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Question 639:



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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.


d. We see that this test-statistics is greater than our critical value of 1.28 so we reject the Null Hypothesis.  By rejecting the null hypothesis we are saying that there is sufficient evidence (greater than what we'd expect from chance alone) to conclude that the proportion of population that can identify CherryBerry Soda is greater than 15%. If we assume this increase is all due to the ad campaign and not other factors, than it is a successful campaign and time to give out bonuses !  

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