Question

In a sample of 150 households in a certain city, 112 had high-speed internet access. We wish to test the following hypotheses: Ho : p-0.70, H,:p> 0.70 a) The p-value for this test would be: 0.016 b) What conclusion should you come to based on this p-value at a significance level of 0.05? Less than 70 % of households in t

Answer 1

P = X / n = 112/150 = 0.7467

Test Statistic :-
Z = ( P - P0 ) / √ ((P0 * q0)/n))
Z = ( 0.7467 - 0.7 ) / √(( 0.7 * 0.3) /150))
Z = 1.2472

Decision based on P value
P value = P ( Z > 1.2472 ) = 0.1062

Looking the value Z = 1.2472 in standard normal table to find the P value.

Reject null hypothesis if P value < α = 0.05
Since P value = 0.1062 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Fail to reject H0, conclude that   

H : p= 0.70,