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Calculate the value based on this information. A new screening test for Lyme disease is developed for use in the general population. The sensitivity (ability to identify infected persons) is 60% and specificity (ability to identify non- infected persons) is 70%. 300 people are screened (assume the true prevalence of Lyme disease is 10%) The predictive value of a positive test is:
Given that Vacationer will travel less than 500 miles, Find out the probability that he/she will drive or take the Bus/Train?
The time required to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions. a) What is the probability of completing the..
What is the probability for the given frequency distribution?
What does confidence interval tell you? What does it mean? What 2 things can a researcher do to decrease width of a confidence interval? What are the effects?
Given the following data, 1) how do you make Excel create a scatter diagram for you? 2) If you were to draw a line through the points, how straight would the line be?
A recent study found that the average life expectancy of a person living in Africa is 53 years with a standard deviation of 7.5 years. If a person in Africa is selected at random
Which of these variables are discrete and which are continuous random variables?
Estimating the individual score using normal distribution - blood pressure higher than what amount?
Suppose that, from a population of 75 bank accounts, we want to take a random sample of six accounts in order to learn about the population. How many different random samples of six accounts are possible?
Use the trend equation to calculate the points for 2003 and 2005. Plot them on the graph and draw the regression line.
Suppose that it is important to demonstrate that the wall thickness exceeds 4.0 mm. Test appropriate hypotheses using these data. Draw conclusions at α= 0.05.
According to study done by Nick Wilson of Otago University, the probability a randomly selected individual will not cover his or her mouth when sneezing was 0.267. Suppose you sit on a mall bench and observe 200 people pass by as they sneeze.
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