Question

Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the population proportion p at the given level of significance alphaα using the given sample statistics. ​Claim: p≠0.26​; α=0.05​; Sample​ statistics: p=0.22​, n=100

Answer 1

If both np and nq >= 5, then we can use normal sampling distribution

np = 100 * 0.26 = 26 > 5

nq = n * ( 1 - P ) = 100 * 0.74 = 74 > 5

Since both are > 5, hence we can use normal sampling distribution

To Test :-

H0 :- P = 0
H1 :- P ≠ 0

P = X / n = 22/100 = 0.22

Test Statistic :-
Z = ( P - P0 ) / ( √((P0 * q0)/n)
Z = ( 0.22 - 0.26 ) / ( √(( 0.26 * 0.74) /100))
Z = -0.9119 ≈ - 0.91

Test Criteria :-
Reject null hypothesis if Z < -Z(α/2)
Critical value   Z(α/2) = Z(0.05/2) = 1.96
Z > -Z(α/2) = -0.9119 > -1.96, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Decision based on P value
P value = 2 * P ( Z > -0.9119 )
P value = 0.3618
Reject null hypothesis if P value < α = 0.05
Since P value = 0.3618 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There is insufficient evidence to support the claim that p≠0.26.