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 levetest 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=0.2H1: p<0.2О В. Но: р#0.2H1:p= 0.2O C. Ho: p= 0.2D. Ho: p>0.2H1:p>0.2H1:p= 0.2E. Ho: p= 0.2OF. Ho: p<0.2H1:p#0.2H1:p=0.2The 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 isthe significance level,the null hypothesis. There isevidence to support the claim that the return rate is less than 20%.greater thanless than

Given data number of success , x= 1343.0 sample size, n = 6972 population…

Answered: In a study of cell phone usage and…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A certain drug is used to treat asthma. In a clinical trial of the drug, 30 of 299 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 12% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. a. Is the test two-tailed, left-tailed, or right-tailed? Left-tailed test Two-tailed test Right tailed test b. What is the test statistic? Z= (Round to two decimal places as needed.) c. What is the P-value? P-value = (Round to four decimal places as needed.) 1-PropZTest prop <0.12 z 1.046429014 p= 0.1476815015 p=0.1003344482 n = 299
A certain drug is used to treat asthma. In a clinical trial of the drug, 15 of 286 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 12% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. a. Is the test two-tailed, left-tailed, or right-tailed? b. What is the test statistic? z= c.What is the P-value? P-value equals d.What is the null hypothesis, and what do you conclude about it? e. What is the final conclusion?
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…
A certain drug is used to treat asthma. In a clinical trial of the drug, 21 of 270 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 9% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. a. Is the test two-tailed, left-tailed, or right-tailed? Left-tailed test Right tailed test Two-tailed test b. What is the test statistic? z= (Round to two decimal places as needed.) c. What is the P-value? P-value= (Round to four decimal places as needed.) d. What is the null hypothesis, and what do you conclude about it? Identify the null hypothesis. A.Upper H 0 : p equals 0.09 B.Upper H 0 : p not equals 0.09 C.Upper H 0 : p less than 0.09 D.Upper H 0 : p greater than 0.09 Decide…
A certain drug is used to treat asthma. In a clinical trial of the drug, 18 of 282 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 11% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level. What is the final conclusion? A. There is sufficient evidence support the claim that less than 11% of treated subjects experienced headaches. B. There is not sufficient evidence to warrant rejection of the claim that less than 11% of treated subjects experienced headaches. C. There is not sufficient evidence support the claim that less than 11% of treated subjects experienced headaches. D. There is sufficient evidence to warrant rejection of the claim that less than 11% of treated subjects experienced headaches
In a study of cell phone usage and brain hemispheric dominance, an Intemet survey was e-mailed to 6992 subjects randomly selected from an online group involved with ears. There were 1294 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 p0.2 H1 p 0.2 O c. Ho p# 0.2 D. Ho: p 0.2 H p 0.2 H1 p 0.2 OE. Ho p 0.2 O F. Ho p 0.2 H1 p>0.2 Hy: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.) the significance level, Because the P-value is 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 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,
A report states that 10% of all babies born in a certain U.S. state are preterm, which has important implications for the cost of health care. If in an SRS of 400 recent births in the state 36 were preterm, what is sampling distribution of the sample proportion? approximately N(0.10, 0.0150) approximately N(40, 6) binomial with n=400 and p=.10
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 certain drug is used to treat asthma. In a clinical trial of the drug, 26 of 261 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 9% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.01 significance level to complete parts (a) through (e) below. a. Is the test two-tailed, left-tailed, or right-tailed? O Left-tailed test Right tailed test Two-tailed test 1-Prop2Test prop<0.09 z = 0.542889594 p=0.7063970895 -0.0996168582 n=261
A certain drug is used to treat asthma. In a clinical trial of the drug, 20 of 289 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than 9% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. a. Is the test two-tailed, left-tailed, or right-tailed? b. What is the test statistic? c. What is the P-value? d. What is the null hypothesis, and what do you conclude about it? Identify the null hypothesis. Decide whether to reject the null hypothesis. Choose the correct answer below. e. What is the final conclusion?

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