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

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

`p_v =P(z<-0.866)=0.193`

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%

Explanation:

Data given and notation

n=98 represent the random sample taken

X=75 represent the number of correct cases

`\hat p=(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{(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{(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<-0.866)=0.193`

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%