In a random sample of ten cans of corn from supplier B the a

In a random sample of ten cans of corn from supplier B, the average weight per can of corn was sample mean or x bar = 9.4 ounces with standard deviation s = 1.8 ounces. Does this sample contain sufficient evidence to indicate the mean weight is less than 10 ounces? Use the p-value approach. Find a 98% confidence interval for population variance

Solution

a)
Set Up Hypothesis
Null, H0: U=10
Alternate,mean weight is less than 10 ounces H1: U<10
Test Statistic
Population Mean(U)=10
Sample X(Mean)=9.4
Standard Deviation(S.D)=1.8
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =9.4-10/(1.8/Sqrt(9))
to =-1.054
| to | =1.054
P-Value :Left Tail -Ha : ( P < -1.0541 ) = 0.15966
Hence Value of P0.02 < 0.15966,Here We Do not Reject Ho

We don\'t have enough evidence to support mean weight is less than 10 ounces

b)
CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.02
^2 right = (1 - Confidence Level)/2 = (1 - 0.98)/2 = 0.02/2 = 0.01
^2 left = 1 - ^2 right = 1 - 0.01 = 0.99
the two critical values ^2 left, ^2 right at 9 df are 21.666 , 2.088
S.D( S^2 )=1.8
Sample Size(n)=10
Confidence Interval for population variance = [ 9 * 3.24/21.666 < ^2 < 9 * 3.24/2.088 ]
= [ 29.16/21.666 < ^2 < 29.16/2.0879 ]
C.I. population variance = [ 1.3459 , 13.9662 ]