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

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 398 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.43 percent, and assume that equals 1.42 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 398 U.S. advertising agencies gives an average percentage share of billing volume from spot television commercials equal to 12.45 percent, and assume that equals 1.51 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.

(Click to select)YesNo , confidence interval in (b) is totally (Click to select)belowabove the confidence interval in (a).

Explanation / Answer

solutiona:

zcrit for 95%=1.96

pop sd=1.42

sample mean=7.43

n=398

95% CI for true mean percentage share of billing volume from network television for the population of all U.S. advertising agencies is

sample mean-z crit(pop sd/sqrt(n),sample mean+zcrit(pop sd/sqrt(n)

7.43-1.96*1.42/sqrt(398),7.43+1.96*1.42/sqrt(398)

7.290,7.570

The 95 percent confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agenciesis  7.290,7.570

solutionb:

sample mean=12.45

pop sd=1.51

n=398

95% confidence interval for the true mean

12.45-1.96*1.51/sqrt(398),12.45+1.96*1.51/sqrt(398)

12.302,12.598

lower limit=12.302

upper limit=12.598

The 95 percent confidence interval for the true mean percentage share of billing volume from spot television commercials for the population of all U.S. advertising agenciesis 12.302,12.598

Solutionc:

yes

confidence interval in (b) is totally above the confidence interval in (a).

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