Question

The sugar content of the syrup is canned peaches is normally distributed. Assumethe can is designed to have standard deviation 5 milligrams. A random sample ofn= 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 chanceof observing the sample standard deviation greater than 4.8 milligrams?

Answer 1

Answer: 0.50477

Explanation:

Given : The sugar content of the syrup is canned peaches is normally distributed.

We assume the can is designed to have standard deviation
`\sigma=5` milligrams.

The sampling distribution of the sample variance is chi-square distribution.

Also,The data yields a sample standard deviation of
`s=4.8` milligrams.

Sample size : n= 10

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

i.e.
`\chi^2=((4.8)^2(10-1))/((5)^2)=8.2944`

Now, P-value =
`P(\chi^2>8.2944)=0.50477` [By using the chi-square distribution table for p-values.]

Hence, the chance of observing the sample standard deviation greater than 4.8 milligrams = 0.50477