Question

A forensic psychologist studying the accuracy of a new type of polygraph (lie detector) test instructed a participant ahead of time to lie about some of the questions asked by the polygraph operator. On average, the current polygraph test is 75% accurate, with a standard deviation of 6.5%. With the new machine, the operator correctly identified 83.5% of the false responses for one participant. Using the.05 level of significance, is the accuracy of the new polygraph different from the current one? Fill in the following information: Assuming an ?-0.05, determine the z-score cutoff for the rejection region. Calculate the test statistic for the given data Zobt Based on the data above, finish the statement about your decision: Based on the observed z-score, we would decide to (accept, reject, fail to reject, fail to accept) hypothesis. the (null, alternative)

Answer 1

Answer:

$z=\frac{83.5-75}{6.5}=1.31$    

The rejection zone for this case would be:

$z> 1.96 \cup Z$

$p_v =2*P(z>1.31)=0.1901$  

If we compare the p value 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 reject the null hypothesis, so we can't conclude that the true mean is significantly different from 75 at 5% of significance.

Step-by-step explanation:

Data given and notation  

$\bar X=83.5$ represent the sample mean

$\sigma=6.5$ represent the population standard deviation for the sample  

$\mu_o =75$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 75, the system of hypothesis would be:  

Null hypothesis:$\mu = 75$  

Alternative hypothesis:$\mu \neq 75$  

For this case we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\sigma}$  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{83.5-75}{6.5}=1.31$    

Cutoff for the rejection regon

Since our significance level is $\alpha=0.05$ and we are conducting a bilateral test we need to find a quantile in the standard normal distribution that accumulates 0.025 of the area on each tail.

And for this case those values are $z_{crit}= \pm 1.96$

So the rejection zone for this case would be:

$z> 1.96 \cup Z$

Our calculated value is not on the rejection zone. So we fail to reject the null hypothesis.

P-value

Since is a two sided test the p value would be given by:  

$p_v =2*P(z>1.31)=0.1901$  

Conclusion  

If we compare the p value 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 reject the null hypothesis, so we can't conclude that the true mean is significantly different from 75 at 5% of significance.