One proportion z test

Apr 2010
I have attempted this question, but im not sure what ive done is right. Could someone please explain how to do it so I can make sure that ive done it right?

“Take the Pepsi Challenge” was a marketing campaign used by the Pepsi-Cola Company. Cola drinkers took the taste test in which they tasted unmarked cups of Pepsi and Coke and were asked to select the one they liked best. Pepsi claimed that in test, more than half the Diet Cola drinkers said they preferred the taste of Diet Pepsi.​
Suppose 100 Diet Cola drinkers took the Pepsi Challenge and 56 preferred the taste of Diet Pepsi.

- Test the hypothesis that more than half of all Diet Cola drinkers will select Diet Pepsi in a blind taste test. Use
α = 0.05

- What is the
P -value and how is it found

- Find the 90% confidence interval for the true proportion of all Diet Cola drinkers who prefer Diet Pepsi and what it means in the context of this problem



MHF Hall of Honor
Feb 2009
\(\displaystyle H_0:p=.5\) vs. \(\displaystyle H_a:p>.5\)

where p is the probability that a Diet Cola drinker will select Diet Pepsi.

No one seems ever use the correction factor when approximating with a nomal
so the test stat is

\(\displaystyle z^*={\hat p-.5\over \sqrt{(.5)(.5)\over 100}}\)

where \(\displaystyle \hat p=.56\)

and your rejection region with alpha equal to .05, is \(\displaystyle (1.645,\infty)\)

Or use the binomial tables.

The p-value is \(\displaystyle P(Z>z^*)\)