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Question 1 of 10 (1 point)

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 amount for randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 7.86 fluid ounces and 0.2 ounces, respectively.

If we assume that the discharge amounts are approximately normally distributed, is there enough evidence to conclude that the population mean 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 [tex]$H_0$[/tex] and the alternative hypothesis [tex]$H_1$[/tex].
[tex]\[
\begin{array}{l}
H_0: \mu = 8 \\
H_1: \mu \neq 8
\end{array}
\][/tex]

(b) Determine the type of test statistic to use.

Degrees of freedom: 12

(c) Find the value of the test statistic. (Round to three or more decimal places.)

[tex]\[2.3094\][/tex]

(d) Find the [tex]$p$[/tex]-value. (Round to three or more decimal places.)



Answer :

GinnyAnswer
Certainly! Let's solve the problem step by step.

(a) State the null and alternative hypotheses

We are testing whether the population mean amount of coffee dispensed by the machine differs from 8 fluid ounces.

- Null Hypothesis [tex]\( H_0 \)[/tex]: [tex]\( \mu = 8 \)[/tex]
- Alternative Hypothesis [tex]\( H_1 \)[/tex]: [tex]\( \mu \neq 8 \)[/tex]

(b) Determine the type of test statistic to use

Since the population standard deviation is unknown and the sample size is relatively small (n = 13), we use the t-statistic for the hypothesis test. The degrees of freedom (df) for this test is [tex]\( n - 1 = 13 - 1 = 12 \)[/tex].

(c) Find the value of the test statistic

To calculate the t-statistic, we use the formula:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
where:
- [tex]\(\bar{x} = 7.86\)[/tex] is the sample mean
- [tex]\(\mu = 8\)[/tex] is the population mean
- [tex]\(s = 0.2\)[/tex] is the sample standard deviation
- [tex]\(n = 13\)[/tex] is the sample size

Plugging in the values, we get:
[tex]\[ t = \frac{7.86 - 8}{\frac{0.2}{\sqrt{13}}} \approx -2.524 \][/tex]
(The test statistic value is approximately -2.524 after rounding to three decimal places.)

(d) Find the p-value

For a two-tailed test, the p-value is calculated as:
[tex]\[ \text{p-value} = 2 \times P(T > |t|) \][/tex]
where [tex]\( T \)[/tex] follows a t-distribution with 12 degrees of freedom.

Given:
[tex]\[ t = -2.524 \][/tex]

Looking up the cumulative distribution function (CDF) for the t-distribution at [tex]\( |t| = 2.524 \)[/tex] with 12 degrees of freedom, we find the CDF value. The p-value is then:
[tex]\[ \text{p-value} = 2 \times (1 - F(|t|)) \][/tex]
[tex]\[ \text{p-value} \approx 0.027 \][/tex]
(Rounding the p-value to three decimal places, we get approximately 0.027.)

Decision:

Using the significance level [tex]\(\alpha = 0.10\)[/tex], compare the p-value to [tex]\(\alpha\)[/tex]:

- If the p-value [tex]\(\leq \alpha\)[/tex], reject the null hypothesis.
- If the p-value [tex]\(> \alpha\)[/tex], do not reject the null hypothesis.

Here, the p-value (0.027) is less than the significance level (0.10), so we reject the null hypothesis.

Conclusion:

There is sufficient evidence at the 0.10 significance level to conclude that the population mean amount of coffee dispensed by the machine differs from 8 fluid ounces.

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