Question

Trials in an experiment with a polygraph include 98 results that include 23 cases of wrong results and 75 cases of correct results. Use a 0.01 significance level to test the claim that such polygraph results are correct less than 80​% of the time. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method. Use the normal distribution as an approximation of the binomial distribution.

Answer 1

Answer:

$z=\frac{0.765 -0.8}{\sqrt{\frac{0.8(1-0.8)}{98}}}=-0.866$  

$p_v =P(z$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ 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 1% of significance the proportion of correct cases is not significantly lower than 0.8 or 80%

Step-by-step explanation:

Data given and notation

n=98 represent the random sample taken

X=75 represent the number of correct cases

$\hat p=\frac{75}{98}=0.765$ estimated proportion of correct cases

$p_o=0.8$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

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 proportions is lower than 0.8 or not.:  

Null hypothesis:$p \geq 0.8$  

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:  

$z=\frac{0.765 -0.8}{\sqrt{\frac{0.8(1-0.8)}{98}}}=-0.866$  

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 $\alpha=0.01$. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

$p_v =P(z$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.01$ 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 1% of significance the proportion of correct cases is not significantly lower than 0.8 or 80%