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Arturiano [62]
2 years ago
5

A certain affects virus 0.8% of the population. A test used to detect the virus in a person is positive 86% of the time if the p

erson has the virus (true positive) and 15% of the time if the person does not have the virus (false postive). Fill out the remainder of the following table and use it to answer the two questions below.
Infected Not Infected Total
Positive Test
Negative Test
Total 800 99,200 100,000
a) Find the probability that a person has the virus given that they have tested positive. Round your answer to the nearest tenth of a percent and do not include a percent sign.
P(Infected | Positive Test)=
%
b) Find the probability that a person does not have the virus given that they test negative. Round your answer to the nearest tenth of a percent and do not include a percent sign.
P(Not Infected | Negative Test) =
%
Mathematics
1 answer:
Klio2033 [76]2 years ago
7 0

Answer:

<h3>I can't help <em><u>sorry</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u> </u></em><em><u>.</u></em><em><u>bru</u></em></h3>
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t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638    

p_v =P(t_{(19)}>1.638)=0.0589    

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 to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.    

Step-by-step explanation:

Data given and notation    

\bar X=7.03 represent the sample mean

s=9.42 represent the sample standard deviation    

n=20 sample size    

\mu_o =3.58 represent the value that we want to test  

\alpha=0.01 represent the significance level for the hypothesis test.    

z 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 higher than 3.58 :    

Null hypothesis:\mu \leq 3.58    

Alternative hypothesis:\mu > 3.58    

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

t-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:    

t=\frac{7.03-3.58}{\frac{9.42}{\sqrt{20}}}=1.638    

P-value    

First we need to calculate the degrees of freedom given by:

df=n-1=20-1=19

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

p_v =P(t_{(19)}>1.638)=0.0589    

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 to reject the null hypothesis, and we can conclude that the true mean is not significantly higher than 3.58 at 5% of signficance.    

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