Question

A coin-operated drink machine was designed to discharge a mean of 7 ounces of coffee per cup. In a test of the machine, the discharge amounts in 18 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 7.04 ounces and 0.17 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.1 level of significance, to conclude that the true mean discharge,u, differs from ounces? Perform a two-tailed test. Then fill in the table below. I need the two critical values at the 0.1 level of significace. Also need the answers to al other questions and whether we accpe for reject.

Answer 1

Using the t-distribution, it is found that since the test statistic is between -1.7341 and 1.7341, we do not reject(accept) the null hypothesis, hence there is not enough evidence to conclude that the true mean discharge differs from 7 ounces.

What are the hypothesis tested?

At the null hypothesis, it is tested if the mean is of 7 ounces, that is:

$H_0: \mu = 7$

At the alternative hypothesis, it is tested if the mean is different of 7 ounces, that is:

$H_1: \mu \neq 7$

What is the test statistic?

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• s is the standard deviation of the sample.

• n is the sample size.

The values of the parameters are given by:

$\overline{x} = 7.04, \mu = 7, s = 0.17, n = 18$

Hence the value of the test statistic is found as follows:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{7.04 - 7}{\frac{0.17}{\sqrt{18}}}$

t = 1

What is the decision?

Considering a two-tailed test, as we are testing if the mean is different of a value, with 18 - 1 = 17 df and a significance level of 0.1, the critical value is of $t^{\ast} = 1.7341$

Since the test statistic is between -1.7341 and 1.7341, we do not reject(accept) the null hypothesis, hence there is not enough evidence to conclude that the true mean discharge differs from 7 ounces.

More can be learned about the t-distribution at brainly.com/question/13873630

#SPJ1