Question

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 1348 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. Hop<02 HP#02 Hyp=02 © C. Hop= 0,2 OD. Ho p=02 H, p<02 Hyp>02 O E. Ho:p>0.2 OF. Họ p = 0.2 H, p=02 H, p=02 The test statistic is Z=-1.41 (Round to two decimal places as needed.) The P-value is 0.079 (Round to three decimal places as needed.) Because the P-value is greater than the significance level, fail to reject the null hypothesis. There is insufficient evidence to support the claim that the return rate is less than 20%

Answer 1

Hyothesis :


H:p < 0.20

Given ,

n = 6973

x = 1348

T 1348 р = 0.193317 n 6973

Test statistics :

$Z_0 =\frac{\hat{p}-p}{\sqrt{\frac{p*(1-p)}{n}}}$

$Z_0 =\frac{0.193317-0.20}{\sqrt{\frac{0.20*(1-0.20)}{6973}}}$

$Z_0 =\frac{-0.006683}{\sqrt{\frac{0.16}{6973}}}$

$Z_0 =\frac{-0.0067}{0.004790162}$

Z0 = -1.40

P value : 0.081

( to get p value check Z table or use excel command =NORMSDIST(-1.40) )

significance level = 0.01

P value greater than significance level, fails to reject null hypothesis and conclude that there is insufficient evidence to support the claim that the return rate is less than 20%