business statistics

Question

Out of 200 people sampled, 136 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids.

Give your answers as decimals, to three places

______ < p < _______

Answer 1

Hi Salma,

Both our successes (people with kids) and failures (people without kids) exceed 10, so we can assume normality i.e. use z. With that in mind, formula for this confidence interval is:

CI=p-hat +/ z*SE

Now, let’s break this down.

p-hat=estimated proportion

z*=z-critical value

SE=standard error, which has its own formula

SE=sqrt(p-hat(1-p-hat))/n)

Let’s compute this first.

p-hat=136/200=0.68

1-p-hat=0.32

n=200

Thus:

SE=sqrt((0.68*0.32)/200)

SE=0.033

Now, back to the original formula:

CI=p-hat +/-z*SE

p-hat=0.68

z*=2.576–may be worth memorizing or adding to formulas sheet if allowed on tests

SE=0.033

CI=0.68 +/- (2.576*0,033)

CI=(0.595, 0.765)

In your problem’s notation:

0.595 < p < 0.765

I hope this helps.