Question

A husband and wife ted and suzanne share a digital music player that has a feature that randomly selects which song to play. A total of 3476 songs have been loaded into the player, some by Ted and some by Susanne. They are interested in determining whether thay have each loaded a different proportion of songs into the player. Supposed that when the player was in the random selection mode, 22 of the first 30 songs selected were songs loaded by Suzanne. Let p denote the proprtion of songs that were loaded by Suzanne. State the null and alternative hypothesis to be tested. How strong is the evidence that Ted and Suzznne have each loaded a different proportion of songs into the player

Answer 1

`z=\frac{0.733 -0.5}{\sqrt{(0.5(1-0.5))/(30)}}=2.552`

`p_v =2*P(z>2.552)=0.0107`

So the p value obtained was a very low value and using the significance level assumed
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of songs selectd at random differs from 0.5 .

There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs from 0.5.

Explanation:

Data given and notation

n=30 represent the random sample taken

X=22 represent the number of songs selectd at random

`\hat p=(22)/(30)=0.733` estimated proportion of songs selectd at random

`p_o=0.5` is the value that we want to test

`\alpha` represent the significance level

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of random songs selected is 0.5 (random) or no.:

Null hypothesis:
`p=0.5`

Alternative hypothesis:
`p \\eq 0.5`

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.733 -0.5}{\sqrt{(0.5(1-0.5))/(30)}}=2.552`

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed for this case is
`\alpha=0.05`. The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

`p_v =2*P(z>2.552)=0.0107`

So the p value obtained was a very low value and using the significance level assumed
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of songs selectd at random differs from 0.5 .

There is strong evidence that the proportion of songs downloaded by Ted and Suzanne differs from 0.5.