Question

A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 80%.
At the .05 level of significance, it can be concluded that the proportion of the population in favor of candidate A is____.

i) significantly greater than 80%
ii) not significantly greater than 80%
iii) significantly greater than 85%
iv) not significantly greater than 85%
v) None of the above

Answer 1

Answer:

$z=\frac{0.85 -0.8}{\sqrt{\frac{0.8(1-0.8)}{100}}}=1.25$  

$p_v =P(z>1.25)=0.106$  

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 who favored candidate A  is not significalty higher than 0.8

ii) not significantly greater than 80%

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

$\hat p=0.85$ estimated proportion of people who favored candidate A

$p_o=0.8$ 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 higher than 0.8.:  

Null hypothesis:$p\leq 0.8$  

Alternative hypothesis:$p > 0.8$  

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

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

$z=\frac{0.85 -0.8}{\sqrt{\frac{0.8(1-0.8)}{100}}}=1.25$  

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 right tailed test the p value would be:  

$p_v =P(z>1.25)=0.106$  

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ 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 who favored candidate A  is not significalty higher than 0.8

ii) not significantly greater than 80%