In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6991 subjects randomly selected from an online group involved with ears. There were 1308 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. Identify the null hypothesis and alternative hypothesis. O A. Ho: p<0.2 O B. Ho: p=0.2 H1:p=0.2 H1:p<0.2 OD. Ho: p=0.2 H:p>0.2 Ос. Но: р> 0.2 H:p=0.2 OF. Ho: p=0.2 O E. Ho: p#0.2 H: p= 0.2 Hạ: 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 v the significance level, v the null hypothesis. There is evidence to support the claim that the return rate is less than 20%.

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In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6,991 subjects randomly selected from an online group involved with ears. There were 1,308 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.

Identify the null hypothesis and alternative hypothesis.

- Option A: \( H_0: p < 0.2 \)  
  \( H_1: p = 0.2 \)

- Option B: \( H_0: p = 0.2 \)  
  \( H_1: p < 0.2 \)

- Option C: \( H_0: p > 0.2 \)  
  \( H_1: p = 0.2 \)

- Option D: \( H_0: p = 0.2 \)  
  \( H_1: p > 0.2 \)

- Option E: \( H_0: p \neq 0.2 \)  
  \( H_1: p = 0.2 \)

- Option F: \( H_0: p = 0.2 \)  
  \( H_1: p \neq 0.2 \)

The test statistic is \( z = \underline{\hspace{2cm}} \)   
(Round to two decimal places as needed.)

The P-value is \(\underline{\hspace{2cm}}\)  
(Round to three decimal places as needed.)

Because the P-value is \(\underline{\hspace{3cm}}\) the significance level, \(\underline{\hspace{3cm}}\) the null hypothesis. There is \(\underline{\hspace{3cm}}\) evidence to support the claim that the return rate is less than 20%.

(Note: This template guides users in identifying the correct hypothesis and calculating the test statistic and P-value in hypothesis testing.)
Transcribed Image Text:In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6,991 subjects randomly selected from an online group involved with ears. There were 1,308 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. Identify the null hypothesis and alternative hypothesis. - Option A: \( H_0: p < 0.2 \) \( H_1: p = 0.2 \) - Option B: \( H_0: p = 0.2 \) \( H_1: p < 0.2 \) - Option C: \( H_0: p > 0.2 \) \( H_1: p = 0.2 \) - Option D: \( H_0: p = 0.2 \) \( H_1: p > 0.2 \) - Option E: \( H_0: p \neq 0.2 \) \( H_1: p = 0.2 \) - Option F: \( H_0: p = 0.2 \) \( H_1: p \neq 0.2 \) The test statistic is \( z = \underline{\hspace{2cm}} \) (Round to two decimal places as needed.) The P-value is \(\underline{\hspace{2cm}}\) (Round to three decimal places as needed.) Because the P-value is \(\underline{\hspace{3cm}}\) the significance level, \(\underline{\hspace{3cm}}\) the null hypothesis. There is \(\underline{\hspace{3cm}}\) evidence to support the claim that the return rate is less than 20%. (Note: This template guides users in identifying the correct hypothesis and calculating the test statistic and P-value in hypothesis testing.)
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