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answer this and refer to the table for the guide

Let's go through each step to solve this **one-sample t-test** using the data provided in the image.

### Information from the problem:
- Hypothesized population mean (\(\mu\)): 175 cm
- Sample mean (\(\bar{x}\)): 177 cm
- Sample size (\(n\)): 30
- Sample standard deviation (\(s\)): 5 cm
- Significance level: \(\alpha = 0.05\) (given significance level corresponds to the t-critical values \(-2.0452\) to \(2.0452\) which matches \(\alpha = 0.05\))
  
### Formula:
The test statistic for a one-sample t-test is calculated as:

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

### Steps:

#### 1. **Define Null and Alternate Hypotheses:**
   - **Null Hypothesis (H₀):** \(\mu = 175\) (The mean height of the population is 175 cm).
   - **Alternative Hypothesis (H₁):** \(\mu \neq 175\) (The mean height of the population is not 175 cm). This is a two-tailed test.

#### 2. **State the Significance Level (Alpha Level):**
   The significance level is \(\alpha = 0.05\). This means that we are willing to accept a 5% chance of rejecting the null hypothesis when it is actually true (Type I error).

#### 3. **Calculate the Degrees of Freedom:**
   The degrees of freedom for a one-sample t-test are given by:
   \[
   \text{df} = n - 1 = 30 - 1 = 29
   \]

#### 4. **State the Decision Rule:**
   From the t-table provided, for a two-tailed test with \(df = 29\) and \(\alpha = 0.05\), the critical t-value is approximately **2.045** (from the table). Therefore:
   - If \( t \) falls between \(-2.045\) and \(2.045\), **do not reject** the null hypothesis.
   - If \( t \) is less than \(-2.045\) or greater than \(2.045\), **reject** the null hypothesis.

#### 5. **Calculate the Test Statistic:**
   Using the formula for the test statistic:

   \[
   t = \frac{\bar{x} - \mu}{s / \sqrt{n}} = \frac{177 - 175}{5 / \sqrt{30}} = \frac{2}{5 / 5.477} = \frac{2}{0.9129} \approx 2.19
   \]

#### 6. **State the Result and Conclusion:**
   The calculated t-statistic is **2.19**.

   - Compare this with the critical value \(\pm 2.045\) (for \(\alpha = 0.05\), two-tailed).
   - Since \(2.19 > 2.045\), the t-statistic falls **outside** the critical range.

Thus, we **reject the null hypothesis**. This means that there is sufficient evidence to suggest that the mean height of adult males in the population is significantly different from 175 cm at the 5% significance level.

---

Would you like further details on any of these steps, or do you have any questions?

### Relative Questions:
1. What is the interpretation of rejecting or failing to reject the null hypothesis in a t-test?
2. How does the choice of alpha level affect the decision in hypothesis testing?
3. What are the assumptions of a one-sample t-test, and how do they influence the results?
4. How would the results change if the sample size were larger or smaller?
5. What is the importance of the degrees of freedom in determining the critical t-value?

### Tip:
Always check whether the assumptions of normality and sample randomness are met before performing a t-test, especially for smaller samples.

You're a researcher investigating the mean height of adult males in a certain population. The hypothesized mean height is 175 cm. A sample of 30 adult males is collected, and the sample mean is 177 cm, with a sample standard deviation of 5 cm. Is the mean height of this population significantly different from 175 cm? The significance level is 0.05.

Learn how to perform a one-sample t-test with a real-world example. In this problem, we investigate whether the mean height of a population differs from a hypothesized value of 175 cm, using a sample of 30 adult males and a significance level of 0.05. The problem covers key concepts in hypothesis testing, including calculating degrees of freedom and test statistics.

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