Question

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

Answer:

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

$p_v =2*P(z>0.264)=0.792$  

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 people said that blue is their favorite color is not differetn from 0.24 or 24% at 5% of significance.

Step-by-step explanation:

Data given and notation

n=75 represent the random sample taken

X=19 represent the number of people said that blue is their favorite color

$\hat p=\frac{19}{75}=0.253$ estimated proportion of people 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 differs from 0.24 nor no.:  

Null hypothesis:$p=0.24$  

Alternative hypothesis:$p \neq 0.24$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

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

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 people said that blue is their favorite color is not differetn from 0.24 or 24% at 5% of significance.