A plant manager is concerned her equipment may need recalibrating. It seems that the actual weight of the $15oz$. cereal boxes it fills have been fluctuating. The standard deviation should be at most $0.5oz$. In order to determine if the machine needs to be recalibrated, $84$randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the $84$ boxes was $0.54$. Does the machine need to be recalibrated?

The null hypothesis is shown below:

$H_0:{\sigma }^2\le (0.5{)}^2$

That is, the variance of cereal boxes should be at most $0.5\mathrm{oz}$.

Against the alternative hypothesis as shown below:

$H_a:{\sigma }^2>(0.5{)}^2$

That is, the variance of cereal boxes is greater than $0.5\mathrm{oz}$.

The degrees of freedom can be calculated as shown below:

$n-1=84-1$

$=83$

From the above calculation, it is clear that the distribution for the test is ${\chi }_{83}^2$.

The test statistics can be calculated by using the formula shown below:

Test statistic $=\frac{(n-1)s^2}{{\sigma }^2}$

Here, $n$is sample size, $s^2$is sample variance and ${\sigma }^2$is population variance. Therefore, the calculation is shown below:

$\text{Test statistic}=\frac{(n-1)s^2}{{\sigma }^2}$

$=\frac{(84-1)(0.54{)}^2}{(0.5{)}^2}$

$=\frac{83\times 0.2916}{0.25}$

$=96.81$

Hence, the test statistic is$96.81$.

The $p$-value can be calculated in excel by using CHIDIST ( ) formula as shown below:

Hence the $p$- value is $0.14$

the Chi-Square sketch is given below: