Question

1.     The produce of a soft drink wants to identify the average age of its consumers. A sample of 55 consumers was taken. The average age in the sample was 21 years with a standard deviation of 4 years.  
A.   Calculate the margin of error for a 95% confidence interval for the average age of the company’s consumers. [2 marks]  
B.    Calculate a 95% confidence interval for the average age of the company’s consumers. [2 marks]
C.    Carefully interpret in your words the 95% confidence interval found in part B. Be very specific. [4 marks]  
D.   Now, suppose the company believes the average age of its consumers is 22. Use your confidence interval from part (c) to challenge that belief using a hypothesis test with significance level ?=0.05. (State the hypotheses, test decision, and conclusion.)

Answer 1

a)

sample mean, xbar = 21
sample standard deviation, s = 4
sample size, n = 55
degrees of freedom, df = n - 1 = 54

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.005

ME = tc * s/sqrt(n)
ME = 2.005 * 4/sqrt(55)
Margin of Error = 1.081

b)

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (21 - 2.005 * 4/sqrt(55) , 21 + 2.005 * 4/sqrt(55))
CI = (19.919 , 22.081)

c)

We are 95% confident atht the average age of the company’s consumers is between 19.919 and 22.081

d)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 22
Alternative Hypothesis, Ha: μ ≠ 22

Rejection Region
This is two tailed test, for α = 0.05 and df = 54
Critical value of t are -2.005 and 2.005.
Hence reject H0 if t < -2.005 or t > 2.005

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (21 - 22)/(4/sqrt(55))
t = -1.854

fail to reject the null hypothesis

confidence interval contains 22 fail to reject H0