Question

Asked May 13, 2019

75 views

Step 1

Given incidence of urinary tract infections = 13.32%

Nitride test for urinary tract infections sensitivity = 26.7%

NItride test for urinary tract infections specificity = 93.3%

The total women with urinary tract infections among 100,000 women = 13320.

The total women without urinary tract infection among 100,000 women = 86680

The total women with urinary infection and test positive = 13320*0.267 = 3556.44

The total women wth urinary nifection and test negative = 13320 - 3556.44 = 9763.56

The total women without Urinary infection and test negative = 86680*0.933 = 80872.44

The total women without Urinary infection and test positive = 86680 - 80872.44 = 5807.56

All the following calculated values are shown in table form below

Step 2

From the below table

Probability of being with Urinary tract infection given that test is positive P(With Infection | Test positive) = 3556.44/9364

Probability of being without Urinary tract infection given that test is positive P( Without Infection | Test positive) = 5807.56/9364

Estimated positive predictive value is nothing but the probability of being with Urinary tract infection given that test is positive

P(With Infection | Test positive) = 3556.44/9364 = 0.379799 = 0.3798 (rounding to four decimal points)

Step 3

From the below table

Probability of being with Urinary tract infection given that test is negative P(With Infection | Test negative) = 9763.56/90636

Probability of being without Urinary tract infection given that test is negative P( Without Infection | Test negative) = 80872....

Tagged in

Find answers to questions asked by student like you

Show more Q&A

Q: A marketing manager for a cell phone company claims that more than 55% of children aged 8-12 have ce...

A: Null and alternative hypotheses:Here, to check whether more than 55% of children aged 8-12 have cell...

Q: A therapist believes that the music a child listens to may have an impact on the number of words on ...

A: Given Null hypothesis H0 : µd = 0 Alternative hypothesis Ha: µd ≠ 0A sample size of 19 students is c...

Q: A sample of 50 women is obtained, and their heights (in inches) and pulse rates (in beats per mi...

A: According to the provided data, the variable x represents the height of the women. The sample size, ...

Q: A simple random sample of 60 items from a population with o 9 resulted in a sample mean of 36. If re...

A: Confidence interval:The (1-α) % confidence interval for population mean µ is given by:

Q: The table below shows the critical reading scores for 14 students the first two times they took a st...

A: Given the data of critical reading scores for 14 students the first two times they took a standardiz...

Q: I am working on constructing a histogram for some data for the 39 US national parks under 900,000 ac...

A: According to the given data41 ,66 ,233 ,775 ,169, 36 ,338 ,233 ,236, 64,183 ,61 ,13 ,308 ,77,520, 77...

Q: In a population of ages where the mean age is 45 and the standard deviation is 5.25, what are the ag...

A: Given mean age = 45Standard deviation = 5.25The Z score with 0.0505 to its left is -1.64 and the Z s...

Q: Let x be a random variable that represents hemoglobin count (HC) in grams per 100 milliliters of who...

A: (A)The sample mean and standard deviation is calculated below:

Q: Two dice are rolled. Find the probability of getting A. A sum of 5,6,7 B. Doubles or sum of 6 or 8 ...

A: Sample space for rolling two dice:It is given that two dice are rolled. The number of possible outco...