Question

The BIG BURGER manager now has the following problem. Nationally 60% of the clientele surveyed respond that BIG BURGER has really good food. The BIG BURGER manager wishes to establish that he is exceeding the national average here in Cleveland (He is up for promotion to Regional Head Burger Person). He randomly samples 300 customers. 195 respond that BIG BURGER has really good food. Can he conclude that the population proportion who think that BIG BURGER has really good food is exceeding the national average in Cleveland? Let alpha = .05

Answer 1

Solution: Here we have the given information are

n=300. =0.05

X= number of customers that BIG BURGER has really good food.

p= Proportion of  customers that BIG BURGER has really good food.= x/n =195/300=0.65

Null hypothesis: H0: P=0.60, i.e. the proportion of  60% of the clientele surveyed respond that BIG BURGER has really good

food.

Alternative hypothesis : H1: P>0.60 ( Right -tail alternative)

Test Statistic, Under H0, the test statistic is:

Now

Z=1.768

Conclusion: Since the alternative hypothesis is one-sided (right-tailed), we shall apply right-tailed test for testing significance

of Z. The significant value of Z at 5% level of significance for right-tail test is 1.645. Since computed value of Z=1.768 is

greater than 1.645, it is significant and we may reject the null hypothesis at 5% level of significance () and accept

alternative hypothesis.

That means manager claim BIG BURGER has really good food is exceeding the national average in Cleveland.