For each of the problems below perform an hypothesis test. State the null and al
ID: 3315344 • Letter: F
Question
For each of the problems below perform an hypothesis test. State the null and alternative hypothesis, the p-value and your conclusion in context of the problem. Perform each test at a .05 significance.
1. A manufacturer of a plasticised line used in home-assembly mobiles advertises that their product has an average tensile strength of 30 kilograms (this is a measure of how strong the product is). You took a sample of 100 sections of the line and tested them. The average tensile strength of this sample was 28 kilos, with a standard deviation of 12 kilos. Does this enable you to dismiss the manufacturer's claims?
2. The EPA reports that the exhaust emissions for a certain car model has a normal distribution with a mean of 1.45 grams of nitrous oxide per mile. The car manufacturer claims their new process reduces the mean level of exhaust emitted for this car model. A random sample of 28 cars is taken and the mean level of exhaust emitted for this sample is 1.21 grams with a standard deviation of 0.4.
3. A credit card company wondered whether giving frequent flyer miles for every purchase would increase card usage, which has a current mean of $2500 per year. They gave free miles to a random sample of 51 credit card customers and found the sample mean to be $2542. The sample standard deviation is $109.
4. Studies conducted in the 1970s indicated that the average age at which children take their first alcoholic drink is 14.6 years old. Sociologists believe that children are starting to drink at a younger age. A random sample of 144 young adults (18 years of age) is selected and the age at which each adult took their first alcoholic drink is recorded. The sample mean age was 13.3 years of age with a standard deviation of 5 years.
Explanation / Answer
Given that,
population mean(u)=30
sample mean, x =28
standard deviation, s =12
number (n)=100
null, Ho: =30
alternate, H1: !=30
level of significance, = 0.05
from standard normal table, two tailed t /2 =1.984
since our test is two-tailed
reject Ho, if to < -1.984 OR if to > 1.984
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =28-30/(12/sqrt(100))
to =-1.6667
| to | =1.6667
critical value
the value of |t | with n-1 = 99 d.f is 1.984
we got |to| =1.6667 & | t | =1.984
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.6667 ) = 0.0987
hence value of p0.05 < 0.0987,here we do not reject Ho
ANSWERS
---------------
null, Ho: =30
alternate, H1: !=30
test statistic: -1.6667
critical value: -1.984 , 1.984
decision: do not reject Ho
p-value: 0.0987
no, we follow the manufacturer's claims
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