A simple random sample of size nequals15 is drawn from a population that is normally distributed. The sample mean is found to be x overbarequals22.2 and the sample standard deviation is found to be sequals6.3. Determine if the population mean is different from 25 at the alpha equals 0.01 level of significance. Complete parts ​(a) through ​(d) below.
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Part 1
​(a) Determine the null and alternative hypotheses.
Upper H 0​:
mu
equals
25
Upper H 1​:
mu
not equals
25
Part 2
​(b) Calculate the​ P-value.
​P-valueequals

0.107
​(Round to three decimal places as​ needed.)
Part 3
​(c) State the conclusion for the test.
A.
Reject Upper H 0 because the​ P-value is greater than the alphaequals0.01 level of significance.
B.
Do not reject Upper H 0 because the​ P-value is less than the alphaequals0.01 level of significance.
C.
Do not reject Upper H 0 because the​ P-value is greater than the alphaequals0.01 level of significance.
D.
Reject Upper H 0 because the​ P-value is less than the alphaequals0.01 level of significance.

To address the problem, we'll follow the standard procedure for hypothesis testing. Let's go through each part step by step.

### Part 1: State the Hypotheses

The null hypothesis (H₀) and the alternative hypothesis (H₁) are given as:

- **Null Hypothesis (H₀)**: \(\mu = 25\)
- **Alternative Hypothesis (H₁)**: \(\mu \neq 25\)

### Part 2: Calculate the P-value

We need to perform a t-test to determine if the sample mean is significantly different from the population mean. The formula for the test statistic (t) is:

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

Where:
- \(\bar{x} = 22.2\) (sample mean)
- \(\mu = 25\) (population mean)
- \(s = 6.3\) (sample standard deviation)
- \(n = 15\) (sample size)

First, we calculate the test statistic \( t \):

\[ t = \frac{22.2 - 25}{6.3 / \sqrt{15}} \]

Calculate the denominator:

\[ \frac{6.3}{\sqrt{15}} = \frac{6.3}{3.87298} \approx 1.6266 \]

Then the test statistic:

\[ t = \frac{22.2 - 25}{1.6266} \approx \frac{-2.8}{1.6266} \approx -1.721 \]

Next, we find the degrees of freedom (df):

\[ df = n - 1 = 15 - 1 = 14 \]

Using a t-distribution table or a calculator for a two-tailed test, we find the p-value for \( t = -1.721 \) with 14 degrees of freedom.

According to the information given in the question, the p-value is approximately 0.107.

### Part 3: State the Conclusion

We compare the p-value with the significance level (\(\alpha\)) to make a decision. Given:

- \(\alpha = 0.01\)
- \( \text{p-value} = 0.107 \)

Since the p-value (0.107) is greater than the significance level (0.01), we do not reject the null hypothesis.

Therefore, the correct conclusion is:

C. **Do not reject \( H_0 \) because the p-value is greater than the \(\alpha = 0.01\) level of significance.**

### Summary
- **Hypotheses**: \( H_0: \mu = 25 \), \( H_1: \mu \neq 25 \)
- **P-value**: 0.107
- **Conclusion**: Do not reject \( H_0 \)

Would you like more details or have any questions?

Here are five questions you can ask next:
1. How is the t-distribution different from the normal distribution?
2. What are the assumptions of the t-test?
3. How can we interpret the p-value in hypothesis testing?
4. Can we use this method if the population is not normally distributed?
5. How does sample size affect the t-test?

**Tip:** When conducting hypothesis tests, always ensure your sample size and data meet the necessary assumptions to validate the results.

Explore a hypothesis testing example where a sample mean from a normally distributed population is tested against a population mean of 25. Discover the calculation of the t-statistic, determination of the p-value, and conclusion based on alpha = 0.01 significance level.

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