In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6965 subjects randomly selected from an online group involved with ears. There were 1290 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. A. $H_0: p = 0.2$ $H_1: p > 0.2$ B. $H_0: p \neq 0.2$ $H_1: p = 0.2$ C. $H_0: p > 0.2$ $H_1: p = 0.2$ D. $H_0: p = 0.2$ $H_1: p \neq 0.2$ E. $H_0: p < 0.2$ $H_1: p = 0.2$ F. $H_0: p = 0.2$ $H_1: p < 0.2$
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The claim to be tested is that "the return rate is less than 20%". Let $p$ be the true population proportion of the return rate. The claim can be written as $p < 0.20$. Step 2: The alternative hypothesis ($H_1$) is a statement that contradicts the null Show more…
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In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6958 subjects randomly selected from an online group involved with ears. There were 1319 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. A. H0: p < 0.2 H1: p = 0.2 B. H0: p ≠0.2 H1: p = 0.2 C. H0: p = 0.2 H1: p < 0.2 D. H0: p = 0.2 H1: p > 0.2 E. H0: p = 0.2 H1: p ≠0.2 F. H0: p > 0.2 H1: p = 0.2 The test statistic is z = [ ]. (Round to two decimal places as needed.)
Qudsiya A.
In a study of cell phone usage and brain hemispheric dominance, an Internet survey was emailed to 6973 subjects randomly selected from an online group involved with ears. There were 1310 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. A. H0: p = 0.2 H1: p < 0.2 B. H0: p > 0.2 H1: p = 0.2 C. H0: p = 0.2 H1: p = 0.2 D. H0: p < 0.2 H1: p = 0.2 E. H0: p = 0.2 H1: p > 0.2 F. 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 greater than/less than the significance level, reject/fail to reject the null hypothesis. There is sufficient/insufficient evidence to support the claim that the return rate is less than 20%.
Chris M.
In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6956subjects randomly selected from an online group involved with ears. There were 1290 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. A. H0: p=0.2 H1: p≠0.2 B. H0: p=0.2 H1: p<0.2 C. H0: p>0.2 H1: p=0.2 D. H0: p≠0.2 H1: p=0.2 E. H0: p=0.2 H1: p>0.2 F. 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 ▼ greater than less than the significance level, ▼ fail to reject reject the null hypothesis. There is ▼ insufficient sufficient evidence to support the claim that the return rate is less than 20%.
Adi S.
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