Question

A husband and wife, Ed and Rena, 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 Ed and the rest by RIna. They are interested in determining whether they have loaded different proportions of songs into the player. Suppose that when the player was in random-selection mode, 34 of the first 50 songs selected songs were loaded by Rena. Let "p" denote the proportion of the songs that were loaded by Rina.

A) State the null and alternative hypothesis to be tested. How strong is the evidence that Ed and Rina have loaded different proportions of songs into the player? Make sure to check the conditions for the use of this test.
B)Are the conditions for the use of the large sample confidence interval met? If so, estimate with 95% confidence the proportions of the songs that were loaded by Rina.

Answer 1

a) Null hypothesis:
`p=0.5`

Alternative hypothesis:
`p \\eq 0.5`

`p_v =2*P(z>2.55)=0.0107`

If we compare the p value obtained and the significance level given
`\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 the songs that were loaded by Rena is different from 0.5.

b) The 95% confidence interval would be given (0.551;0.809).

Explanation:

1) Data given and notation

n=50 represent the random sample taken

X=34 represent the songs that were loaded by Rena

`\hat p=(34)/(50)=0.68` estimated proportion of songs that were loaded by Rena

`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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is not equal to 0.5.:

Null hypothesis:
`p=0.5`

Alternative hypothesis:
`p \\eq 0.5`

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

`np_o =50*0.5=25>10`

`n(1-p_o)=50*(1-0.5)=25>10`

So the conditions are meeted.

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`.

3) Calculate the statistic

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

`z=\frac{0.68 -0.5}{\sqrt{(0.5(1-0.5))/(50)}}=2.55`

4) 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 is not provided but we can assume
`\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>2.55)=0.0107`

If we compare the p value obtained and the significance level given
`\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 the songs that were loaded by Rena is different from 0.5.

5) Confidence interval

The conditions ar satisfied and explained before.

The confidence interval would be given by this formula

`\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}`

For the 95% confidence interval the value of
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`0.68 - 1.96 \sqrt{(0.68(1-0.68))/(50)}=0.551`

`0.68 + 1.96 \sqrt{(0.68(1-0.68))/(50)}=0.809`

And the 95% confidence interval would be given (0.551;0.809).