2. Estimate with $$ 80 \% $$ confidence, the percentage of drivers use hand-held cell phones while driving. In a study of distracting driving, out of 1050 randomly selected drivers, 28 were observed driving while holding their cell phones. a. Define the random variable, $$ X $$. Classify it as categorical or numerical. $$ X= $$ Number of drivert use hand-held cell (Circle) fhones while driving Categorical Numerical b. What is the value of the critical value, $$ z_{\alpha / 2} $$, for the required confidence interval. Illustrate your answer. $$ \frac{1-0.8}{2}=0.1-1.28 $$ c. Compute the confidence interval. (4 decimal places.) $$ \begin{array}{l} \frac{28}{1050} \pm 1.28 \sqrt{\frac{28}{1050}\left(1-\frac{28}{1050}\right)}(0.0329) \ 0.0266 \pm 1.28(0.0049658) \ 0.0266 \pm 0.0063562 \end{array} $$ d. Interpret your confidence interval following the class guidelines. $$ 3.2 c_{1} \cdot 1 \text { and } 2.02 .1 $$ e. State the conditions for the confidence interval that you computed. Explair how the conditions were satisfied.

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### a. Define the random variable, $$ X $$. Classify it as categorical or numerical.

The random variable $$ X $$ represents the number of drivers who use hand-held cell phones while driving.

- **Classification**: $$ X $$ is a **numerical** variable because it counts the number of occurrences (drivers using hand-held cell phones).

### b. What is the value of the critical value, $$ z_{\alpha / 2} $$, for the required confidence interval?

To find the critical value $$ z_{\alpha / 2} $$ for an 80% confidence interval, we first calculate $$ \alpha $$:

$$
\alpha = 1 - 0.8 = 0.2
$$

Then, we find $$ \frac{\alpha}{2} $$:

$$
\frac{\alpha}{2} = \frac{0.2}{2} = 0.1
$$

Now, we look for the critical value $$ z_{0.1} $$. The critical value for $$ z_{0.1} $$ corresponds to the 90th percentile of the standard normal distribution, which is approximately:

$$
z_{0.1} \approx 1.28
$$

### c. Compute the confidence interval. (4 decimal places.)

1. Calculate the sample proportion $$ \hat{p} $$:

$$
\hat{p} = \frac{X}{n} = \frac{28}{1050} \approx 0.0266667
$$

2. Calculate the standard error (SE):

$$
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.0266667(1 - 0.0266667)}{1050}} \approx 0.0049658
$$

3. Compute the margin of error (ME):

$$
ME = z_{\alpha/2} \cdot SE = 1.28 \cdot 0.0049658 \approx 0.0063562
$$

4. Finally, compute the confidence interval:

$$
\hat{p} \pm ME = 0.0266667 \pm 0.0063562
$$

Calculating the lower and upper bounds:

- Lower bound: $$ 0.0266667 - 0.0063562 \approx 0.0203105 $$
- Upper bound: $$ 0.0266667 + 0.0063562 \approx 0.032973 $$

Thus, the confidence interval is:

$$
(0.0203, 0.0329)
$$

### d. Interpret your confidence interval following the class guidelines.

The interpretation of the confidence interval $$ (0.0203, 0.0329) $$ is as follows:

We are 80% confident that the true proportion of drivers who use hand-held cell phones while driving is between 2.03% and 3.29%.

### e. State the conditions for the confidence interval that you computed. Explain how the conditions were satisfied.

The conditions for constructing a confidence interval for a proportion are:

1. **Random Sampling**: The sample of drivers was randomly selected, which helps ensure that the sample is representative of the population.

2. **Normality Condition**: The sample size should be large enough for the sampling distribution of the proportion to be approximately normal. This is typically satisfied if both $$ np $$ and $$ n(1-p) $$ are greater than 5.

- Here, $$ n = 1050 $$ and $$ p = \frac{28}{1050} \approx 0.0266667 $$.
   - Calculate $$ np $$ and $$ n(1-p) $$:
     - $$ np = 1050 \cdot 0.0266667 \approx 28 $$ (satisfied)
     - $$ n(1-p) = 1050 \cdot (1 - 0.0266667) \approx 1022 $$ (satisfied)

Since both conditions are satisfied, we can confidently use the normal approximation to construct the confidence interval.
a. $$ X $$ is the number of drivers using hand-held cell phones while driving. It is a **numerical** variable. b. The critical value $$ z_{\alpha/2} $$ for an 80% confidence interval is **1.28**. c. The 80% confidence interval for the proportion of drivers using hand-held cell phones while driving is **(0.0203, 0.0329)**. d. We are 80% confident that the true proportion of drivers who use hand-held cell phones while driving is between **2.03% and 3.29%**. e. The conditions for the confidence interval are: - **Random Sampling**: The drivers were randomly selected. - **Normality Condition**: Both $$ np $$ and $$ n(1-p) $$ are greater than 5, satisfying the normality requirement. These conditions were satisfied in the study.

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