Question

an automatic coffee machine dispenses cups of coffee whose volume per cup varies normally with the mean μ = 10 oz. a quality - control researcher randomly selects 8 cups of coffee from the machine and finds that in this sample the mean volume is 9.92 oz. and the standard deviation is 0.23 oz. we want to determine if these data provide enough evidence to conclude that the mean volume per cup is below the target level. which one of the two outputs represents the correct way to conduct this test? a. one - sample t test of μ = 10 vs < 10 n mean stdev se mean 95% upper bound t p 8 9.92000 0.23000 0.08132 10.07406 - 0.98 0.179 b. one - sample z test of μ = 10 vs < 10 the assumed standard deviation = 0.23 n mean se mean 95% upper bound z p 8 9.92000 0.08132 10.05376 - 0.98 0.163

Explanation:

Step1: Identify the test - statistic situation

We have a small sample ($n = 8$) and the population standard - deviation is unknown. We only know the sample standard deviation ($s=0.23$). In such cases, a one - sample t - test is appropriate. A one - sample z - test is used when the population standard deviation $\sigma$ is known or when the sample size $n$ is large (usually $n\geq30$).

Step2: Analyze the options

Option A is a one - sample t - test output, and option B is a one - sample z - test output. Since we should use a one - sample t - test for the given situation, option A is correct.

Answer:

A. One - Sample T
Test of $\mu = 10$ vs $<10$
N Mean StDev SE Mean 95% Upper Bound T P
8 9.92000 0.23000 0.08132 10.07406 - 0.98 0.179