Question

Example 3: 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.

Null Hypothesis:

Alternative Hypothesis:

Test Static=

P-Value=

Conclusion:

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 12
Alternative Hypothesis, Ha: μ ≠ 12

Rejection Region
This is two tailed test, for α = 0.05 and df = 7
Critical value of t are -2.365 and 2.365.
Hence reject H0 if t < -2.365 or t > 2.365

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (12.025 - 12)/(0.0727/sqrt(8))
t = 0.973

P-value Approach
P-value = 0.363
As P-value >= 0.05, fail to reject null hypothesis.

sample mean, xbar = 12.025
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

ME = tc * s/sqrt(n)
ME = 3.499 * 0.0727/sqrt(8)
ME = 0.0899

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (12.025 - 3.499 * 0.0727/sqrt(8) , 12.025 + 3.499 * 0.0727/sqrt(8))
CI = (11.9351 , 12.1149)