Ted and Suzanne load 3476 songs into a player that randomly selects which song to play.

When in random-selection mode, 22 of the first 30 songs were loaded by Suzanne.
T and S want to determine if each have loaded a different proportion of songs.
Let p denote proportion of songs loaded by Suzanne.

A) State null and alternative hypotheses

B) How strong is the evidence that Ted and Suzanne have each loaded a different proportion of songs?

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A) Null hypothesis (H0): The proportion of songs loaded by Suzanne is equal to the proportion of songs loaded by Ted (p = 0.5).

Alternative hypothesis (Ha): The proportion of songs loaded by Suzanne is different from the proportion of songs loaded by Ted (p ≠ 0.5).

B) To determine the strength of the evidence, we can use a hypothesis test. In this case, we will use a two-proportion z-test.

First, we count the number of songs loaded by Suzanne (22) and the number of songs loaded by Ted (30 - 22 = 8) in the first 30 songs played.

Given:
Number of successes for Suzanne (x1) = 22
Number of trials for Suzanne (n1) = 30
Number of successes for Ted (x2) = 8
Number of trials for Ted (n2) = 30

We can calculate the sample proportions for Suzanne (p̂1) and Ted (p̂2):

p̂1 = x1 / n1 = 22 / 30 = 0.7333
p̂2 = x2 / n2 = 8 / 30 = 0.2667

Next, we calculate the standard error (SE) of the difference in proportions:

SE = sqrt( (p̂1 * (1 - p̂1)) / n1 + (p̂2 * (1 - p̂2)) / n2 )

SE = sqrt( (0.7333 * (1 - 0.7333)) / 30 + (0.2667 * (1 - 0.2667)) / 30 )
≈ sqrt(0.0213 + 0.01830)
≈ sqrt(0.0396)
≈ 0.1987

Now, we calculate the z-score using the formula:

z = (p̂1 - p̂2) / SE

z = (0.7333 - 0.2667) / 0.1987 ≈ 2.3612

We then compare the z-score to the critical value at a chosen significance level (e.g., α = 0.05) to determine if we reject or fail to reject the null hypothesis.

Looking up the critical value of z at α = 0.05, we find it to be approximately ±1.96 for a two-tailed test.

Since the calculated z-score of 2.3612 is greater than 1.96, we reject the null hypothesis. This means there is evidence to suggest that Ted and Suzanne have loaded a different proportion of songs into the player.

The strength of the evidence is determined by the magnitude of the z-score. A larger z-score indicates stronger evidence against the null hypothesis. In this case, the z-score of 2.3612 suggests relatively strong evidence that the proportions are different.