91影视

Given that telephone interviews were conducted on a representative sample of adults living in the United States. For this sample,adults admitted that they have paid to download music.

The true proportion of the point estimate is given by

$CI=\stackrel{炉}{X}卤z脳\frac{蟽}{\sqrt{n}}$

Forconfidence interval, the interval is given by

$CI=\stackrel{炉}{X}卤1.645脳\frac{蟽}{\sqrt{n}}$

Here the z score is$z=1.645$

For$90\%$confidence interval, the interval is given by

$CI=\stackrel{炉}{X}卤1.645脳\frac{蟽}{\sqrt{n}}$

Here the Z score is$z=1.645$

Here the total sample size $n=1003$ and the mean is given by

$\begin{array}{rcl}\stackrel{炉}{X}& =& \frac{{鈭�}_{i=1}^n{X}_i}{n}\\ & =& \frac{506}{1003}\\ & =& 0.504\\ & & \end{array}$

And the variance is unknown. So here we have to find the variance. Then it is given by

$\begin{array}{rcl}0.90& =& 0.504卤1.645脳\frac{蟽}{\sqrt{1003}}\\ 28.503& =& 15.96+1.645蟽\\ 蟽& =& \frac{12.543}{1.645}\\ 蟽& =& 7.624\\ & & \end{array}$

So, here in this problem, to interpret that for confidence interval, it is applied for paid to download music.

The meaning of confidence interval is that it is interpreted for paid to download music users.

For $90\%$confidence interval, the interval estimate is given by

$\begin{array}{rcl}CI& =& \stackrel{炉}{X}卤1.645脳\frac{蟽}{\sqrt{n}}\\ & =& 0.504卤1.645脳\frac{7.624}{31.67}\\ & =& 0.504+0.394\\ & =& 0.907\\ & & \end{array}$

And

$\begin{array}{rcl}CI& =& 0.504-0.394\\ & =& 0.11\\ & & \end{array}$

So, there are$90\%$adults need to be sampled to reduce the margin of error.