A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean of the population. Use the P-Value Approach.

x-bar=40.9, s=6.5, n=15, Ho:μ=32.6, Ha:μ≠32.6, α=0.05

Question 8 options:

P-Value = 0.2200, Since P-Value is greater than Alpha, we fail to reject H0 in favor of Ha


P-Value = 0.0495, Since P-Value is less or equal to Alpha, we  reject H0 in favor of Ha


P-Value = 0.00022, Since P-Value is less than Alpha, we reject H0 in favor of Ha


P-Value = 4.95, Since P-Value is greater than Alpha, we fail to reject H0 in favor of Ha



We are conducting a one-sample **t-test** for the population mean using the sample data provided:

- Sample mean (\(\bar{x}\)) = 40.9
- Sample standard deviation (\(s\)) = 6.5
- Sample size (\(n\)) = 15
- Null hypothesis (\(H_0\)): \(\mu = 32.6\)
- Alternative hypothesis (\(H_a\)): \(\mu \neq 32.6\)
- Significance level (\(\alpha\)) = 0.05

### Steps to follow:

1. **State the hypotheses**:
   - \( H_0: \mu = 32.6 \)
   - \( H_a: \mu \neq 32.6 \)

2. **Test statistic**:
   We calculate the t-statistic using the formula:
   \[
   t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}
   \]
   Where:
   - \(\bar{x}\) = 40.9 (sample mean)
   - \(\mu_0\) = 32.6 (population mean under the null hypothesis)
   - \(s\) = 6.5 (sample standard deviation)
   - \(n\) = 15 (sample size)

   Substituting the values:
   \[
   t = \frac{40.9 - 32.6}{\frac{6.5}{\sqrt{15}}}
   \]

3. **Degrees of freedom**:
   The degrees of freedom for the t-test is:
   \[
   df = n - 1 = 15 - 1 = 14
   \]

4. **P-value**:
   Using the t-distribution, we'll find the P-value based on the calculated t-statistic.

Let's calculate the t-statistic and P-value.The calculated t-statistic is approximately **4.95**, and the P-value is **0.00022**.

### Conclusion:
Since the P-value (0.00022) is less than the significance level (\(\alpha = 0.05\)), we **reject the null hypothesis** \(H_0\) in favor of the alternative hypothesis \(H_a\).

Thus, the correct answer is:
- **P-Value = 0.00022, Since P-Value is less than Alpha, we reject \(H_0\) in favor of \(H_a\).**

Would you like more details or clarification?

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Here are 5 related questions for further exploration:
1. How is a one-tailed t-test different from a two-tailed t-test?
2. How does increasing the sample size affect the P-value?
3. What is the relationship between the t-statistic and the P-value?
4. What assumptions must hold true for a one-sample t-test?
5. How would a change in significance level (\(\alpha\)) affect the hypothesis test outcome?

**Tip:** In hypothesis testing, always check if the sample size is large enough for a t-test to be valid under the assumption of normality.

A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean of the population. Use the P-Value Approach.

x-bar=40.9, s=6.5, n=15, Ho:μ=32.6, Ha:μ≠32.6, α=0.05

Question 8 options:

P-Value = 0.2200, Since P-Value is greater than Alpha, we fail to reject H0 in favor of Ha

P-Value = 0.0495, Since P-Value is less or equal to Alpha, we reject H0 in favor of Ha

P-Value = 0.00022, Since P-Value is less than Alpha, we reject H0 in favor of Ha

P-Value = 4.95, Since P-Value is greater than Alpha, we fail to reject H0 in favor of Ha

Use the one-mean t-test to determine if the population mean is different from the sample mean of 40.9, with a population mean hypothesis of 32.6 and a sample size of 15. This example includes a step-by-step guide using the P-Value approach and significance level α = 0.05.

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