Question

a coin - operated drink machine was designed to discharge a mean of 6 fluid ounces of coffee per cup. in a test of the machine, the discharge amounts in 14 randomly chosen cups of coffee from the machine were recorded. the sample mean and sample standard deviation were 6.17 fluid ounces and 0.25 fluid ounces, respectively.
if we assume that the discharge amounts are approximately normally distributed, is there enough evidence, to conclude that the population mean discharge, μ, differs from 6 fluid ounces? use the 0.05 level of significance.
perform a two - tailed test. then complete the parts below.
carry your intermediate computations to three or more decimal places. (if necessary, consult a list of formulas.)
(a) state the null hypothesis (h_0) and the alternative hypothesis (h_1).
(h_0:)
(h_1:)
(b) determine the type of test statistic to use.
select
(c) find the value of the test statistic. (round to three or more decimal places.)
(d) find the p - value. (round to three or more decimal places.)
(e) can we conclude that the mean discharge differs from 6 fluid ounces?
yes no

Explanation:

Step1: State the hypotheses

The null hypothesis $H_0$ is that the population mean $\mu$ is equal to the hypothesized mean. The alternative hypothesis $H_1$ for a two - tailed test is that the population mean $\mu$ is not equal to the hypothesized mean.
$H_0:\mu = 6$
$H_1:\mu
eq6$

Step2: Determine the test statistic type

Since the population standard deviation $\sigma$ is unknown and we have a sample standard deviation $s$, and the population is approximately normally distributed, we use a t - test statistic. The formula for the t - test statistic is $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the hypothesized population mean, $s$ is the sample standard deviation, and $n$ is the sample size.

Step3: Calculate the test statistic

We are given that $\bar{x} = 6.17$, $\mu = 6$, $s=0.25$, and $n = 14$.
$t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=\frac{6.17 - 6}{0.25/\sqrt{14}}\approx\frac{0.17}{0.25/3.742}\approx\frac{0.17}{0.067}\approx2.537$

Step4: Calculate the p - value

The degrees of freedom is $df=n - 1=14 - 1 = 13$. For a two - tailed t - test with $t = 2.537$ and $df=13$, using a t - distribution table or a calculator, the p - value is $2\times P(t_{13}>2.537)$. Using a calculator, the p - value is approximately $0.024$.

Step5: Make a decision

The level of significance $\alpha = 0.05$. Since the p - value ($0.024$) is less than $\alpha=0.05$, we reject the null hypothesis.

Answer:

(a) $H_0:\mu = 6$, $H_1:\mu
eq6$
(b) t - test statistic
(c) $2.537$
(d) $0.024$
(e) Yes