Stat 1 Quiz 5 Show work fov full credit. Late submission is Not accepted. SP I. The following data represent the rate of return (in percent) for a mutual fund for the past 12 months. $$ \begin{array}{llllll}13.8 & 15.9 & 10.0 & 12.4 & 11.3 & 6.6 \ 9.6 & 12.4 & 10.3 & 8.7 & 14.9 & 6.7\end{array} $$ Construct a $$ 90 \% $$ confidence interval for the population variance of the rate of return.

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To construct a $$ 90\% $$ confidence interval for the population variance of the rate of return, we'll follow these steps:

1. **List the Data and Determine Sample Size:**

The given data (in percent) for the past 12 months are:
   $$
   13.8,\ 15.9,\ 10.0,\ 12.4,\ 11.3,\ 6.6,\ 9.6,\ 12.4,\ 10.3,\ 8.7,\ 14.9,\ 6.7
   $$
   - **Sample Size ($$ n $$)**: 12

2. **Calculate the Sample Mean ($$ \bar{x} $$):**

$$
   \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}
   $$
   
   $$
   \sum x_i = 13.8 + 15.9 + 10.0 + 12.4 + 11.3 + 6.6 + 9.6 + 12.4 + 10.3 + 8.7 + 14.9 + 6.7 = 123.5
   $$
   
   $$
   \bar{x} = \frac{123.5}{12} = 10.2917\ \text{(rounded to 4 decimal places)}
   $$

3. **Compute the Sample Variance ($$ s^2 $$):**

$$
   s^2 = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n - 1}
   $$
   
   First, calculate each $$ (x_i - \bar{x})^2 $$:

$$
   \begin{align*}
   (13.8 - 10.2917)^2 &= (3.5083)^2 = 12.2923 \
   (15.9 - 10.2917)^2 &= (5.6083)^2 = 31.4750 \
   (10.0 - 10.2917)^2 &= (-0.2917)^2 = 0.0852 \
   (12.4 - 10.2917)^2 &= (2.1083)^2 = 4.4393 \
   (11.3 - 10.2917)^2 &= (1.0083)^2 = 1.0168 \
   (6.6 - 10.2917)^2 &= (-3.6917)^2 = 13.6063 \
   (9.6 - 10.2917)^2 &= (-0.6917)^2 = 0.4784 \
   (12.4 - 10.2917)^2 &= (2.1083)^2 = 4.4393 \
   (10.3 - 10.2917)^2 &= (0.0083)^2 = 0.0001 \
   (8.7 - 10.2917)^2 &= (-1.5917)^2 = 2.5302 \
   (14.9 - 10.2917)^2 &= (4.6083)^2 = 21.1375 \
   (6.7 - 10.2917)^2 &= (-3.5917)^2 = 12.9291 \
   \end{align*}
   $$
   
   $$
   \sum (x_i - \bar{x})^2 = 12.2923 + 31.4750 + 0.0852 + 4.4393 + 1.0168 + 13.6063 + 0.4784 + 4.4393 + 0.0001 + 2.5302 + 21.1375 + 12.9291 = 103.4330
   $$
   
   $$
   s^2 = \frac{103.4330}{12 - 1} = \frac{103.4330}{11} \approx 9.4030
   $$

4. **Determine the Degrees of Freedom ($$ df $$):**

$$
   df = n - 1 = 12 - 1 = 11
   $$

5. **Find the Critical Chi-Squared Values ($$ \chi^2_{\alpha/2, df} $$ and $$ \chi^2_{1 - \alpha/2, df} $$):**

For a $$ 90\% $$ confidence interval, the significance level $$ \alpha = 1 - 0.90 = 0.10 $$. Thus, $$ \alpha/2 = 0.05 $$.

Using the Chi-Squared distribution table or a calculator:

- $$ \chi^2_{0.05, 11} \approx 19.675 $$
   - $$ \chi^2_{0.95, 11} \approx 3.816 $$

*(Note: These values might slightly vary based on the table or calculator used.)*

6. **Construct the $$ 90\% $$ Confidence Interval for the Population Variance ($$ \sigma^2 $$):**

The confidence interval formula for variance is:
   
   $$
   \left( \frac{(n - 1)s^2}{\chi^2_{1 - \alpha/2, df}},\ \frac{(n - 1)s^2}{\chi^2_{\alpha/2, df}} \right)
   $$
   
   Plugging in the values:

$$
   \left( \frac{11 \times 9.4030}{19.675},\ \frac{11 \times 9.4030}{3.816} \right)
   $$
   
   Calculate the lower bound:

$$
   \frac{11 \times 9.4030}{19.675} = \frac{103.433}{19.675} \approx 5.258
   $$
   
   Calculate the upper bound:

$$
   \frac{11 \times 9.4030}{3.816} = \frac{103.433}{3.816} \approx 27.103
   $$
   
   So, the $$ 90\% $$ confidence interval for the population variance is approximately:

$$
   (5.258,\ 27.103)
   $$

7. **Conclusion:**

We are $$ 90\% $$ confident that the true population variance of the rate of return for the mutual fund lies between **5.258** and **27.103** (in percentage squared).
The $$ 90\% $$ confidence interval for the population variance of the rate of return is approximately **5.258** to **27.103**.

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