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

Step-by-step 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 :$\chi^2=\dfrac{s^2(n-1)}{\sigma^2}$

i.e. $\chi^2=\dfrac{(4.8)^2(10-1)}{(5)^2}=8.2944$

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