Question

Alood say 90% confidence 058 is that the mo nth should be below 1 part perm o m Listed below are the amount of mercury om found in una sushi sabiedt den stores a major city contra estimate of the mean amount of mercury in the population Deest appear that there is too much mercury in una sust? 10 000 0 005 What is the confidence to the populationen? Trompel som Round to the decimal places need

Answer 1

$\begin{tabular}{rrr} \hline X & X - mean & (X-mean)\^{}2 \\ \hline 0.56 & -0.1729 & 0.0299 \\ 0.68 & -0.0529 & 0.0028 \\ 0.1 & -0.6329 & 0.4006 \\ 0.99 & 0.2571 & 0.0661 \\ 1.38 & 0.6471 & 0.4187 \\ 0.57 & -0.1629 & 0.0265 \\ 0.85 & 0.1171 & 0.0137 \\ \hline \end{tabular}$

Since we know that
$\\Mean(\bar{x}) = \frac{\sum_{i=1}^n x_i}{n} \\Where\ n\ is\ the\ number\ of\ data\ points \\Now \\\sum_{i=1}^n x_i = 5.13 \\and\ n = 7 \\This\ implies\ that \\Mean(\bar{x}) = \frac{5.13}{7} \\Mean(\bar{x}) = 0.7329$

$\\Since\ we\ know\ that \\Variance(s^2) = \frac{(\sum{x_i - \bar{x}})^2}{n-1} \\(\sum{x_i - \bar{x}})^2 = 0.9583 \\n = 7 \\Variance(s^2) = \frac{0.9583}{6} \\Variance(s^2) = 0.1597 \\Standard\;Deviation(s) = \sqrt{Variance} \\Standard\;Deviation(s) = 0.3996$

$\\Mean (\bar{x}) = 0.7329$
Sample size (n) = 7
Standard deviation (s) = 0.3996
Confidence interval(in %) = 90
$\\t_{\alpha/2, n-1} = 1.9432 \\Since\ we\ know\ that \\Confidence\; interval = \bar{x} \pm t_{\alpha/2, n-1}\frac{s}{\sqrt{n}} \\Required\ confidence\ interval = (0.7329-1.9432\frac{0.3996}{\sqrt{7}}, 0.7329+1.9432\frac{0.3996}{\sqrt{7}})$
Required confidence interval = (0.733-0.294, 0.733+0.294)
Required confidence interval = (0.439, 1.027)
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