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 [tex]\sigma=5[/tex] milligrams.
The sampling distribution of the sample variance is chi-square distribution.
Also,The data yields a sample standard deviation of [tex]s=4.8[/tex] milligrams.
Sample size : n= 10
Test statistic for chi-square :[tex]\chi^2=\dfrac{s^2(n-1)}{\sigma^2}[/tex]
i.e. [tex]\chi^2=\dfrac{(4.8)^2(10-1)}{(5)^2}=8.2944[/tex]
Now, P-value = [tex]P(\chi^2>8.2944)=0.50477[/tex] [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
