Question

A lumber mill is tested for consistency by measuring the variance of board thickness. The target accuracy is a variance of 0.0035 square inches or less. If 28 measurements are made and their variance is 0.006 square inches, is there enough evidence to reject the claim that the standard deviation is within the limit at a = 0.01? No, since the x test valle 46.29 is less than the critical value 46.963. Yes, since the x test value 8.91 is less than the critical value 46.963. No, since the x test value 46.29 is less than the critical value 48.278. Yes, since the y test value 8.91 is less than the critical value 48.278.

Answer 1

The null and alternative hypotheses

Ho : $\LARGE \sigma$ = 0.0592

H1 : $\LARGE \sigma$$\LARGE \leq$ 0.0592

Test statistic

  $\LARGE \chi$² = (n-1)*s^{​​​​​​2}/$\LARGE \sigma$²

$\LARGE \chi$² = 27*0.006/0.0035

  $\LARGE \chi$² = 46.29

critical value for a = 0.01 and d.f = n -1 = 27

  $\LARGE \chi$²critical = $\LARGE \chi$²0.01 , 27

$\LARGE \chi$²critical = 46.963

Decision rule : if $\LARGE \chi$² > $\LARGE \chi$²critical , we reject the null hypothesis, otherwise we fail to reject the null hypothesis.

Our 46.29 < 46.963

So we fail to reject the null hypothesis.

Answer is

No, since chi square test value 46.29 is less than the critical value 46.963