Answer: [tex]=(0.06573,\ 0.09427)[/tex]
Step-by-step explanation:
The confidence interval for population proportion (p) is given by :_
[tex]\hat{p}\pm z^* \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where n= sample size
z* = Critical value.
[tex]\hat{p}[/tex] = Sample proportion.
Let p be the true population proportion of visitors that click on the advertisement.
As per given , we have
n= 978
[tex]\hat{p}=0.08[/tex]
Critical value for 90% confidence interval : z* = 1.645 [ By z-table]
Now , 90% confidence interval for the population proportion of visitors that click on the advertisement:
[tex]0.08\pm (1.645) \sqrt{\dfrac{0.08(1-0.08)}{978}}[/tex]
[tex]0.08\pm (1.645) \sqrt{0.0000752556237219}[/tex]
[tex]0.08\pm (1.645) (0.00867499992633)[/tex]
[tex]\approx0.08\pm(0.01427)[/tex]
[tex]=(0.08-0.01427,\ 0.08+0.01427)=(0.06573,\ 0.09427)[/tex]
Hence, a 90% confidence interval for the population proportion of visitors that click on the advertisement. [tex]=(0.06573,\ 0.09427)[/tex]