867

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =7 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 98% of confidence, our significance level would be given by
`\alpha=1-0.98=0.02` and
`\alpha/2 =0.01`. The degrees of freedom are given by:

`df=n-1=7-1=6`

We can find the critical values in excel using the following formulas:

"=T.INV(0.01,6)" for
`t_(\alpha/2)=-3.14`

"=T.INV(1-0.01,6)" for
`t_(1-\alpha/2)=3.14`

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))`

If we replace we got:

`0.704 - 3.14(0.3867)/(√(7))=0.245`

`0.704 + 3.14(0.3867)/(√(7))=1.163`

So the 98% confidence interval would be given by (0.245;1.163)

Does it appear that there is too much mercury in tuna? sushi?

Since the upper value for the confidence interval is higher than 1 we can't conclude that the specification required is meeting at 2% of significance.