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.

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....

Many antacid tablets contain calcium carbonate
(CaCO3). Calcium carbonate reacts with HCl in the
stomach to produce water, calcium chloride, and carbon
dioxide. At 37°C and 1 atm, what volume (in L) of
CO2 is produced when a person consumes a Rolaids antacid
tablet that contains 550.0 mg of CaCO3 to eliminate
excess stomach acid?
Assume acid is in excess of the carbonate. (R = 0.08206
L.atm/K. mol)
1 mg = 0.001 g

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 3 minutes ago

asked 14 minutes ago

asked 27 minutes ago

asked 29 minutes ago

asked 35 minutes ago

asked 35 minutes ago

asked 45 minutes ago

asked 57 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago