Respuesta :
Answer: C.I = 0.67523 or 0.6747
Explanation: please find the attached file for the solution.

The hypothesis for the test have been written as H1 and H0 in the answer below
Null hypothesis, H0: P = 0.50
Alternative hypothesis, H1: P ≠ 0.50
The significance level of the test is 0.05. The one-sample z-test is what is going to be used here.
How to test for the standard deviation
Standard deviation = sqrt[ P * ( 1 - P ) / n ]
= 0.00911
Find the z test statistic
z = (p - P) /S.D
z = 19.21
P is the population proportion , p is the sample proportion, and n is the number of sample.
We have a two-tailed test, the P-value gives probability of z-score being less than -19.21 or greater than 19.21.
Thus, the P-value < 0.0001.
Given that the P value is less than 0.05, we have to reject the null. There is enough evidence that they have loaded different proportions of the song.
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