Question

11. A manufacturer of building irrigation systems claims that their systems activate at an average temperature of 130 degrees Fahrenheit, but affected customers claim that the systems have not been activated in time and conclude that the average activation temperature is higher than claimed by the manufacturer, which puts people's lives at risk. 9 systems were tested and the average activation temperature was 131.8 degrees. If we assume that the activation temperature follows a normal distribution with a known variance of 1.5 degrees:

a. Determine if the evidence supports the customer's claim. Clearly identify the null and alternative hypothesis, as well as the areas of acceptance and rejection and conclusion. Use a 90% confidence level.

b. If the true mean activation temperature were 131 degrees, what would be the probability of making a type II error? Explain what this means.

c. What sample size is required so that the probability of making a type II error for the deviation described in the previous question is not greater than 5%, keeping the probability of making a type I error at the same level?

Answer 1

given that

sample size=n=9

sample mean =m=131.8

population variance =V=1.5

a)

we have to test that if true temperature is more than 130

Hence

confidence level =95% so

let rejection region is

we can find "c" by equationg

now

from Z table

P(Z>1.645)=0.05

so

so we reject H0 if  

fail to reject H0 if

Here sample mean =m=131.8

so m> c=130.6716

Hence we reject H0 hence there is enough evidence to conclude that true temperature mean is more than 130.

B)

we have to find P(type 2 error)

P(type 2 error ) is defined as Probability of fail to reject the claim that mean temperature is 130 while in actual mean temperature is 131.

C)

we have to find "n" for P(type 2 error) =0.05

now

from Z table

P(Z<-1.645) =0.05

Hence

n~38