Question

A local brewery distributes beer in bottles labeled 16.9 fluid ounces. A government agency thinks that the brewery is cheating its customers. The agency selects 31 of these bottles, measures their contents, and obtains a sample mean of 16.71 fluid ounces and a standard deviation of 0.51 fluid ounce. Does the sample show that the government agency is correct in thinking that the mean amount of beer is less than 16.9 fluid ounces? Use a = 0.05.

a. State the appropriate hypotheses.




b. Assuming the null hypothesis is true, the relevant sampling distribution has an approximate normal distribution with:

Mean =

Standard deviation =


c. The value of the standardized test statistic is __________________.


d. The P-value for this significance test is __________________.


e. Summarize the final conclusion of this significance test in context of the problem.






f. Give the 90% confidence interval for the relevant parameter.

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 16.9
Alternative Hypothesis, Ha: μ < 16.9

b)

mean = 16.9

std,deviation = 0.51/sqrt(31) = 0.0916

c)

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (16.71 - 16.9)/(0.51/sqrt(31))
t = -2.074

d)

P-value Approach
P-value = 0.0234

e)

As P-value < 0.05, reject the null hypothesis.
There is sufficient evidence to conclude that the mean amount of beer is less than 16.9 fluid ounces

f)

sample mean, xbar = 16.71
sample standard deviation, s = 0.51
sample size, n = 31
degrees of freedom, df = n - 1 = 30

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.697

ME = tc * s/sqrt(n)
ME = 1.697 * 0.51/sqrt(31)
ME = 0.155

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (16.71 - 1.697 * 0.51/sqrt(31) , 16.71 + 1.697 * 0.51/sqrt(31))
CI = (16.55 , 16.87)