Question

his emplyer informed him that no more than 60% of customers prefer soft shell tacos, while the rest prefer hard one. he decided to observe the next 120 customers and observed that 80 of them ordered soft one

a set up and test the appropriate hypotheses using a significance level of 0.02 assuming that all customers are independent of each other

b if he wished to use a significance level of 0.05, determine an appropriate rejection region for his test

c find a 90% confidence interval for the proportion of customers that prefer soft one

Answer 1

a .We have to test, H0: p 0.6 against H1: p > 0.6,

where, p is the proportion of customers preferring soft shell tacos.

The test-statistic is given by, Z = , where, = 80/120 = 0.6667, p = 0.6, n = 120

Thus, Z = 1.4915.

Under H0, Z ~ N(0,1).

The p-value is = P(Z > 1.4915) = 1 - (1.4915) [(.) is the cdf of N(0,1)] = 1 - 0.9321 = 0.0679.

Since, p-value > level of significance = 0.05, we fail to reject H0.

b. If the significance level is 0.05, then we reject H0 if the observed test-statistic is greater than i.e. 1.64.

Since, the test-statistic = 1.4915 < 1.64, we fail to reject H0.

c. The 90% confidence interval of the proportion of customers who prefer soft tacos is:

[, ], where, = 80/120 = 0.6667, n = 120, = 1.64

= [0.6667 - 0.0706, 0.6667 + 0.0706] = [0.5961, 0.7373]. (Ans).