The manager of "Frenchies" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most $1.5$oz., but the manager thinks that it may be higher. He randomly weighs$49$ orders of fries, which yields a mean of $11$ oz. and a standard deviation of two oz.

The null hypothesis is shown below:

$H_0:{\sigma }^2\le (1.5{)}^2$

That is, the standard deviation for a ten-ounce order of fries is at most $1.5\mathrm{oz}$.

Against the alternative hypothesis as shown below:

$H_a:{\sigma }^2>(1.5{)}^2$

That is, the standard deviation for a ten-ounce order of fries is greater than$1.5oz$.

The degrees of freedom can be calculated as shown below:

$n-1=49-1$

$48$

From the above calculation, it is clear that the distribution for the test is ${\chi }_{48}^2$.

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

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

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

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

$=\frac{(49-1)(2{)}^2}{(1.5{)}^2}$

$=\frac{48\times 4}{2.25}$

$=85.33$

Hence, the test statistic is$85.33$

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

Hence the $p$-value is $0.0007$

The Chi-Square sketch is given below: