Question

A sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters. Give a point estimate for the population variance. Round your answer to three decimal places. 1.15,1.24,1.15,1.27,1.13

Answer 1

Answer:

Point estimate for the population variance = 3.92 * $10^{-3}$ .

Step-by-step explanation:

We are given that a sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters ;

       X                       X - $Xbar$                                         $(X-Xbar)^{2}$

      1.13            1.13 - 1.188 = -0.058                                 3.364 * $10^{-3}$

      1.15            1.15 - 1.188 = -0.038                                 1.444 * $10^{-3}$    

      1.15            1.15 - 1.188 = -0.038                                 1.444 * $10^{-3}$

      1.24           1.24 - 1.188 = 0.052                                 2.704 * $10^{-3}$

      1.27           1.27 - 1.188 = 0.082                                6.724 * $10^{-3}$    

                                                                    $\sum (X-Xbar)^{2}$ = 0.01568  

Firstly, Mean of above data, $Xbar$ = $\frac{\sum X}{n}$ = $\frac{1.15+1.24+1.15+1.27+1.13}{5}$ = 1.188

Point estimate of Population Variance = Sample variance

                                                               = $\frac{\sum (X-Xbar)^{2}}{n-1}$ = $\frac{0.01568}{4}$ = 3.92 * $10^{-3}$ .

Therefore, point estimate for the population variance = 3.92 * $10^{-3}$ .

       

Answer 2

Answer:

S² = 0.004

Step-by-step explanation:

Point estimate for the population variance is S²

$S^{2}=\frac{\sum (x_{i}-\bar{x})^{2}}{n-1}$

S² = Sample Variance

∑ = Sum

$x_{i}$ = Term in data set

$\bar{x}$ = Sample mean

n = Sample size

Sample mean ($\bar{x}$)  =

$\bar{x}=\frac{\sum x}{n}$

   = $\frac{1.15+1.24+1.15+1.27+1.13}{5}$

   = $\frac{5,94}{5}$

  = 1.188

$x_{i}$     -     $\bar{x}$             $(x_{i}-\bar{x})$       $(x_{i}-\bar{x})^{2}$

1.15  -   1.88  =       0.038        0.001444

1.24  -  1.88  =       0.052        0.002704

1.15  -  1.88   =       0.038        0.001444

1.27 -   1.88  =       0.082        0.006724

1.13  -   1.88  =       0.058        0.003364

5.94                      0.268          0.01568  

$(x_{i}-\bar{x})^{2}$  =  0.0157

S² = $\frac{0.0157}{5-1}$   =  0.0039

S² = 0.004