Question

Questions: I) 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

Answer 1

a)

The hypothesis

H0: p1 = p2

H1: p1 ≠ p2

He test procedure, called the two-proportion z-test.

Then the following conditions

• Each population is at least 20 times as big as its sample.
• The samples are independent.
• Each sample includes at least 10 successes and 10 failures.
• The sampling method for each population is simple random sampling.

We don't have the sampling proportion for Stan , so the conditions are not met.

(b)

Yes, the conditions are met.

p = 27/40 = 0.675

n = 40

95% confidence interval will be:

= p ± z*(√p(1-p)/n

= 0.675 ± 1.965*(√0.675(1-0.675)/40

= (0.5295, 0.8205)

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