In an article in Marketing Science, Silk and Berndt investigate the output of ad

Question

In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspapers, radio, and so forth. (a) Suppose that a random sample of 390 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.45 percent, and assume that sigma equals 1.48 percent. Calculate a 95 percent confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agencies. The 95 percent confidence Interval is (b) Suppose that a random sample of 390 U.S. advertising agencies gives an average percentage share of billing volume from spot television commercials equal to 12.42 percent, and assume that sigma equals 1.57 percent. Calculate a 95 percent confidence interval for the mean percentage share of billing volume from spot television commercials for the population of all U.S. advertising agencies.

Explanation / Answer

a. Here n=390, so as per central limit theorem we will use normal distribution for average

Hence z=1.96 for 95% CI

Now margin of error=z*sd/sqrt(n)=1.96*1.48/sqrt(390)=0.147

So CI=mean+/-E=7.45+/-0.147=(7.303,7.597)

b. Here again n=390 so as per a we will use normal distribution

So margin of error=z*sd/sqrt9n)=1.96*1.57/sqrt9390)=0.156

Hence CI=12.42+/-0.156=(12.264,12.576)

c. As width of b is greater than a we can conclude that mean percentage share of billing vloume from spot television commercials for US advertising agencies is greater than the mean percentage share of billing volume from network television.

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