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

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

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=(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{(\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{(0.633(1-0.633))/(7000)}=0.622`

`0.633 + 1.96 \sqrt{(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