Question

A random sample of 4000 college students yielded 2250 who are in favor of banning Hawaiian shirts. Estimate the true proportion of all college students who are in favor of banning Hawaiian shirts using a 90% confidence interval.

Answer 1

Answer: (0.5496, 0.5754)

Step-by-step explanation:

The confidence interval for population proportion (p) is given by :-

$\hat{p}\pm z*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$  , where $\hat{p}$= Sample proportion , n= sample size , z*= Critical z-value.

Let p be the  proportion of all college students who are in favor of banning Hawaiian shirt.

Given, A random sample of 4000 college students yielded 2250 who are in favor of banning Hawaiian shirts.

i.e. n=4000

$\hat{p}=\dfrac{2250}{4000}=0.5625$

z-value for 90% confidence level is 1.645

Now , 90% confidence interval for p would be :

$0.5625\pm (1.645)(\sqrt{\dfrac{0.5625(1-0.5625)}{4000}})$

$=0.5625\pm (1.645)(\sqrt{0.0000615234375})\\\\=0.5625\pm (1.645)(0.00784368774876)\\\\\approx0.5625\pm0.0129\\\\=(0.5625-0.0129, \ 0.5625+0.0129)\\\\=(0.5496,\ 0.5754)$

Hence, the required 90% interval = (0.5496, 0.5754)