An electric scale gives a reading equal to the true weight plus a random error thatis normally distributed with mean 0 and standard deviation s = .1 mg. Supposethat the results of five successive weighings of the same object are as follows: 3.142,3.163, 3.155, 3.150, 3.141.(a) Determine a 95 percent confidence interval estimate of the true weight.(b) Determine a 99 percent confidence interval estimate of the true weight.

Answer:a) (3.1388,3.1616) b) (3.1313,3.1691)  Step-by-step explanation:We are given the following data set: 3.142,  3.163, 3.155, 3.150, 3.141Formula:[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{15.751}{5} = 3.1502[/tex]Sum of squares of differences = [tex]=0.00006724 + 0.00016384 + 0.00002304 + 4.000000000e^{-8} + 0.00008464 \\=0.0003388[/tex][tex]S.D = \sqrt{\displaystyle\frac{0.0003388}{4}} = 0.0092[/tex]a) 95% 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 4 and}~\alpha_{0.05} = \pm 2.776[/tex][tex]3.1502 \pm 2.776(\displaystyle\frac{0.0092}{\sqrt{5}} ) = 3.1502 \pm 0.0114 = (3.1388,3.1616)[/tex]b) 99% 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 4 and}~\alpha_{0.01} = \pm 4.6022[/tex][tex]3.1502 \pm 4.6022(\displaystyle\frac{0.0092}{\sqrt{5}} ) = 3.1502 \pm 0.0189 = (3.1313,3.1691)[/tex]