Question

divorce rates the number of years that first-time married couples stay together before they divorce is shown. if the sample is random and \\(\sigma\\) is 2.1, find the 99% confidence interval for the mean. round your answers to one decimal place as needed.

\\
\

$$\begin{array}{cccccccccc} 8 & 11 & 4 & 5 & 3 & 1 & 4 & 11 & 8 & 9 \\\\ 3 & 12 & 6 & 10 & 5 & 3 & 2 & 7 & 9 & 11 \\\\ 6 & 3 & 5 & 5 & 7 & 8 & 9 & 6 & 4 & 10 \\end{array}$$

\\

\\(\square < \mu < \square\\)

Explanation:

Calculate the sample mean

First, we sum the given data values and divide by the sample size \(n\).
The dataset contains the following 30 values:
Row 1: 8, 11, 4, 5, 3, 1, 4, 11, 8, 9
Row 2: 3, 12, 6, 10, 5, 3, 2, 7, 9, 11
Row 3: 6, 3, 5, 5, 7, 8, 9, 6, 4, 10

The sample size is:

$$n = 30$$

The sum of the values is:

$$\sum x = 8+11+4+5+3+1+4+11+8+9+3+12+6+10+5+3+2+7+9+11+6+3+5+5+7+8+9+6+4+10 = 190$$

The sample mean is:

$$\bar{x} = \frac{190}{30} \approx 6.3333$$

Find the critical value

For a \(99\%\) confidence level, the significance level is \(\alpha = 0.01\).
The critical value \(z_{\alpha/2}\) corresponding to a \(99\%\) confidence level is:

$$z_{0.005} \approx 2.576$$

Calculate the margin of error

The population standard deviation is given as \(\sigma = 2.1\).
The formula for the margin of error \(E\) is:

$$E = z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$$

Substituting the values:

$$E = 2.576 \cdot \frac{2.1}{\sqrt{30}} \approx 2.576 \cdot 0.3834 \approx 0.9876$$

Determine the confidence interval

The lower and upper limits of the confidence interval are:

$$\text{Lower limit} = \bar{x} - E = 6.3333 - 0.9876 \approx 5.3457$$

$$\text{Upper limit} = \bar{x} + E = 6.3333 + 0.9876 \approx 7.3209$$

Rounding to one decimal place as requested:

$$\text{Lower limit} \approx 5.3$$

$$\text{Upper limit} \approx 7.3$$

Answer:

<blank>5.3</blank> \(< \mu <\) <blank>7.3</blank>