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 and the upper limit in the second blank

Answer 1

Answer:

$0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286$

$0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562$

The 99% confidence interval would be given by (0.286;0.562)

Step-by-step explanation:

Information given:

$X= 36$ represent the families owned at least one DVD player

$n= 85$ represent the total number of families

$\hat p=\frac{36}{85}= 0.424$ represent the estimated proportion of families owned at least one DVD player

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.05$. And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:  

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

If we replace the values obtained we got:

$0.424 - 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.286$

$0.424 + 2.58\sqrt{\frac{0.424(1-0.424)}{85}}=0.562$

The 99% confidence interval would be given by (0.286;0.562)