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 391 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.41 percent, and assume that equals 1.46 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. (Round your answers to 3 decimal places.)

The 95 percent confidence interval is [ , ].

(b) Suppose that a random sample of 391 U.S. advertising agencies gives an average percentage share of billing volume from spot television commercials equal to 12.46 percent, and assume that equals 1.55 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. (Round your answers to 3 decimal places.)

The 95 percent confidence interval is [ , ].

(c) Compare the confidence intervals in parts a and b. Does it appear that the mean percentage share of billing volume from spot television commercials for U.S. advertising agencies is greater than the mean percentage share of billing volume from network television? Explain.

Explanation / Answer

a. Here  random sample of 391 U.S. advertising agencies re consider with mean=7.41 and sd=1.46

Now as per central limit theorem if n is sufficient large distribution of sample mean is normal with mean=7.41 and sd=sd/sqrt(n)=1.46/sqrt(391)=0.074. Hence we will consider z distribution and value for z is 1.96 as from standard normal distribution P(-1.96<z<1.96)=0.95

Now Margin of Error=z*sd/sqrt(n)=1.96*0.074=0.145

Hence CI=mean+/-E=7.41+/-0.145=(7.265,7.555) is 95 percent confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agencies.

b. Here  random sample of 391 U.S. advertising agencies re consider with mean=12.46 and sd=1.55

Now as per central limit theorem if n is sufficient large distribution of sample mean is normal with mean=12.46 and sd=sd/sqrt(n)=1.55/sqrt(391)=0.0784. Hence we will consider z distribution and value for z is 1.96 as from standard normal distribution P(-1.96<z<1.96)=0.95

Now Margin of Error=z*sd/sqrt(n)=1.96*0.0784=0.154

Hence CI=mean+/-E=12.46+/-0.154=(12.306,12.614) is 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.

c. Width of 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 is 0.308

And With of 95 percent confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agencies is 0.29

As mean of share of billing volume from spot television commercials is greater than mean of of billing volume from network television for the population of all U.S. advertising agencies so its percentage is greater.

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