Answer: (0.0654, 0.1184)
Step-by-step explanation:
The confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^* \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] (1)
, where [tex]\hat{p}[/tex] = Sample proportion
n= Sample size
z* = Critical z-value.
Let p be the population proportion of orders that are filled accurately in the ethnic fast.
As per given , we have
n= 457
[tex]\hat{p}=\dfrac{42}{ 457 }=0.0919[/tex]
Critical z-value for 95% confidence interval = 1.96
Put all values in (1) , we get
[tex]0.0919\pm(1.96)\sqrt{\dfrac{0.0919(1-0.0919)}{457}}[/tex]
[tex]0.0919\pm(1.96)\sqrt{0.0001826}[/tex]
[tex]0.0919\pm(1.96)(0.013513)[/tex]
[tex]0.0919\pm0.02648548[/tex]
[tex]=(0.0919-0.02648548,\ 0.0919+0.02648548) =(0.06541452,\ 0.11838548)\\\\=\approx(0.0654,\ 0.1184)[/tex]
Hence, the 95% confidence interval for the population proportion of orders that are filled accurately in the ethnic fast is (0.0654, 0.1184) .