Question

The company who makes Chips Ahoy cookies states that there is an average of 23 chocolate chips per cookie. You take a sample of cookies and count the chips. For your sample of 30 cookies, the average # of chips in the cookies is 23.6 chips with a standard deviation of 2 chips. Use this data to test the claim that the company makes. Use a 95% significance level but you may use a two tailed OR a one tailed test. You decide. Does your hypothesis test support the claim that Chip Ahoy is making?

Provide a 95% confidence interval to estimate the true number of chips in Chips Ahoy cookies. Does your interval support the claim of the company?

Answer 1

Step 1:

Ho: = 23

Ha: 23

The Null hypothesis states that the average chocolate chips per cookie is 23.

Step 2: Test statistics

n = 30

sample mean = 23.6

sample sd = 2

Assuming that the data is normally distributed, also as the population sd is not given, we will use t statistics.

P value = TDIST (t statistics, df, 2) = TDIST(1.643, 29, 2)= 0.1112

Step 3:

df = 29

level of significance= 0.05

t critical = +/- 2.04522964

As the t stat (1.643) does not fall in the rejection area, we fail to reject the Null hypothesis.

Also as the p value (0.112) is greater than level of significance (0.05), we fail to reject the null hypothesis.

Step 4:

t value for 95% of confidence interval at 29 df is TINV(0.05,29) = 2.045

CI = mean + / - E = 23.6 +/- 0.747

CI = (22.85 , 24.35)

As 23 falls in the CI range, we fail to reject the Null hypothesis. Yes the interval supports the claim of the company.