Question

business statistics 8-10 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 400 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.46 percent, and assume that the population standard deviation 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.

Answer 1

Answer: $(7.321\ , \ 7.599)$

Step-by-step explanation:

Given : Sample size : n= 400  () large sample

Significance level : $\alpha: 1-0.95=0.05$

By using Standard normal table , Critical value : $z_{\alpha/2}=1.96$

Sample mean : $\overline{x}= 7.46$

Standard deviation: $\sigma=1.42$

The confidence interval for population means is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

$=7.46\pm(1.96)\dfrac{1.42}{\sqrt{400}}$

$\approx 7.46\pm0.139=( 7.46-0.139,\ 7.46+0.139)\\\\=(7.321\ , \ 7.599)$

Hence, the 95 percent confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agencies = $(7.321\ , \ 7.599)$