## Question 639:

1## Answer:

No answer provided yet.For this question we have a proportion and a sample size which is relatively large (1000). We can use the Normal Approximation to the Binomial instead of doing the exact binomial calculations. It's usually Ok to use the normal approximation to the binomial when n*p and n*q > 5, where p is the proportion identifying with the ad campaign and q is the proportion not. This gets us .17*1000 =170 and .83*1000 = 830 so we're fine using this approach.

a. The null hypothesis is that the percent of the population who identify its products is 15 percent or less. The alternative hypothesis is that the percent who identify the products is greater than 15%.

b. We will reject the null hypothesis if the test statistic is greater than the critical value of 1.28. We find the critical value from a table of normal values or using the percentile to z-score calculator

for 1-sided area of .10.

c. The test statistic is found as the difference between the two proportions .17-.15 = .02 divided by the standard error of the proportion(SE). The SE is found as:

The test proportion (.15) times 1-test proportion .85 = .15*.85 = .1275 and we divide this by the sample size = .1275/1000 = .000128. Now we take the square root of this value to get the .0112916. Finally or SE is the difference divided by this value = .02/.0112916 = 1.7712.