Q:

# 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.

Accepted Solution

A:
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:$$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$$  where $$x_i$$ are data points, $$\bar{x}$$ is the mean and n is the number of observations.  $$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$$$$Mean =\displaystyle\frac{15.751}{5} = 3.1502$$Sum of squares of differences = $$=0.00006724 + 0.00016384 + 0.00002304 + 4.000000000e^{-8} + 0.00008464 \\=0.0003388$$$$S.D = \sqrt{\displaystyle\frac{0.0003388}{4}} = 0.0092$$a) 95% Confidence interval:$$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$$Putting the values, we get,$$t_{critical}\text{ at degree of freedom 4 and}~\alpha_{0.05} = \pm 2.776$$$$3.1502 \pm 2.776(\displaystyle\frac{0.0092}{\sqrt{5}} ) = 3.1502 \pm 0.0114 = (3.1388,3.1616)$$b) 99% Confidence interval:$$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$$Putting the values, we get,$$t_{critical}\text{ at degree of freedom 4 and}~\alpha_{0.01} = \pm 4.6022$$$$3.1502 \pm 4.6022(\displaystyle\frac{0.0092}{\sqrt{5}} ) = 3.1502 \pm 0.0189 = (3.1313,3.1691)$$