Questions: Claim: Fewer than 11.7% of homes have only a landline telephone and no wireless phone. Sample data: A survey by the National Center for Health Statistics showed that among 14,941 homes 5.79% had landline phones without wireless phones. Complete parts (a) and (b). a. Express the original claim in symbolic form. Let the parameter represent a value with respect to homes that have only a landline telephone and no wireless phone.

Claim: Fewer than 11.7% of homes have only a landline telephone and no wireless phone. Sample data: A survey by the National Center for Health Statistics showed that among 14,941 homes 5.79% had landline phones without wireless phones. Complete parts (a) and (b).
a. Express the original claim in symbolic form. Let the parameter represent a value with respect to homes that have only a landline telephone and no wireless phone.

Transcript text: Claim: Fewer than $11.7 \%$ of homes have only a landline telephone and no wireless phone. Sample data: A survey by the National Center for Health Statistics showed that among 14,941 homes $5.79 \%$ had landline phones without wireless phones. Complete parts (a) and (b). a. Express the original claim in symbolic form. Let the parameter represent a value with respect to homes that have only a landline telephone and no wireless phone.

Solution

Solution Steps

Step 1: State the Hypotheses

We are testing the claim that fewer than $11.7\%$ of homes have only a landline telephone and no wireless phone. We can express this in terms of hypotheses:

Null Hypothesis ($H_0$): $p \geq 0.117$
Alternative Hypothesis ($H_a$): $p < 0.117$

Step 2: Calculate the Test Statistic

The test statistic for the hypothesis test for a population proportion is calculated using the formula:

\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

Substituting the values:

$\hat{p} = 0.0579$
$p_0 = 0.117$
$n = 14941$

We find:

\[ Z = \frac{0.0579 - 0.117}{\sqrt{\frac{0.117(1 - 0.117)}{14941}}} = -22.4752 \]

Step 3: Determine the P-value

The P-value associated with the calculated test statistic $Z = -22.4752$ is $0.0$. This indicates that the probability of observing a sample proportion as extreme as $0.0579$ or lower, given that the null hypothesis is true, is extremely low.

Step 4: Compare with the Critical Value

For a significance level of $\alpha = 0.05$ in a one-tailed test, the critical value for $Z$ is approximately $-1.6449$. Since our calculated test statistic $Z = -22.4752$ is less than $-1.6449$, we fall into the critical region.

Step 5: Conclusion

Since the P-value $0.0$ is less than $\alpha = 0.05$ and the test statistic falls into the critical region, we reject the null hypothesis.