Statistics question..help me answer!?
A person picked at random has 8% chance of the illness before the test is applied. A 92% change of being healthy. We are told that with an ill patient the test is positive 96% of the time. So it is negative 4% of the time with an ill patient. I.E. 4% of a false negative. We are told it is gives a false positive in 7% of healthy patients so it gives a correct negative in 93% of healthy patients. There are 4 cases.. Ill person & correct positive => 8% * 96% => 7.68% Healthy person & false positive => 92% * 7% => 6.44% Ill person & false negative => 8% * 4% => 0.32% Healthy person & correct negative 92% * 93% => 85.56% So the answer is 7.68%. To check the result.. These 4 cases should add up to 100% 7.68 + 6.44+0.32+85.56 = 100 So the numbers are correct. -- adding and multiplying percentages can be confusing if you don't understand them. Percents are hundredths. So 80% is really a proportion of 0.8. So 80% * 20% is really 0.8*0.2 = 0.16 = 16%
Statistics help?
Let X and Y be two independent random variables with variances σx² and σy². Let a, b, and c be constants. Let W = aX + bY + c The mean of W is: E(W) = E(aX + bY + c) = aE(X) + bE(Y) + c The variance of W = aX + bY + c is: Var(W) = Var(aX + bY + c) = Var(aX + bY) {because additive constants do not affect the spread of the data} = Var(aX) + Var(bY) {This is only true when X and Y are independent} = a²Var(X) + b²Var(Y) = (aσx)² + (bσy)² === ==== === ==== === ==== W = Y - X E(W) = E(Y) - E(X) = 2.001 - 2.000 = 0.001 Var(W) = Var(Y) + Var(X) = 0.001^2 + 0.002^2 = 0.000005 StdDev(W) = 0.002236068 E(Z) = E(X + Y) /2 = 1/2 * (E(X) + E(Y)) = 1/2 * (2.000 + 2.001) = 2.0005 Var(Z) = 1/4 * (Var(X) + Var(Y)) = 1/4 * (0.002^2 + 0.001^2) = 0.00000125 StdDev(Z) = 0.001118034
STATISTICS GIVE-OR-TAKE NUMBER HELP?!?!?!?!?
In a random sample of 292 hospitalizations with a diagnosis of staph infection, 126 cases were caused by drug-resistant staph. Estimate the proportion of all staph hospitalizations that are caused by drug-resistant staph (Round to four decimal places). 0.4315 (correct) Put a give-or-take number on your answer to the previous question (Round to four decimal places). (CAN'T FIND THE ANSWER)
Statistics help plz will give 10pts?
In engineering and product design, it is important to consider the weights of people so that airplanes or elevators aren't overloaded. Based on data from the National Health Survey, assume the weight of adult males in the US follows a normal distribution with a mean weight of 173 pounds and standard deviation of 31 pounds. a. If one US adult male is randomly selected, what is the probability that his weight will be greater than 180 pounds? b. How much does a US adult male have to weigh in order for his weight to be in the top 10%? c. If 10 different US adult males are randomly selected from the population above, what is the probability that their sample mean weight will be greater than 180 pounds? d. If 34 different US adult males are randomly selected from the population above, what is the probability that their sample mean weight will be greater than 180 pounds?
Can anyone help me with my statistics homework?
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STATISTICS HELP ON HW. PLEASEEEEEEEEE ANSWERR?
A drug test is accurate 97% of the time. If the test is given to 1900 people who have not taken drugs, what is the probability that at least 60 will test positive? Probability = Give your answers to at least 3 decimal places.
Statistics 101..help! will give points ASAP if right?
A report in Time magazine (April 15, 2002) stated that the average age for women to marry in the United States is now 25. If the standard deviation is assumed to be 3.2 years, find the probability that a random sample of 40 U.S. women would show a mean age at marriage of less than or equal to 24 years. Round to 4 decimal places with zero in the units place (i.e. 0.xxxx)
Statistics question? plz help.?
Mental measurements of young children are often made by giving them blocks and telling them to build a tower as tall as possible and the time taken for the block building is recorded. One experiment of block building was repeated a month later, with the times (in seconds) listed below. a) Is there sufficient evidence to support the claim that there is a difference between the two times (first versus second trial)? Use a 0.01 significance level. b) Construct a 99% confidence interval of differences. Do the confidence interval limits contain 0, indicating there is not a significant difference between the first and second trials? Child A B C D E F G H I J K L M N O Trial 1 30 19 19 23 29 178 42 20 12 39 14 81 17 31 52 Trial 2 30 6 14 8 14 52 14 22 17 8 11 30 14 17 15
HELP!! Statistics: Two-Sample Problem (Multiple Choice) **Giving 10 pts.**?
Got stuck w/ these 3 multiple choice problems! Please help me out, thanks! 1. Option 2 for conducting inference on population means is considered conservative because: A. it requires less computational work than Option 1 does. B. it provides higher P-values and wider confidence intervals than Option 1. C. it provides lower P-values and narrower confidence intervals than Option 1. D. None of the above. 2. The rationale for avoiding the pooled procedures for inference is that: A. testing for the equality of variances is an unreliable procedure that is not robust. B. the two-sample t, or “unequal variances procedure,” is valid regardless of whether or not the two variances are actually unequal. C. the two-sample t, or “unequal procedure,” is almost always more accurate than the pooled procedure. D. All of the above. 3. Which of the following procedures is not robust to non-normality? A. The one-sample t test B. The t test for matched pairs C. The two-sample t test D. The F test for comparing two population standard deviations.
Need help with Statistics Wordproblem. Its giving me hell.?
The aerage height of males in the population is 68 inches with sd=2.4inches. in a sample of 75 men from Peru the aveage height is 64inches. Is the height of men from peru significantly different from the population of men?