ading music. A husband and wife, Stan and Lucretia, share a digital music player that has a feature that randomly selects which song to play. A total of 2643 songs have been loaded into the player, some by Stan and the rest by Lucretia. They are interested in determining whether they have loaded different proportions of songs into the player. Suppose that when the player was in the random-selection mode, 27 of the first 40 songs selected were songs loaded by Lucretia. Let p denote the proportion of songs that were loaded by Lucretia. State the null and alternative hypotheses to be tested. How strong is the evidence that Stan and Lucretia have loaded different proportions of songs into the player? Make sure to check the conditions for the use of this test. Are the conditions for the use of the large-sample confidence interval met? If so, estimate with 95% confidence the proportion of songs that were loaded by Lucretia.

Respuesta :

Answer: C.I = 0.67523 or 0.6747

Explanation: please find the attached file for the solution.

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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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