In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6972 subjects randomly selected from an online group involved with ears. There were 1295 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. OA. Ho: p=0.2 H₁: p=0.2 OC. Ho: p 0.2 H₁: p=0.2 O E. Ho: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, OB. Ho: p=0.2 H₁: p<0.2 O D. Ho: p=0.2 H₁: p>0.2 OF. Ho: p<0.2 H₁: p=0.2 the null hypothesis. There is evidence to support the claim that the return rate is less than 20%.

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6976subjects randomly selected from an online group involved with ears. There were 1301 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. test statistic___ pvalue___ Because the p value is______the significance level,____the null hypothesis.There is ____edivence to support the claim that the return rate is less than 20%

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6989 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: p0.2 H: p= 0.2 H4: 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%. Question Viewer

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6972 subjects randomly selected from an online group involved with ears. There were 1343 surveys returned. Use a 0.01 significance leve 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 H1: p0.2 H1:p>0.2 H1:p= 0.2 E. Ho: p= 0.2 OF. Ho: p<0.2 H1: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%. greater than less than

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6990 subjects randomly selected from an online group involved with ears. There were 1347 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 diştribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis. O A. Ho p>02 H p=02 О В. Но р302 H p0.2 The test statistic is z= (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.) the null hypothesis. There is evidence to support the claim that the return rate is less than 20%. Because the P-value is the significance level,

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6968subjects randomly selected from an online group involved with ears. There were 1335 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.)

Childhood participation in sports, cultural groups, and youth groups appears to be related to improved self-esteem for adolescents (McGee, Williams, Howden-Chapman, Martin, & Kawachi, 2006). In a representative study, a sample of n = 100 adolescents with a history of group participation is given a stan- dardized self-esteem questionnaire. For the general population of adolescents, scores on this questionnaire form a normal distribution with a mean of p %3D 50 and %3D 15. The sample of group- a standard deviation of o %3D participation adolescents had an average of M = 53.8.

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6973 subjects randomly selected from an online group involved with ears. There were 1343 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. А. Но: Р0.2 ОЕ. Но: р> 0.2 OF. Ho:p=0.2 H4: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%.

A researcher claims that 45% of U.S. adults think that the IRS is not aggressive enough in pursuing people who cheat on their taxes. You believe the true percentage is more than 45%. To test this, you take a simple random sample of 400 U.S. adults and find that 208 of them say the IRS is not aggressive enough. Test at .05 significance. Round to the fourth Ho: P HA: P How many successes were there? Test Statistic: = ४ .45 .45 P-value: Did something significant happen? Significance Happened Select the Decision Rule: Reject the Null There is Oenough evidence to conclude that more than 45% of U.S. adults think that the IRS is not aggressive enough in pursuing people who cheat on their taxes. ✓

A study is comparing the taste of instant coffee with fresh-brewed coffee. 50 individuals(random;y sampled) take part in the study. Each individual tastes both kinds of coffee, and declares which kind he or she prefers. The order in which the two kinds of coffee are given to each individual is randomized. 32 individuals in the sample prefer the fresh-brewed coffee. Use a suitable normal approximation to find a 95% confidence interval for the proportion of all individuals who would prefer fresh-brewed coffee. What is the lower limit of this confidence interval?