Question

in a survey of 7000 women 4431 say they change their nail polish once a week.Construct a 95% confidence interval for the population proportion of women who change their nail polish once a week.A 95% confidence interval for the population proportion is​

Answer 1

Answer:

The 95% confidence interval would be given (0.622;0.644).

We are confident (95%) that the true proportion of people that said that they change their nail polish once a week  is between 0.622 and 0.644

Step-by-step explanation:

Data given and notation  

n=7000 represent the random sample taken    

X=4431 represent the people that said that they change their nail polish once a week

$\hat p=\frac{4431}{7000}=0.633$ estimated proportion of people that said that they change their nail polish once a week

$\alpha=0.05$ represent the significance level

Confidence =0.95 or 95%

p= population proportion of people that said that they change their nail polish once a week

Solution to the problem

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.633 - 1.96 \sqrt{\frac{0.633(1-0.633)}{7000}}=0.622$

$0.633 + 1.96 \sqrt{\frac{0.633(1-0.633)}{7000}}=0.644$

And the 95% confidence interval would be given (0.622;0.644).

We are confident (95%) that the true proportion of people that said that they change their nail polish once a week  is between 0.622 and 0.644