Question

In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6987 subjects randomly selected from an online group involved with ears. There were 1500 surveys returned. Use a 0.01 significance level to test the claim that the return rate is greater than 20%. Use the P-Value method and use the normal distribution as an approximation to the binomial distribution. (Hint: may use the appropriate calculator function).

(I have a TI-84 calculator).

Answer 1

given data and necessary calculations are:-

sample size (n) = 6987

sample proportion () = 1500/6987 = 0.214684

hypothesis:-

[ claim ]

the test statistic be:-

the p value is:-

[ TI 84 calculator function : =normalcdf (3.0685,1E99,0,1) ]

decision:-

p value = 0.0011 < 0.01 (alpha)

we reject the null hypothesis and conclude that, there is enough evidence to  claim that the return rate is greater than 20%.

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