Number of results: 32
college math - biostatistics
hey... first time user and really lost. wondering if someone can help: Based on data from the Statistical Abstract of the United States, 112th Edition, only about 14% of senior citizens (65 years old or older) get the flu each year. However, about 24% of the people under 65 ...
Wednesday, February 18, 2009 at 2:54am by will
college math - biostatistics
a)(0.125) x (0.14) = ? b)(0.875) x (0.24) = c)(0.50)x(0.14) + (0.50)x(0.24) = ?
Wednesday, February 18, 2009 at 2:54am by drwls
if we select a birth at random, what is the probability the mother is 19 years of age or younger
Tuesday, March 23, 2010 at 3:14pm by Brenda
What groups are you considering? What country? What areas of the country? What socioeconomic levels? More data needed.
Tuesday, March 23, 2010 at 3:14pm by PsyDAG
Age(10-14)(15-19)(20-29)(30-39) (40-44) Mid 12.5 17.5 25 35 42.5 P(X=x) .003 .100 .500 .300 .097 If we select a birth at random, what is the probability the mother is 19 years or younger?
Sunday, March 28, 2010 at 9:25pm by Louise
I'm not sure what you mean by "P(X=x). If it is probability, add the probabilities for the two lower categories to indicate the probability that it is either (10-14) or (15-19). Does "Mid" refer to median? I hope this helps. If not, repost your question in clearer terms.
Sunday, March 28, 2010 at 9:25pm by PsyDAG
if the mean bp is 110 and the standard deviation is 15 where would 95% of population be
Sunday, October 3, 2010 at 10:29am by donna
http://davidmlane.com/hyperstat/z_table.html Try in the region of 135
Sunday, October 3, 2010 at 10:29am by bobpursley
Also you can find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion indicated. Insert Z score value in the formula below and solve for the score. Z = (score-mean)/SD = (score-110)/15 This method does not ...
Sunday, October 3, 2010 at 10:29am by PsyDAG
17. A research biologist has carried out an experiment on a random sample of 15 experimental plots in a field. Following the collection of data, a test of significance was conducted under appropriate null and alternative hypotheses and the P-value was determined to be ...
Wednesday, June 1, 2011 at 3:23pm by Anonymous
among coffee drinkers, men drink a mean of 3.2 cups per day with a standard deviation of 0.8 cups. Assume the number of coffee drinks per day follows a normal distribution. a. What proportion drinks 2 cups per day or more? b. What proportion drink no more than 4 cups per day? ...
Friday, November 4, 2011 at 2:29pm by natasha
Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores. For c, work backwards from the Z score for that proportion.
Friday, November 4, 2011 at 2:29pm by PsyDAG
A studyis run to estimate the mean total cholesterol level in children 2 to 6 years of age. A sample of 9 participants is selected and their total cholesterol levels are measured as follows: 185 225 240 196 175 180 194 147 223 Generate a 95% confidence interval for the true ...
Monday, November 7, 2011 at 10:00am by natasha
95% confidence interval = mean ± 1.96 SEm SEm = SD/√n Find the mean first = sum of scores/number of scores Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance. Standard ...
Monday, November 7, 2011 at 10:00am by PsyDAG
Suppose we want to design a new placebo-controlled trial to evaluate an experimental medication to increase lung capacity. The primary outcome is peak expiratory flow rate, a continuous variable measured in liters per minute. the primary outcome will be measured after 6 months...
Saturday, November 19, 2011 at 6:58pm by Natasha
8. The following data were collected in a clinical trial to compare a new drug to a placebo for its effectiveness in lowering total serum cholesterol. Generate a 95% confidence interval for the difference in mean total cholesterol levels between treatments. New Drug (n=75) ...
Saturday, November 26, 2011 at 10:42am by natasha
7. Consider again the data in problem #6. Suppose that in the aerobic exercise group we also measured the number of hours of aerobic exercise per week and the mean is 5.2 hours with a standard deviation of 2.1 hours. The sample correlation is -0.42. a)Is there evidence of a ...
Saturday, November 26, 2011 at 10:51am by natasha
A clinical trial is run to assess the effects of different forms of regular exercise on HDL levels in persons between the ages of 18 and 29. Participants in the study are randomly assigned to one of three exercise groups - Weight training, Aerobic exercise or Stretching/Yoga...
Saturday, November 26, 2011 at 11:50am by natasha
Everybody in my class got their paper work they have to do by the end of the week.I got mine too,so i would like you to help me with this :D? I got a paper with instructions: There are 8 students,each with different IQ.Those students recieved 10 medical books,(medical because ...
Tuesday, January 10, 2012 at 5:15am by Danijel
Why are the results of an independent samples test different from the results of a Tukey HSD test on the same data?
Monday, March 19, 2012 at 4:15pm by Lindsey