Questions: In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6964 subjects randomly selected from an online group involved with ears. There were 1299 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than 20%. Use the P-value method and use the normal distribution as an approximation to the binomial distribution. H0: p=0.2 H1: p<0.2 The test statistic is z=-2.81

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6964 subjects randomly selected from an online group involved with ears. There were 1299 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than 20%. Use the P-value method and use the normal distribution as an approximation to the binomial distribution.
H0: p=0.2
H1: p<0.2

The test statistic is z=-2.81

Transcript text: In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6964 subjects randomly selected from an online group involved with ears. There were 1299 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than $20 \%$. Use the P -value method and use the normal distribution as an approximation to the binomial distribution. $H_{0}: p=0.2$ $\mathrm{H}_{1}: \mathrm{p}<0.2$ The test statistic is $\mathrm{z}=-2.81$

Solution

Solution Steps

Step 1: Define Hypotheses

We are testing the claim that the return rate is less than $20\%$. Thus, we set up our hypotheses as follows:

Null Hypothesis: $H_0: p = 0.2$
Alternative Hypothesis: $H_1: p < 0.2$

Step 2: Calculate Sample Proportion

The sample proportion $\hat{p}$ is calculated as: \[ \hat{p} = \frac{x}{n} = \frac{1299}{6964} \approx 0.1865 \]

Step 3: Calculate Test Statistic

The test statistic $Z$ is calculated using the formula: \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Substituting the values: \[ Z = \frac{0.1865 - 0.2}{\sqrt{\frac{0.2(1 - 0.2)}{6964}}} \approx -2.81 \]

Step 4: Calculate P-value

The P-value associated with the test statistic $Z = -2.81$ is found to be: \[ \text{P-value} \approx 0.0025 \]

Step 5: Decision Rule

At a significance level of $\alpha = 0.01$, we compare the P-value to $\alpha$: \[ 0.0025 < 0.01 \] Since the P-value is less than $\alpha$, we reject the null hypothesis $H_0$.