Question

You have been hired by the Coca-Cola Company to determine if students at Oregon State University prefer Coke or Pepsi. A taste test was performed where students were given two identical cups and were asked to taste both drinks. They had to report which drink they prefer. It was found that 69 out of 125 students indicated they preferred cup that contained Coke.

1. (4 pts) What is the random variable in this problem? Does the random variable have a binomial distribution? Explain. (Recall, there are 4 checks for a discrete random variable to have a binomial distribution – make sure you list all 4. Discuss in detail if you think the “observations are independent of each other”.)

2. (1 pt) What does ?? represent in the context of this study?

3. (1 pt) Calculate the sample proportion, ??̂, of students in the sample prefer Coke. Show work.

4. Perform a hypothesis test to determine if OSU students prefer one brand over the other by answering the following questions:

a. (3 pts) State the null and alternative hypotheses in statistical notation. Define any notation used. (Hint: if there really is no preference, would be expect 50% to prefer Coke and 50% to prefer Pepsi?)

b. (3 pts) Report the p-value and state a conclusion in a complete sentence in the context of the problem.

c. (3 pts) Report a 95% confidence interval for the proportion of all OSU students who prefer Coke. Interpret this confidence interval in the context of the problem.

5. (2 pts) Do you believe it is legitimate to use the results of this hypothesis test and confidence interval to make a conclusion about all OSU students? Why or why not?

Answer 1

1:

Each of 125 students are independent from each other.

There is only two outcomes of each trial : Coke or Pepsi

Each time the probability of success remain same and equal to 0.5.

There are fixed number od trials.

2:

Since Coke and Pepsi are equally likely so

p = 0.50

3:

4:

(a)

Hypotheses are;

(b)

Standard deviation of the proportion is:

Test statistics will be:

The p-value is:

p-value =2[1 - P(z<= 1.16)] = 0.246

Since p-value is greater than 0.05 so we fail to reject the null hypothesis.

c)

For 95% confidence interval, using excel function "=NORMSINV(0.975)", critical value of z will be

Confidence interval for population proportion will be

We are 95% confident that true proportion of student who prefer coke lies in the above interval.

5:

Sample is large so it should be good representative of populaiton. So it is legitimate to use the results of this hypothesis test and confidence interval to make a conclusion about all OSU students