Question

an automatic coffee machine dispenses cups of coffee whose volume per cup varies normally with the mean \\( \mu = 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 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 b. one - sample z test of mu = 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: Determine the appropriate test

Since the population standard deviation is unknown and we are using the sample standard deviation, a \(t -\)test is appropriate. A \(z -\)test is used when the population standard deviation \(\sigma\) is known.

Step2: Analyze the outputs

Output A is a One - Sample \(T\) test (which uses the sample standard deviation \(s\)) and Output B is a One - Sample \(Z\) test (which assumes the population standard deviation \(\sigma\) is known). In our problem, we are given the sample standard deviation \(s = 0.23\) (not the population standard deviation).

Answer:

A. One - Sample T