A coffee-dispensing machine is supposed to deliver 8 ounces of liquid into each paper cup,but a consumer believes that the actual mean amount is less.The consumer obtained a sample of 49 cups of the dispensed liquid with average of 7.75 ounces.If the sample variance of the dispensed liquid delivered per cup is 0.81 ounces,and α = 0.05,the appropriate decision is ________.
A) increase the sample size
B) reduce the sample size
C) fail to reject the 8-ounces claim
D) maintain status quo
E) reject the 8-ounces claim

Question

Quizplus · Accepted Answer

We need to conduct a one-sample t-test to determine whether the mean amount of liquid delivered per cup is significantly different from 8 ounces.

The null hypothesis is that the mean amount of liquid delivered per cup is 8 ounces. The alternative hypothesis is that the mean amount of liquid delivered per cup is less than 8 ounces.

We can use the t-test formula: t = (x̄ - μ) / (s / √n), where x̄ is the sample mean (7.75 ounces), μ is the hypothesized mean (8 ounces), s is the sample standard deviation (0.9 ounces), and n is the sample size (49).

Plugging in the numbers, we get: t = (7.75 - 8) / (0.9 / √49) = -2.78.

The degrees of freedom (df) for this test is n-1 = 48.

Using a t-table or a calculator with a t-distribution function and df=48, we find the critical t-value for α = 0.05, one-tailed test is -1.677.

Because our calculated t-value (-2.78) is less than the critical t-value (-1.677), we reject the null hypothesis.

Therefore, we can conclude that there is enough evidence to reject the claim that the machine delivers 8-ounce cups. The appropriate decision is to reject the 8-ounces claim (E).

A Coffee-Dispensing Machine Is Supposed to Deliver 8 Ounces of Liquid