Question

1. A worldwide fast food chain was bought out in a hostile takeover. Longtime customers are suspicious that the new owners have changed the chain’s famous “Big Bucket o’Fries” and don’t include the same number of fries. Prior to the takeover, the average number of fries included was 40 (SD = 3). These loyal customers with too much time on their hands purchased 9 buckets and counted the number of fries in each (before scarfing them down), here are their numbers: 38 42 37 49 36 39 43 36 35
   1. What is the null hypothesis?
   2. What is the alternative hypothesis?
   3. What test statistic will you use? Why?
   4. What alpha level will you set?
   5. What is the set up for the test statistic (i.e., write out the formulas needed with the appropriate numbers from the problem/show the initial steps)?
   6. What is the value of the test statistic?
   7. What is the p-value?
   8. What is your decision about the null hypothesis?
   1. What is your interpretation of the results?

Answer 1

What is the null hypothesis?

Ho:he average number of fries included was 40

Ho:mu=40

What is the alternative hypothesis?

Ha:the average number of fries included was 40

Ha:mu not =40

What test statistic will you use? Why?

since population standard deviation (sigma=3) is known use z statistic

What alpha level will you set?

alpha=0.05

What is the set up for the test statistic (i.e., write out the formulas needed with the appropriate numbers from the problem/show the initial steps)?

z=xbar-mu/sigma/sqrt(n)

xbar=sample mean=355/9=39.44444

mu=population mean=40

sigma=standard deviation=3

n=sample size=9

What is the value of the test statistic?

z=(39.44444-40)/(3/sqrt(9))

=-0.55556

What is the p-value?

right tail prob==NORMSDIST(0.55556)=0.710744159

left tail prob=1-right tail prob=1-0.710744159=0.289255841  

two tail p=2*left tail=2*0.289255841=0.578511682

p=0.5785

What is your decision about the null hypothesis?

p=0.5785

p>0.05

Do not reject null hypothesis

What is your interpretation of the results?

There is no sufficient statistical evidence at 5% level of significance to conclude that  the average number of fries changed from 40