Question

A) A slow cooker manufacturer claims that the true average ’Low’ temperature setting for their prod- ucts is 130◦ C. A sample of 9 slow cookers are tested and the average ’Low’ temperature setting is 131.08◦ C. If the distribution is normal with standard deviation 1.5◦C , does the data contradict the manufacturer’s claim at significance level α = 0.01 ?

B) Suppose that in an AB test, we test two website design for an online retailer. The first design leads to an average of 10 purchases per day fro a sample of 100 days, while the other leads to 11 purchases per day, also for a sample of 100 days. Assuming a common standard deviation of 4 purchases per day. If the groups are independent and the days are independent and identically distributed, calculate the z-test statistic. Give a p-value for the test. Do you reject at the 5% level?

Answer 1

Ans:

A)

Test statistic:

z=(131.08-130)/(1.5/SQRT(9))

z=2.16

p-value(2 tailed)=2*P(z>2.16)=0.0308

critical z values=+/-2.576

As,test statistic does not fall in rejection region,we fail to reject the null hypothesis.

There is not sufficient evidence to contradict the manufacturer’s claim.

B)

standard error=4*SQRT((1/100)+(1/100))=0.5657

Test statistic:

z=(10-11)/0.5657

z=-1.768

p-value(one tailed)=P(z<-1.768)=0.0386

As,p-value<0.05,we reject the null hypothesis.