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OLga [1]
2 years ago
13

A polygraph (lie detector) is an instrument used to determine if the individual is telling the truth. These tests are considered

to be 86% reliable. In other words, if an individual lies, there is a 0.86 probability that the test will detect a lie. Let there also be a 0.070 probability that the test erroneously detects a lie even when the individual is actually telling the truth. Consider the null hypothesis, "the individual is telling the truth," to answer the following questions.
a. What is the probability of Type I error? (Round your answer to 3 decimal places.)


Probability


b. What is the probability of Type II error? (Round your answer to 2 decimal places.)


Probability
Mathematics
2 answers:
zhannawk [14.2K]2 years ago
7 0

Answer:

Step-by-step explanation:

a) The probability of a Type I error in a lie detection test would be the probability that the lie detection machine incorrectly detected lie for the truth tellers. This is already given in the problem as 0.07.

Therefore,

P(Type-I) = 0.07

Therefore 0.07 is the required probability here.

b) The probability of a Type II error in a lie detection test would be the probability that the lie detection machine incorrectly detected truth for the the people who are actually liars. This is thus 1 - reliability.

P(Type-II) = 1 - Reliability = 1- 0.86 = 0.14

Therefore 0.14 is the required probability here.

STatiana [176]2 years ago
6 0

Answer:

a) 0.070

b) 0.14

Step-by-step explanation:

Given that the tests are 86% reliable, i.e a probability of 0.86 a lie would be detected.

Probability of error = 0.070

a) For type I error, we have:

The probability of a type I error in this lie detector is the probability that the test erroneously detects a lie even when the individual is actually telling the truth, i.e

P(type I error) = P(rejecting true null)

= 0.070

b) The probability of a Type II error this lie detectot is the probability that the test erroneously detected truth insteax of lie.

i.e = 1 - reliability

P (Type II error) = P(Failing to reject false Null)

= P(Not detecting a lie)

= 1-0.86

= 0.14

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x = [-b +-sq root (b² -4ac)] / 2a
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Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad
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Answer:

17-2.262\frac{1.9}{\sqrt{10}}=15.641    

17+2.262\frac{1.9}{\sqrt{10}}=18.359    

So on this case the 95% confidence interval would be given by (15.641;18.359)    

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=17 represent the sample mean

\mu population mean (variable of interest)

s=1.9 represent the sample standard deviation

n=10 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=10-1=9

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that t_{\alpha/2}=2.262

Now we have everything in order to replace into formula (1):

17-2.262\frac{1.9}{\sqrt{10}}=15.641    

17+2.262\frac{1.9}{\sqrt{10}}=18.359    

So on this case the 95% confidence interval would be given by (15.641;18.359)    

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

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