Question

A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true proportion of families who own at least one DVD player. Place your limits, rounded to 3 decimal places, in the blanks. Place the lower limit in the first blank

Answer 1

Answer:   $(0.367,\ 0.473)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ is the sample proportion, n is the sample size , $z_{\alpha/2}$ is the critical z-value.

Given : Significance level : $\alpha:1-0.99=0.01$

Sample size : n= 85

Critical value : $z_{\alpha/2}=2.576$

Sample proportion: $\hat{p}=\dfrac{36}{85}\approx0.42$

Now, the  99% confidence level will be :

$\hat{p}\pmz_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.42\pm(2.576)\sqrt{\dfrac{0.42(1-0.42)}{85}}\\\\\approx0.42\pm0.053\\\\=(0.42-0.053,\ 0.42+0.053)=(0.367,\ 0.473)$

Hence, the  99% confidence interval estimate of the true proportion of families who own at least one DVD player is  $(0.367,\ 0.473)$