Question

a coin - operated drink machine was designed to discharge a mean of 8 fluid 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 8.17 fluid ounces and 0.27 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 8 fluid ounces? use the 0.10 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 h₀ and the alternative hypothesis h₁. h₀:□ h₁:□ (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 8 fluid ounces? oyes ono

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the population mean $\mu$ is equal to the designed - mean. The alternative hypothesis $H_1$ is that the population mean $\mu$ is not equal to the designed - mean.
$H_0:\mu = 8$
$H_1:\mu
eq8$

Step2: Determine test statistic type

Since the population standard deviation $\sigma$ is unknown and we have a sample standard deviation $s$, and the sample size $n = 15$ (small sample, $n<30$) with a normal - distributed population, we use the $t$ - test statistic.
$t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$

Step3: Calculate test - statistic value

We are given $\bar{x}=8.17$, $\mu = 8$, $s = 0.27$, and $n = 15$.
$t=\frac{8.17 - 8}{0.27/\sqrt{15}}\approx2.477$

Step4: Calculate p - value

The degrees of freedom is $df=n - 1=15 - 1 = 14$. Since it is a two - tailed test, the p - value is $2P(T>|t|)$. Using a $t$ - distribution table or calculator, $P(T>2.477)$ with $df = 14$, and the p - value is $2\times(1 - P(T\leq2.477))$. Looking up in the $t$ - distribution table or using a calculator, $P(T\leq2.477)\approx0.99$, so the p - value is $2\times(1 - 0.99)=0.026$.

Step5: Make a decision

The level of significance $\alpha = 0.10$. Since the p - value $0.026<0.10$, we reject the null hypothesis.

Answer:

(a) $H_0:\mu = 8$; $H_1:\mu
eq8$
(b) $t$ - test statistic
(c) $2.477$
(d) $0.026$
(e) Yes