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

S² = 0.004

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}=(\sum x)/(n)`

=
`(1.15+1.24+1.15+1.27+1.13)/(5)`

=
`(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 =
`(0.0157)/(5-1)` = 0.0039

S² = 0.004

Answer 2

Point estimate for the population variance = 3.92 *
`10^(-3)` .

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` =
`(\sum X)/(n)` =
`(1.15+1.24+1.15+1.27+1.13)/(5)` = 1.188

Point estimate of Population Variance = Sample variance

=
`(\sum (X-Xbar)^(2))/(n-1)` =
`(0.01568)/(4)` = 3.92 *
`10^(-3)` .

Therefore, point estimate for the population variance = 3.92 *
`10^(-3)` .