A manufacturer of mixed nuts promises:“At least 20% cashews in every can.” Aconsumer-research agency tests 150 cansof nuts and ﬁnds a mean of 22% cashewswith a standard deviation of 1.5%....
A manufacturer of mixed nuts promises:
“At least 20% cashews in every can.” A
consumer-research agency tests 150 cans
of nuts and ﬁnds a mean of 22% cashews
with a standard deviation of 1.5%. The
proportions of cashews are normally
a) What is the probability that the
population mean is less than 20%
b) Based on the sample, what proportion of
cans have between 15% and 30%
c) The company must stop making their
claim if more than 3% of the cans
contain less than 20% cashews. Write a
brief report to the company outlining
whether they need a new motto
We are given a sample with n=150,`bar(x)=22`, and s=1.5
(a) We can use a T-test to determine the likelihood of `mu<20` :
The test value is `t=(22-20)/(1.5/sqrt(150))=16.33`
Consulting a student's T-table we find that if the degrees of freedom are greater than 100 any value greater than 2.58 is in the critical region.
The probability that a random sample has a t-value of 16.33 is vanishingly small. My calculator gives a p-value of 1; that is the probability of selecting a random sample of 150 cans with a mean greater than 20 is 1; virtually certain.
Therefore the probability of the population mean being less than 20 is zero.
(b) Using the sample statistics, with the understanding that the proportions are approximately normal, then the probability that the percentage is between 15 and 30 can be found by converting the raw scores to z-scores.
`x=15 ==> z=(15-22)/(1.5)=-4.bar(6)`
`P(-4.bar(6)<z<5.bar(3))=.9999984193` from a calculator. These values are off of the standard normal table and you would be instructed to use .9999
The probability that a can has between 15 and 30 percent is virtually 1.
(c) There is certainly no need for a new motto as it is virtually impossible that any cans will have less than 20%