Question

According to a survey conducted by the Association for Ranch Dressings and Sauces, 20% of American adults eat pizza once a week. A nutritionist suspects that this percentage is not accurate. She conducts a survey of 700 American adults and finds that 77 of them eat pizza once a week. Use a 0.05 significance level to test the claim that the proportion of American adults who eat pizza once a week is different from 20%.

Claim:

Opposite:

The test is: (Right, left, two tailed)

Z= (Test statistic)

The critical values are:

Based on this information we?

Conclusion:

Answer 1

To Test :-

H0 :- P = 0.20
H1 :- P ≠ 0.20

It is two tailed test

P = X / n = 77/700 = 0.11

Test Statistic :-
Z = ( P - P0 ) / ( √((P0 * q0)/n)
Z = ( 0.11 - 0.2 ) / ( √(( 0.2 * 0.8) /700))
Z = - 5.9529

Critical values are   Z(α/2) = Z(0.05/2) = ±1.960
Z < - 1.960 OR Z > 1.960

Test Criteria :-
Reject null hypothesis if Z < -Z(α/2)
Z < -Z(α/2) = -5.9529 < -1.96, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that the proportion of American adults who eat pizza once a week is different from 20%.