Answer: (0.205905, 0.242095)
Step-by-step explanation:
We know that the confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where [tex]\hat{p}[/tex]= sample proportion
n= sample size
Let p be the proportion of complaints of identity theft .
As per given , we have
n= 1432
[tex]\hat{p}=\dfrac{321}{1432}\approx0.224[/tex]
By standard normal table , z* = 1.645 for 90% confidence level.
Now , 90% confidence interval for p will be :
[tex]0.224\pm (1.645)\sqrt{\dfrac{0.224(1-0.224)}{1432}}\\\\=0.224\pm (1.645)(0.011)\\\\=0.224\pm 0.018095\\\\=(0.224-0.018095,\ 0.224+0.018095)\\\\=(0.205905,\ 0.242095)[/tex]
Hence, required confidence interval = (0.205905, 0.242095)