Question

Among a survey of 50 students at B.C ,48 of the respondents stated yes to texting and driving.

Test the claim at 0.05 level of significance that the proportion of students who text and drive is greater than 80%..

Show work

Answer 1

Solution ;

Given that,

n = 50

x = 48

= x / n = 48 / 50 = 0.96

P₀ = 80 %  

1 - P₀ = 1 - 0.80 = 0.20

z = - P₀ / [P_{0 *} (1 - P₀ ) / n]

= 0.96 - 0.80 / [(0.80* 0.20) / 50]

= 2.828

Test statistic = 2.828  

This is the right tailed test .

P(z > 2.828 ) = 1 - P(z < 2.828 ) = 1 - 0.9977 = 0.0023

P-value = 0.0023

= 0.05

P-value <

Reject the null hypothesis .