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 7.92 fluid ounces and 0.13 fold 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-talled 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 Ho and the alternative hypothesis f

H0

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

(Choose one)▼

(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?

Yes No

To address this problem, we'll perform a hypothesis test step-by-step. Here’s the process:

### (a) State the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \):

- **Null Hypothesis** \( H_0 \): The population mean discharge is equal to 8 fluid ounces.
  \[
  H_0: \mu = 8
  \]
- **Alternative Hypothesis** \( H_a \): The population mean discharge is different from 8 fluid ounces.
  \[
  H_a: \mu \neq 8
  \]

This is a two-tailed test because we're interested in whether the mean is different from 8 fluid ounces (either higher or lower).

### (b) Determine the type of test statistic to use:

Since the sample size is small (n = 15) and the population standard deviation is unknown, we use the **t-test** statistic.

### (c) Find the value of the test statistic:

The test statistic is calculated using the formula:

\[
t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
\]

Where:
- \(\bar{x}\) = sample mean = 7.92 fluid ounces
- \(\mu_0\) = hypothesized population mean = 8 fluid ounces
- \(s\) = sample standard deviation = 0.13 fluid ounces
- \(n\) = sample size = 15

Substituting the values:

\[
t = \frac{7.92 - 8}{\frac{0.13}{\sqrt{15}}} \approx \frac{-0.08}{\frac{0.13}{3.872}} \approx \frac{-0.08}{0.0336} \approx -2.381
\]

So, the test statistic \( t \approx -2.381 \).

### (d) Find the p-value:

The degrees of freedom (df) for this t-test is \( n - 1 = 15 - 1 = 14 \).

Using a t-distribution table or calculator, we find the p-value for \( t \approx -2.381 \) with 14 degrees of freedom in a two-tailed test.

The p-value is approximately \( 0.032 \).

### (e) Can we conclude that the mean discharge differs from 8 fluid ounces?

To determine this, compare the p-value to the significance level \(\alpha = 0.10\):

- If \( p \leq \alpha \), we reject the null hypothesis.
- If \( p > \alpha \), we fail to reject the null hypothesis.

In this case, the p-value (0.032) is less than the significance level (0.10). Therefore, **we reject the null hypothesis** and conclude that there is enough evidence to suggest that the mean discharge differs from 8 fluid ounces.

### Summary:
- **Null Hypothesis \( H_0 \):** \(\mu = 8\)
- **Alternative Hypothesis \( H_a \):** \(\mu \neq 8\)
- **Test Statistic:** \( t \approx -2.381 \)
- **p-value:** \( \approx 0.032 \)
- **Conclusion:** Yes, there is enough evidence to conclude that the mean discharge differs from 8 fluid ounces.

Would you like more details on any part of the solution or have any questions?

Here are 5 related questions to consider:
1. How does the choice of significance level affect the conclusion of a hypothesis test?
2. What would the conclusion be if the p-value were greater than 0.10?
3. How would the test result change if the sample mean were exactly 8.0 fluid ounces?
4. Why is the t-test used instead of a z-test in this scenario?
5. What assumptions are necessary for the validity of this t-test?

**Tip:** Always verify that the sample data meets the assumptions for the test you are using to ensure accurate results.

Learn how to perform a hypothesis test to determine if the mean discharge from a coffee machine differs from 8 fluid ounces. Detailed step-by-step solution using a t-test statistic with sample mean and standard deviation provided.

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