Consider the graph of the linear function () -

A screening test for a newly discovered disease is being evaluated for its performance as a screening test. In order to determin

Elena-2011 [213]

Answer:

- Sensitivity = 62.7 %

- specificity = 74.9 %

- positive predictive value = 45.2 %

- negative predictive value = 85.9 %

- Accuracy of the test = 71.9 %

Step-by-step explanation:

Given the data in th question;

CONDITION POSITIVE CONDITION NEGATIVE TOTAL

TEST POSITIVE 42 51 93

TEST NEGATIVE 25 152 177

TOTAL 67 203 270

so; A=42, B=51, C=25 and D=152

Sensitivity = [A / ( A + C) ]×100= [42 /( 42 + 25)]×100 = [42 / 67] 100=

Sensitivity = 62.7 %

specificity = [D / ( D + B) ]×100= [152 /( 152 + 51)]×100 = [152 / 203] 100=

specificity = 74.9 %

positive predictive value = [A / ( A + B) ]×100 = [42 /( 42 + 51)]×100 =[42/93] 100

positive predictive value = 45.2 %

negative predictive value = [D / ( D + C) ]×100 = [152 /( 152 + 25)]×100 =[152/177] 100

negative predictive value = 85.9 %

Accuracy of the test = [A + D / ( A+B+C+D)] = [42 + 152 / ( 42+51+25+152)]

= [ 194 / 270 ]

Accuracy of the test = 71.9 %

3 0

2 years ago

The data given to the right includes data from candies, and of them are red. The company that makes the candy claims that % of

Dafna1 [17]

Using the z-distribution, the 95% confidence interval for the percentage of red candies is of (7.84%, 33.18%). Since 33% is part of the interval, there is not enough evidence to conclude that the claim is wrong.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

Researching this problem on the internet, 8 out of 39 candies are red, hence the sample size and the estimate are given by:

$n = 39, \pi = \frac{8}{39} = 0.2051$

Hence the bounds of the interval are:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2051 - 1.96\sqrt{\frac{0.2051(0.7949)}{39}} = 0.0784$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2051 + 1.96\sqrt{\frac{0.2051(0.7949)}{39}} = 0.3318$

As a percentage, the 95% confidence interval for the percentage of red candies is of (7.84%, 33.18%). Since 33% is part of the interval, there is not enough evidence to conclude that the claim is wrong.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

7 0

2 years ago

What is 3 cups doubled ...?

Jobisdone [24]

<span> 3 cups doubled would mean to multiply by 2 so it's 6 cups</span>

3 0

3 years ago

Read 2 more answers

Mike scored 10 points less than twice the

vladimir1956 [14]