Question

Bottles of antacid tablets include a printed label claiming that they contain 1000 mg of calcium carbonate. In a simple random sample of 21 bottles of tablets made by the Medassist Pharmaceutical Company, the amounts of calcium carbonate are measured and the average amount of calcium carbonate is found to be 985 mg. Suppose that the amounts of calcium carbonate in bottles of antacid tablets produced by the company follows a normal distribution with a standard deviation of 50 mg. Is there sufficient evidence to sup-port the claim that consumers are being cheated at the significance level of α=0.01. Use the p-value and confidence interval approach.

Answer 1

H0: mu = 1000
Ha: mu < 1000

As the CI level is 0.01 and this is one tailed test, we should use CL = 100 - 2*1 = 98%

sample mean, xbar = 985
sample standard deviation, σ = 50
sample size, n = 21

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, Zc = Z(α/2) = 2.33

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (985 - 2.33 * 50/sqrt(21) , 985 + 2.33 * 50/sqrt(21))
CI = (959.58 , 1010.42)

As 1000 is included in the CI, there are not significant evidence to conclude that the customers are being cheated.