Question

According to a survey by the Association for Dressings and Sauces (this is a real association!), 85% of American adults eat salad at least once a week. A nutritionist suspects that the percentage is lower than this. Conduct a survey of 200 American adults and discover that 160 of them eat salad at least once a week. Conduct a test that addresses the nutritionist's suspicions. Use a significance level of 0.10.

Answer the following:

Hypothesis test: Ho: p = Answer versus H1: p

Using the p-value method: The test statistic (to two decimal places) is.

The value - p (to four decimal places) is:

The decision is: Is H0 rejected or not rejected

Conclusion:

Answer 1

Given : p =0.85 , n=200 , r = 160

The population probability of success of each trial = r / n   where r is number of success.

                q = 1-p = 1-0.85 = 0.15

As sample size is large(n >30) n =200 and np = 200*0.85 = 170 ; nq = 200*0.15 =30

That is np and nq both greater than 5 ,

The distribution of = r/n values approximately follow normal distribution with mean µ = p and

σ =   

µ = 0.85 and σ =     = 0.0252

Step 1) Define hypothesis:

Claim: The percentage of American adults eats salad at least once a week is less than 85%.

Hypothesis test is:

H0 : p= 0.85                Vs   H1 : P < 0.85

Step 2)

Calculating test statistics.

n = 200, p =0.85 ,q =0.15 , = 160/ 200 = 0.80 , σ = 0.0252 , α = 0.10

Z =     

= (0.80- 0.85) / 0.0252 As σ =        =     = 0.0252

Z = -1.98

Test statistic = -1.98

Step 3) P-value

P-value = P( Z < -1.98) = 0.0239                     Using standard normal table

P- Value = 0.0239

You can also find this probabilities using excel command “=NORMSDIST(-1.98) “

Step 4) Decision rule:

If P – value < level of significance ,we reject H0 .

If P value > Level of significance . we fail to reject H0 .

As P- Value 0.0239 < level of significance α = 0.10 , we reject H0 .

Step 5) Conclusion: At 10 % level of significance, the evidence shows that percentage of American adults eats salad at least once a week is less than 85%.