Question

Trinity takes a sample of 50 songs and finds that 20 are by a female artist. Based on this problem, which of the following is a 99% confidence interval for the proportion of songs on her phone by a female artist?

Answer 1

Answer: (0.2215, 0.5785)

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 :-

`\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}` ...(i)

As per given , Sample size : n= 50

Sample proportion of songs on her phone by a female artist.
`\hat{p}=(20)/(50)=0.4`

For 99% confidence level , critical z-value =2.576

Substituting values in (i), we get

`0.4\pm 2.576(\sqrt{(0.4(1-0.4))/(50)})\\\\=0.4\pm 2.576√(0.0048)\\\\\approx0.4\pm0.1785\\\\=(0.4-0.1785,\ 0.4+0.1785)\\\\=(0.2215,\ 0.5785)`

Hence, a 99% confidence interval for the proportion of songs on her phone by a female artist : (0.2215, 0.5785)