Question

1. According to a survey conducted by the Association for Dressing and Sauces, 85% of American adults eat salad at least once a week. A nutritionist suspects that the percentage is higher than this. She conducts a survey of 200 American adults and finds that 171 of them eat salad at least once a week. Conduct the appropriate test that addresses the nutritionist's suspicions use a significance level of 0.05.

Answer 1

n = 200, x = 171

p̄ = x/n = 0.855

α = 0.05

Null and Alternative hypothesis:

Ho : p ≤ 0.85

H1 : p > 0.85

Test statistic:

z = (p̄ -p)/√(p*(1-p)/n) = (0.855 - 0.85)/√(0.85 * 0.15/200) = 0.1980

Critical value :

Right tailed critical value, z crit = ABS(NORM.S.INV(0.05)) = 1.645

Reject Ho if z > 1.645

or p-value :

p-value = 1- NORM.S.DIST(0.198, 1) = 0.4215

Decision:

p-value > α, Do not reject the null hypothesis

Conclusion:

There is not enough evidence to conclude that the percentage is higher than 0.85 at 0.05 significance level.