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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​?

Respuesta :

Answer:

[tex]CI=(0.431,1.0376)[/tex]

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:

[tex]S.D. = \sqrt {\frac {\sum_{i=1}^{i=7}(x_i-\bar{x})^2}{n-1}}[/tex]

Calculating SD as:

[tex]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}}[/tex]

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,

[tex]90 \% \ confidence\ interval=Mean\pm Z\times \frac {SD}{\sqrt {n}}[/tex]

So, applying values , we get:

[tex]CI=0.7343\pm 2.043\times \frac {0.3928}{\sqrt {7}}[/tex]

[tex]CI=(0.431,1.0376)[/tex]

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