Question

A coin-operated drink machine was designed to discharge a mean of 8 ounces of coffee per cup. In a test of the machine, the discharge amounts in 21 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 7.87 ounces and 0.23 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, ?, differs from 8 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 8 The alternative hypothesis:I The type of test statistic: (Choose one) ' The value of the test statistic: (Round to at least three decimal places.) The two at the 0.05 level of significance: (Round to at least three decimal places.) and At the 0.05 level of significance, can we conclude that the true mean discharge differs from 8 ounces? YesNo

Answer 1

As we are testing here the claim that the true mean discharge differs from 8 ounces, therefore the null and alternate hypothesis here are given as:

$\LARGE H_0: \mu = 8$

$\LARGE H_1: \mu \neq 8$

The type of test statistic here used would be a t test statistic as we are not given the population standard deviation but only the sample standard deviation.

The test statistic here is computed as:

$\LARGE t^* = \frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{7.87 - 8}{\frac{0.23}{\sqrt{21}}} =-2.590$

Therefore -2.590 is the test statistic value here.

For 0.05 level of significance and n - 1 = 20 degrees of freedom, we get from the t distribution tables:

P( -2.086 < t₂₀ < 2.086 ) = 0.95

Therefore the critical values here are: -2.086 and +2.086

As the critical value here is -2.086 > -2.590, therefore the test statistic lies in the rejection region and therefore we conclude that the test is significant and we can reject the null hypothesis here. Therefore we can conclude here that the true mean differs from 8. Therefore Yes.