Question

A coin-operated drink machine was designed to discharge a mean of

9

ounces of coffee per cup. In a test of the machine, the discharge amounts in

15

randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were

8.82

ounces and

0.28

ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the

0.05

level of significance, to conclude that the true mean discharge,

μ

, differs from

9

ounces?

Perform a two-tailed test. Then fill in the table below.

Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)

The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: (Choose one)ZtChi squareF Degrees of freedom: The value of the test statistic: (Round to at least three decimal places.) The p-value: (Round to at least three decimal places.) At the 0.05 level of significance, can we conclude that the true mean discharge differs from 9 ounces? Yes No

Answer 1

This is the two tailed test .

The null and alternative hypothesis is

H₀ : = 9

H_a : 9

Test statistic = t

= ( - ) / s / n

= (8.82 - 9) / 0.28 / 15

= -2.490

Test statistic = -2.490

degrees of freedom = 14

P-value = 0.026

= 0.05

P-value <

Reject the null hypothesis .

Yes