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Topic 5 Homework (Nonadaptive)
Question 12 of 21 (1 point)|Question Attempt: 1 of Unlimited

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Question 12
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 14 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 8.04 fluid ounces and 0.12 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.
▼(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

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This problem involves hypothesis testing, specifically a two-tailed test for a population mean when the population standard deviation is unknown. Since the sample size is 14 (small), and we assume the data follows a normal distribution, we'll use a **t-test** for this hypothesis.

### Step-by-Step Solution:

#### Given:
- Sample mean (\(\bar{x}\)) = 8.04 fluid ounces
- Hypothesized population mean (\(\mu_0\)) = 8 fluid ounces
- Sample standard deviation (\(s\)) = 0.12 fluid ounces
- Sample size (\(n\)) = 14
- Significance level (\(\alpha\)) = 0.10

#### (a) Null and Alternative Hypothesis
We are performing a two-tailed test because we are checking whether the population mean differs (either higher or lower) from 8 fluid ounces.

- **Null Hypothesis (H₀):** \(\mu = 8\) fluid ounces (the population mean discharge is 8 oz).
- **Alternative Hypothesis (H₁):** \(\mu \neq 8\) fluid ounces (the population mean discharge is not 8 oz).

#### (b) Type of Test Statistic
Since the population standard deviation is unknown, and we are working with a small sample size (n = 14), we use the **t-test** statistic.

#### (c) Test Statistic Formula
The t-statistic is calculated as:

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

Where:
- \(\bar{x} = 8.04\)
- \(\mu_0 = 8\)
- \(s = 0.12\)
- \(n = 14\)

Now, let's calculate the test statistic:

\[
t = \frac{8.04 - 8}{\frac{0.12}{\sqrt{14}}}
\]

\[
t = \frac{0.04}{\frac{0.12}{3.7417}} = \frac{0.04}{0.0321} \approx 1.246
\]

So, the value of the test statistic is approximately **1.246**.

#### (d) Finding the p-value
For a two-tailed test, we need to find the p-value for \(t = 1.246\) with \(n - 1 = 13\) degrees of freedom. Using a t-distribution table or software, we can look up the p-value.

Looking up the t-value or using a calculator, we get a p-value of approximately **0.234**.

#### (e) Conclusion
Since the p-value (0.234) is **greater than** the significance level (\(\alpha = 0.10\)), we **fail to reject** the null hypothesis.

Thus, there is not enough evidence to conclude that the mean discharge differs from 8 fluid ounces.

### Final Answer:
- **(a) H₀: \(\mu = 8\), H₁: \(\mu \neq 8\)**
- **(b) t-test**
- **(c) Test statistic: \(t = 1.246\)**
- **(d) p-value: 0.234**
- **(e) Conclusion: No, we cannot conclude that the mean discharge differs from 8 fluid ounces.**

Would you like any further details or clarifications?

---

Here are some additional questions to explore:
1. How would the conclusion change if the significance level was 0.05?
2. What assumptions are made when using the t-test?
3. How would the test change if the sample size were larger?
4. What is the effect of a larger standard deviation on the result?
5. How do you calculate the confidence interval for the mean discharge?

**Tip:** The p-value helps you assess the strength of the evidence against the null hypothesis. If it's smaller than the significance level, reject the null hypothesis.

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 14 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 8.04 fluid ounces and 0.12 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.

Explore how to perform a two-tailed t-test on a coffee machine's discharge using sample data (mean = 8.04, standard deviation = 0.12) to test if the population mean differs from 8 ounces. Follow step-by-step computations, using a 0.10 significance level and statistical concepts such as hypothesis testing and the t-distribution.

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