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 95% upper n mean stdev se mean 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 95% upper n mean se mean bound z p 8 9.92000 0.08132 10.05376 - 0.98 0.163

Explanation:

Step1: Determine the appropriate test

When the population standard deviation \(\sigma\) is unknown (as in this case, we are given the sample standard deviation \(s = 0.23\)), we use the \(t -\)test. The \(z -\)test is used when \(\sigma\) is known.

Step2: Analyze the outputs

Output A is a One - Sample \(T\) test (which is appropriate when \(\sigma\) is unknown) with the correct hypotheses \(H_0:\mu = 10\) vs \(H_1:\mu<10\). Output B is a One - Sample \(Z\) test (which is inappropriate here since \(\sigma\) is unknown).

Answer:

A. One - Sample T test output.