91Ó°ÊÓ

The null hypothesis is shown below:

$H_0:{\mathrm{σ}}^2≤(0.5{)}^2$

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

Against the alternative hypothesis as shown below:

$H_a:{\mathrm{σ}}^2>(0.5{)}^2$

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 ${\mathrm{χ}}_{83}^2$.

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

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

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

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

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

$=\frac{83×0.2916}{0.25}$

$=96.81$

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: