Question

] It is claimed that 42% of US college graduates had a mentor in college. For a sample of college graduates in Colorado, it was found that 502 out of 1045 had a mentor in college. Test the claim that the proportion of college grads in Colorado who had a mentor is greater than that of all US college grads. Set up a sampling distribution of proportions.

Answer 1

Answer:

$z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930$  

The p value for this case would be given by:

$p_v =P(z>3.930)=0.0000443$  

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42

Step-by-step explanation:

Information given

n=1045 represent the random sample selected

X=502 represent the college graduates with a mentor

$\hat p=\frac{502}{1045}=0.480$ estimated proportion of college graduates with a mentor

$p_o=0.42$ is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true proportion is higher than 0.42, the system of hypothesis are.:  

Null hypothesis:$p \leq 0.42$  

Alternative hypothesis:$p > 0.42$  

The statistic is given by:

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

Replacing the info we got:

$z=\frac{0.480 -0.42}{\sqrt{\frac{0.42(1-0.42)}{1045}}}=3.930$  

The p value for this case would be given by:

$p_v =P(z>3.930)=0.0000443$  

For this case the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis so then we can conclude that the true proportion is significantly higher than 0.42