Mathematics: An electric scale gives a reading equal to the true weight plus a random error that is normally distributed with mean 0 and standard deviation s = .1 mg. suppose that 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.

z = 0.25 Step-by-step explanation:I could only get one you set the questions confusing can you number them so I can solve all of them individually?...

An electric scale gives a reading equal to the true

Ответ:

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10.01.2022

Mathematics

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, 3.141

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

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,

$t_{critical}\text{ at degree of freedom 4 and}~\alpha_{0.01} = \pm 4.6022$

$3.1502 \pm 4.6022(\displaystyle\frac{0.0092}{\sqrt{5}} ) = 3.1502 \pm 0.0189 = (3.1313,3.1691)$

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