Answer: (0.00306,0.00334)
Step-by-step explanation:
Given : Sample size : n= 420000
The number of users developed cancer of the brain or nervous system =134
Then , the proportion of users developed cancer of the brain or nervous system :
[tex]p=\dfrac{134}{420000}\approx0.0032[/tex]
Significance level : [tex]1-0.90=0.1[/tex]
Critical value : [tex]z_{\alpha/2}=1.645[/tex]
The confidence interval for population proportion is given by :-
[tex]p\ \pm\ z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.0032\pm (1.645)\sqrt{\dfrac{(0.0032)(1-0.0032)}{420000}}\approx0.0032\pm0.00014\\\\=(0.00306,0.00334)[/tex]
Hence, the 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system is (0.00306,0.00334).
