Question

please indicate all answers

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-maled to 6983 subjects randomly selected from an online group involved with ears. There were 1309 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<02 H:p0.2 O C. HỌ p = 0 2 Hyp>0.2 O E. Ho:p>0.2 Hp0.2 OB. Ho: p=02 Hyp#02 OD. Ho: p=02 Hp02 OF HOP 0.2 Hyp=0.2 The best statistic is 2.0 (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, is less than 20% 0 the null hypothesis. There is evidence to support the claim that the return rate Click to select answers

Answer 1

Sample size, n= 6983

sample proportion, $\hat{p}$ = 1309/6983 = 0.1874

a)  D. H0: p=0.2

HA: p<0.2

b) Test statistic, z= $\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$ = $\frac{0.1874-0.2}{\sqrt{\frac{0.2(1-0.2)}{6983}}}$ = -2.63

c) P- value = P( Z < -2.63) = 1- P( Z < 2.63) = 1- 0.99573=0.004

d) Because the P-value is Less than the significance level. Reject   the null hypothesis . There is Sufficient evidence to support the claim that the return rate is lass than 20%.