Question

What is your favorite color? A larger survey of countries, including the United States, China, Russia, France, Turkey, Kenya, and others, indicated that most people prefer the color blue. In fact, about 24% of the population claim blue as their favorite color. Suppose a random sample of n = 75 college students were surveyed and x = 19 of them said that blue is their favorite color. Does this information imply that the proportion of college students who prefer blue differs from that of the general population? Use ???? = 0.05.

Answer 1

0.792

Explanation:

Answer 2

`z=\frac{0.253 -0.24}{\sqrt{(0.24(1-0.24))/(75)}}=0.264`

`p_v =2*P(z>0.264)=0.792`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of students said that blue is their favorite color is not different from 0.24

Explanation:

Data given and notation

n=75 represent the random sample taken

X=19 represent the students said that blue is their favorite color

`\hat p=(19)/(75)=0.253` estimated proportion of students said that blue is their favorite color

`p_o=0.24` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

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 is different from 0.24.:

Null hypothesis:
`p=0.24`

Alternative hypothesis:
`p \\eq 0.24`

When we conduct a proportion test we need to use the z statisitc, 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{0.253 -0.24}{\sqrt{(0.24(1-0.24))/(75)}}=0.264`

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 provided
`\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>0.264)=0.792`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of students said that blue is their favorite color is not different from 0.24