In an article in the Journal of Advertising, Weinberger and Spotts compare the u

Question

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor.

(a) Suppose that a random sample of 375 television ads in the United Kingdom reveals that 141 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. television ads that use humor. (Round your answers to 3 decimal places.)

The 95 percent confidence interval is [ , ].

(b) Suppose a random sample of 533 television ads in the United States reveals that 117 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.S. television ads that use humor. (Round your answers to 3 decimal places.)

The 95 percent confidence interval is [ , ].

Explanation / Answer

.a. Here n=375 and x=141, So point estimte of proportion phat=x/n=0.376

Now we SE(phat)=sqrt(phat(1-phat)/n)=sqrt(0.376*0.624/375)=0.025

So Margin of error=E=z*SE(pht), now z value for 95% CI is 1.96 as from standard normal table P(-1.96<z<1.96)=0.95

So E=1.96*0.025=0.049

Hence CI=phat+/-E=0.376+/-0.049=(0.327,0.425)

b. Here n=533 and x=117

So phat=117/533=0.220

Hence SE(phat)=sqrt(0.220*0.780/533)=0.018

So Mrgin of Error=z*SE(pht)=1.96*0.018=0.0353

hence CI=phat+/-E=0.220+/-0.035=(0.185,0.255)

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