Questions: In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6988 subjects randomly selected from an online group involved with ears. There were 1303 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. H1: p<0.2 H0: p=0.2 H1: p>0.2 The test statistic is z= . (Round to two decimal places as needed.) The P -value is . (Round to three decimal places as needed.) Because the P-value is the significance level, the null hypothesis. There is evidence to support the claim that the return rate is less than 20%.

Solution Steps

We are testing the claim that the return rate of surveys is less than \(20\%\). The hypotheses are formulated as follows:

• Null Hypothesis: \(H_0: p = 0.2\)
• Alternative Hypothesis: \(H_1: p < 0.2\)

The test statistic for the proportion is calculated using the formula:

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

Where:

• \(\hat{p} = \frac{x}{n} = \frac{1303}{6988} \approx 0.1863\)
• \(p_0 = 0.2\)
• \(n = 6988\)

Substituting the values, we have:

\[ Z = \frac{0.1863 - 0.2}{\sqrt{\frac{0.2(1 - 0.2)}{6988}}} \approx -2.8291 \]

Thus, the test statistic is:

\[ Z \approx -2.83 \]

The P-value associated with the test statistic \(Z = -2.83\) is calculated to be:

\[ \text{P-value} \approx 0.0023 \]

We compare the P-value with the significance level \(\alpha = 0.01\):

• Since \(0.0023 < 0.01\), we reject the null hypothesis.

There is sufficient evidence to support the claim that the return rate is less than \(20\%\).

Final Answer