Question 1165670
```python?code_reference&code_event_index=2
import numpy as np
from scipy.stats import norm, chi2_contingency

# --- Part (a) Calculations: Standard Error and 95% CI ---
n_a = 250
x_a = 100
p_hat = x_a / n_a

# 1. Standard Error (SE)
se_p_hat = np.sqrt(p_hat * (1 - p_hat) / n_a)

# 2. Z-score for 95% CI
# Z = 1.96 for 95% confidence
Z = 1.96

# 3. Margin of Error (ME)
me = Z * se_p_hat

# 4. Confidence Interval Bounds
ci_lower = p_hat - me
ci_upper = p_hat + me

print(f"Sample Proportion (p_hat): {p_hat}")
print(f"Standard Error (SE): {se_p_hat}")
print(f"Margin of Error (ME): {me}")
print(f"95% CI Lower Bound: {ci_lower}")
print(f"95% CI Upper Bound: {ci_upper}")

# --- Part (b) Calculations: Chi-square Test of Independence ---
# Observed data (O)
# Rows: No HIV, Yes HIV
# Cols: No STDs, Yes STDs
observed = np.array([[92, 58],
                     [70, 280]])

# Perform Chi-square test of independence
chi2_stat, p_value, dof, expected = chi2_contingency(observed)

print("\nChi-square Test Results:")
print(f"Chi-square statistic: {chi2_stat}")
print(f"P-value: {p_value}")
print(f"Degrees of Freedom (dof): {dof}")
print(f"Expected Frequencies:\n{expected}")

```

```text?code_stdout&code_event_index=2
Sample Proportion (p_hat): 0.4
Standard Error (SE): 0.030983866769659335
Margin of Error (ME): 0.060728378868532294
95% CI Lower Bound: 0.33927162113146775
95% CI Upper Bound: 0.4607283788685323

Chi-square Test Results:
Chi-square statistic: 80.02645502645503
P-value: 3.694302007277756e-19
Degrees of Freedom (dof): 1
Expected Frequencies:
[[ 48.6 101.4]
 [113.4 236.6]]

```

### (a) Standard Error and Confidence Bounds

This calculation pertains to the population proportion ($\hat{p}$) of adults who drink alcohol daily.

**Given:**

  * Sample size ($n$): 250
  * Number who drink daily ($x$): 100
  * Sample proportion ($\hat{p}$): $\hat{p} = \frac{100}{250} = \mathbf{0.40}$

#### 1\. Standard Error (SE)

The standard error of the sample proportion measures the typical distance that the sample proportion ($\hat{p}$) falls from the true population proportion.

$$SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.40 \times 0.60}{250}} = \sqrt{0.00096}$$

$$\mathbf{SE_{\hat{p}} \approx 0.0310}$$

#### 2\. Error Bounds at 95% Confidence Level

The error bounds (Margin of Error, ME) are calculated using the Z-score for 95% confidence, which is $Z_{0.025} = 1.96$.

$$\text{ME} = Z \times SE_{\hat{p}} = 1.96 \times 0.03098$$

$$\mathbf{\text{ME} \approx 0.0607}$$

The 95% confidence interval (CI) is $\hat{p} \pm ME$:
$$\text{CI} = 0.40 \pm 0.0607$$

$$\text{Lower Bound} \approx \mathbf{0.3393}$$
$$\text{Upper Bound} \approx \mathbf{0.4607}$$

**Interpretation:** We are 95% confident that the true percentage of the adult population that drinks alcohol daily is between **33.93% and 46.07%**.

-----

### (b) Chi-square Test of Independence

We use the Chi-square ($\chi^2$) test of independence to determine if there is a statistically significant relationship between HIV status and history of STDs.

**Observed Frequencies ($O$):**

| | No STDs | Yes STDs | Total |
| :---: | :---: | :---: | :---: |
| **No HIV** | 92 | 58 | 150 |
| **Yes HIV** | 70 | 280 | 350 |
| **Total** | 162 | 338 | 500 |

#### 1\. Hypotheses

  * $H_0$: HIV status and History of STDs are **independent** (no relationship).
  * $H_a$: HIV status and History of STDs are **dependent** (there is a relationship).
  * Significance level ($\alpha$): 0.05

#### 2\. Expected Frequencies ($E$) and Test Statistic

The test compares the observed frequencies ($O$) to the expected frequencies ($E$), which are calculated assuming the null hypothesis of independence is true.

**Expected Frequencies:**

| | No STDs | Yes STDs |
| :---: | :---: | :---: |
| **No HIV** | 48.6 | 101.4 |
| **Yes HIV** | 113.4 | 236.6 |

The **Chi-square test statistic ($\chi^2$)** is calculated based on the differences between observed and expected values:
$$\chi^2 = \sum \frac{(O - E)^2}{E}$$

The calculated $\chi^2$ statistic is:
$$\mathbf{\chi^2 \approx 80.026}$$

#### 3\. Degrees of Freedom (dof) and P-value

  * **Degrees of Freedom (dof):** $\text{dof} = (\text{Rows} - 1)(\text{Columns} - 1) = (2 - 1)(2 - 1) = \mathbf{1}$
  * **P-value:** The probability of observing a $\chi^2$ value as extreme as $80.026$ (or more extreme) under the null hypothesis is:
    $$\mathbf{p\text{-value} \approx 3.69 \times 10^{-19}}$$

#### 4\. Conclusion

We compare the p-value to the significance level $\alpha$:
$$\text{P-value } (3.69 \times 10^{-19}) \ll \alpha (0.05)$$

Since the p-value is extremely small (far less than $0.05$), we **reject the null hypothesis ($H_0$)**.

**Conclusion:**
At the $5\%$ significance level, there is **strong evidence of a statistically significant relationship** between HIV status and history of STDs. The two variables are dependent.