Answer: (0.2215, 0.5785)
Step-by-step explanation:
Given, Trinity takes a sample of 50 songs and finds that 20 are by a female artist.
Let p be the proportion of songs on her phone by a female artist.
Then , the 99% confidence interval for p would be :-
[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] ...(i)
As per given , Sample size : n= 50
Sample proportion of songs on her phone by a female artist. [tex]\hat{p}=\dfrac{20}{50}=0.4[/tex]
For 99% confidence level , critical z-value =2.576
Substituting values in (i), we get
[tex]0.4\pm 2.576(\sqrt{\dfrac{0.4(1-0.4)}{50}})\\\\=0.4\pm 2.576\sqrt{0.0048}\\\\\approx0.4\pm0.1785\\\\=(0.4-0.1785,\ 0.4+0.1785)\\\\=(0.2215,\ 0.5785)[/tex]
Hence, a 99% confidence interval for the proportion of songs on her phone by a female artist : (0.2215, 0.5785)
