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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 1.5oz., but the manager thinks that it may be higher. He randomly weighs49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz.

Short Answer

Expert verified

Decision: Reject the null hypothesis H0

Conclusion: There is sufficient evidence to ensure that the standard deviation for a ten-ounce order of fries is greater than 1.5oz.

Step by step solution

01

Given Information

The null hypothesis is shown below:

H0:σ2(1.5)2

That is, the standard deviation for a ten-ounce order of fries is at most 1.5oz.

Against the alternative hypothesis as shown below:

Ha:σ2>(1.5)2

That is, the standard deviation for a ten-ounce order of fries is greater than1.5oz.

02

Calculation

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 χ482.

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

Teststatistic=(n-1)s2σ2

Here, nis sample size, s2is sample variance and σ2is population variance. Therefore, the calculation is shown below:

Teststatistic=(n-1)s2σ2

=(49-1)(2)2(1.5)2

=48×42.25

=85.33

Hence, the test statistic is85.33

03

Graph

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:

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