In: Statistics and Probability
A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. Following are the amounts measured in a simple random sample of eight cans. 11.96 12.10 12.04 12.13 11.98 12.05 11.91 12.03
Perform a hypothesis test to determine whether the mean volume differs from 12 ounces. Use a 0.05 level of significance.
Also construct a 99% confidence interval for the data.
A)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 12
Alternative Hypothesis, Ha: μ ≠ 12
Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (12.03 - 12)/(0.0727/sqrt(8))
t = 1.167
P-value Approach
P-value = 0.2814
As P-value >= 0.05, fail to reject null hypothesis.
mean is 12 ounces
B)
sample mean, xbar = 12.03
sample standard deviation, s = 0.0727
sample size, n = 8
degrees of freedom, df = n - 1 = 7
Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 3.499
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (12.03 - 3.499 * 0.0727/sqrt(8) , 12.03 + 3.499 *
0.0727/sqrt(8))
CI = (11.94 , 12.12)