Respuesta :
Answer:
98% Confidence interval: (0.25,1.16)
Step-by-step 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:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{4.93}{7} = 0.704[/tex]
Sum of squares of differences = 0.897
[tex]S.D = \sqrt{\frac{0.897}{6}} = 0.386[/tex]
98% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 6 and}~\alpha_{0.02} = \pm 3.142[/tex]
[tex]0.704 \pm 3.142(\frac{0.386}{\sqrt{7}} ) = 0.704 \pm 0.458 = (0.25,1.16)[/tex]
No, it does not appear that there is too much mercury in the tuna.
