Question

A simple random sample of 28 Lego sets is obtained and the number of pieces in each set was counted.The sample has a standard deviation of 12.65. Use a 0.05 significance level to test the claim that the number of pieces in a set has a standard deviation different from 11.53.

Answer 1

Explanation:

Given that:

A simple random sample n = 28

sample standard deviation S = 12.65

standard deviation
`\sigma` = 11.53

Level of significance ∝ = 0.05

The objective is to test the claim that the number of pieces in a set has a standard deviation different from 11.53.

The null hypothesis and the alternative hypothesis can be computed as follows:

Null hypothesis:

`H_0: \sigma^2 = \sigma_0^2`

Alternative hypothesis:

`H_1: \sigma^2 \\eq \sigma_0^2`

The test statistics can be determined by using the following formula in order to test if the claim is statistically significant or not.

`X_0^2 = ((n-1)S^2)/(\sigma_0^2)`

`X_0^2 = ((28-1)(12.65)^2)/((11.53)^2)`

`X_0^2 = ((27)(160.0225))/(132.9409)`

`X_0^2 = (4320.6075)/(132.9409)`

`X_0^2 = 32.5002125`

`X^2_(1- \alpha/2 , df) = X^2_(1- 0.05/2 , n-1)`

`X^2_(1- \alpha/2 , df) = X^2_(1- 0.025 , 28-1)`

From the chi-square probabilities table at 0.975 and degree of freedom 27;

`X^2_(0.975 , 27)` = 14.573

`X^2_(\alpha/2 , df) = X^2_( 0.05/2 , n-1)`

`X^2_(\alpha/2 , df) = X^2_(0.025 , 28-1)`

From the chi-square probabilities table at 0.975 and degree of freedom 27;

`X^2_(0.025 , 27)=` 43.195

Decision Rule: To reject the null hypothesis if
`X^2_0 \ > \ X^2_(\alpha/2 , df) \ \ \ or \ \ \ X^2_0 \ < \ X^2_(1- \alpha/2 , df)` ; otherwise , do not reject the null hypothesis:

The rejection region is
`X^2_0 \ > 43.195 \ \ \ or \ \ \ X^2_0 \ < \ 14.573`

Conclusion:

We fail to reject the null hypothesis since test statistic value 32.5002125 lies between 14.573 and 43.195.