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using this file, answer the quiz on the image

Let's go through the quiz step by step, based on the scenario given:

1. **Null and Alternate Hypotheses**:
   - **Null Hypothesis (H₀)**: The mean height of adult males in this population is 175 cm, i.e., \( H_0: \mu = 175 \, \text{cm} \).
   - **Alternative Hypothesis (H₁)**: The mean height of adult males is not 175 cm, i.e., \( H_1: \mu \neq 175 \, \text{cm} \) (this is a two-tailed test).

2. **Significance Level**:
   - The significance level is not directly given, but typically it's \( \alpha = 0.05 \), corresponding to a 95% confidence level. From the t-critical values given (-2.0452 to 2.0452), this aligns with a two-tailed test where \( \alpha = 0.05 \).

3. **Degrees of Freedom (df)**:
   - The degrees of freedom for a one-sample t-test is calculated as:
     \[
     df = n - 1
     \]
     where \( n \) is the sample size. Since \( n = 30 \), the degrees of freedom are:
     \[
     df = 30 - 1 = 29
     \]

4. **Decision Rule**:
   - The decision rule involves comparing the calculated t-statistic to the critical values. The critical t-values at \( \alpha = 0.05 \) (two-tailed) are given as -2.0452 and 2.0452. 
     - If \( t \)-statistic < -2.0452 or \( t \)-statistic > 2.0452, reject the null hypothesis.
     - Otherwise, do not reject the null hypothesis.

5. **Calculate the Test Statistic**:
   - The t-statistic is calculated using the formula:
     \[
     t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
     \]
     where:
     - \( \bar{x} = 177 \, \text{cm} \) (sample mean),
     - \( \mu = 175 \, \text{cm} \) (hypothesized population mean),
     - \( s = 5 \, \text{cm} \) (sample standard deviation),
     - \( n = 30 \) (sample size).

     Substituting the values:
     \[
     t = \frac{177 - 175}{\frac{5}{\sqrt{30}}} = \frac{2}{\frac{5}{5.477}} = \frac{2}{0.9129} = 2.19
     \]

6. **Result and Conclusion**:
   - The calculated t-statistic is 2.19, which is greater than the critical value of 2.0452. Therefore, the test statistic falls in the rejection region.
   - **Conclusion**: Reject the null hypothesis. There is significant evidence to suggest that the mean height of the population is different from 175 cm at the 5% significance level.

---

Would you like further clarification or details?

Here are some questions to expand upon this topic:
1. What are the differences between one-sample t-tests and paired t-tests?
2. How does increasing the sample size affect the t-statistic in hypothesis testing?
3. What is the impact of using a one-tailed test versus a two-tailed test?
4. How would you calculate a confidence interval for the population mean in this scenario?
5. What assumptions must be met for a one-sample t-test to be valid?

**Tip**: Always ensure that the sample data is approximately normally distributed when using a one-sample t-test, especially for smaller sample sizes.

You're a researcher investigating the mean height of adult males in a certain population. You have a hypothesis that the mean height is 175 cm. You collect a sample of 30 adult males, with a sample mean of 177 cm and a standard deviation of 5 cm. Test if this mean differs from 175 cm, using a significance level of -2.0452 to 2.0452.

Perform a one-sample t-test to determine whether the mean height of adult males in a sample of 30 differs from the hypothesized population mean of 175 cm. Given a sample mean of 177 cm, standard deviation of 5 cm, and a significance level of 0.05.

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