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.


0.60 0.74 0.09 0.89 1.31 0.51 0.94


What is the confidence interval estimate of the population mean?


______ppm < u < ______ppm

(Round to three decimal places as needed)


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

Answer 1

Answer:

Confidence Interval: (0.44,1.00)

Step-by-step explanation:

We are given the following data set:

0.60, 0.74, 0.09, 0.89, 1.31, 0.51, 0.94

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.  

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

$Mean =\displaystyle\frac{:5.08}{7} = 0.725$

Sum of squares of differences = 0.8809

$S.D = \sqrt{\frac{0.8809}{6}} = 0.383$

90% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 6 and}~\alpha_{0.10} = \pm 1.943$  

$0.725 \pm 1.943(\frac{0.383}{\sqrt{7}} ) =0.725 \pm 0.2812 = (0.44,1.00)$

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