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The problem presented in the image involves calculating a confidence interval for the mean expenditure of families, given a sample size larger than 30 (n > 30). The formula for the confidence interval is provided, which assumes the population variance is unknown but the sample size is large enough to apply the Central Limit Theorem.

### **Step-by-Step Solution:**

#### **1. Given Data:**
- **Sample size (n):** 300 families
- **Sample data (expenditure in hundreds of thousands of Rupiah):** 

  20, 17, 25, 19, 14, 15, 13, 16, 24, 21, 26, 25, 24, 23, 20, 18, 15, 22, 24, 19, 25, 22, 26, 23, 24, 20, 18, 23, 25, 24

#### **2. Calculate the sample mean (\(\bar{x}\)):**

The sample mean is calculated by summing all the data points and dividing by the number of data points.

\[
\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}
\]

Using the data provided:

\[
\bar{x} = \frac{20 + 17 + 25 + 19 + 14 + 15 + 13 + 16 + 24 + 21 + 26 + 25 + 24 + 23 + 20 + 18 + 15 + 22 + 24 + 19 + 25 + 22 + 26 + 23 + 24 + 20 + 18 + 23 + 25 + 24}{30}
\]

#### **3. Calculate the standard deviation (s):**

The standard deviation is calculated using the formula:

\[
s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n - 1}}
\]

Where \(x_i\) are the data points and \(\bar{x}\) is the sample mean.

#### **4. Determine the Z-value for 95% Confidence Interval:**

For a 95% confidence interval, the critical value \(Z_{\alpha/2}\) is approximately 1.96.

#### **5. Apply the confidence interval formula:**

\[
\bar{x} - Z_{\alpha/2} \times \frac{s}{\sqrt{n}} \leq \mu \leq \bar{x} + Z_{\alpha/2} \times \frac{s}{\sqrt{n}}
\]

Where:
- \(\bar{x}\) is the sample mean
- \(Z_{\alpha/2}\) is the Z-value corresponding to the desired confidence level (1.96 for 95%)
- \(s\) is the sample standard deviation
- \(n\) is the sample size

#### **6. Final Calculation:**

Substitute the values obtained for \(\bar{x}\), \(s\), and \(n\) into the formula to calculate the lower and upper bounds of the confidence interval.

### **Explanation and Summary:**

- **Sample Mean (\(\bar{x}\)):** This gives us the average expenditure of the sampled families.
- **Standard Deviation (s):** This measures the variability in the expenditure among the families.
- **95% Confidence Interval:** This interval provides a range within which we can be 95% confident that the true population mean expenditure lies.

### **Questions for Further Practice:**

1. How would the confidence interval change if the sample size were smaller than 30?
2. What impact does increasing the confidence level (e.g., 99%) have on the confidence interval?
3. If the standard deviation was known for the entire population, how would the formula change?
4. Can you calculate a 90% confidence interval for the same data set?
5. What assumptions are we making when using the Z-distribution for this confidence interval?

**Tip:** Always check if the sample size is large enough to use the Z-distribution or if you should use the t-distribution instead, especially when dealing with smaller samples (n < 30).

Diketahui ada 600 kepala keluarga sebagai populasi yang membeli pakaian. Hasil penelitian terhadap 30 kepala keluarga ternyata jumlah pengeluaran untuk pakaian yang dibeli (ratusan ribuan rupiah) sebagai berikut: 20, 17, 25, 19, 14, 15, 13, 16, 24, 21, 26, 25, 24, 23, 20, 18, 15, 22, 24, 19, 25, 22, 26, 23, 24, 20, 18, 23, 25, 24. Hitunglah nilai estimasi selang 95% dari data tersebut.

Calculate the 95% confidence interval for the mean expenditure of families based on sample data. This problem involves applying statistical concepts like the sample mean, standard deviation, and the Central Limit Theorem.

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