Question

The sugar content of the syrup is canned peaches is normally distributed. Assume the can is designed to have standard deviation 5 milligrams. A random sample of n = 10 cans is studied. What is the sampling distribution of the sample variance? The data yields a sample standard deviation of 4.8 milligrams. What is the chance of observing the sample standard deviation greater than 4.8 milligrams?

Answer 1

Answer: 0.504772

Explanation:

Given : The sugar content of the syrup is canned peaches is normally distributed. Assume the can is designed to have standard deviation
`\sigma=5` milligrams.

A random sample of n = 10 cans is studied.

Then, the sampling distribution of the sample variance is chi-square (
`\chi^2`) distribution witth
`n-1` degrees of freedom.

Sample standard deviation:
`s=4.8` milligrams.

Test statistic for chi-square =
`\chi^2=((n-1)s^2)/(\sigma^2)`

`\\\\=((10-1)(4.8)^2)/((5)^2)\\\\=8.2944`

By using the chi-square distribution table , the chance of observing the sample standard deviation greater than 4.8 milligrams will be :-

P-value =
`P(\chi^2>8.2944)=1-P(\chi^2\leq8.2944)`

`=1-0.495228= 0.504772`

Hence, the chance of observing the sample standard deviation greater than 4.8 milligrams = 0.504772