In: Statistics and Probability
In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to
6978 subjects randomly selected from an online group involved with ears. There were
1331 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.
Ho : p = 0.2
H1 : p < 0.2
(Left tail test)
Level of Significance, α =
0.01
Number of Items of Interest, x =
1331
Sample Size, n = 6978
Sample Proportion , p̂ = x/n =
0.1907
Standard Error , SE = √( p(1-p)/n ) =
0.0048
Z Test Statistic = ( p̂-p)/SE = ( 0.1907
- 0.2 ) / 0.0048
= -1.9333
p-Value = 0.0266 [excel
function =NORMSDIST(z)]
Decision: p value>α ,do not reject null
hypothesis
There is not enough evidence to conclude that return rate is less
than 20%