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marin [14]
2 years ago
14

A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in deter

mining 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
Mathematics
1 answer:
lbvjy [14]2 years ago
8 0

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%

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