Statistical Analysis: Confidence Intervals and Hypothesis Testing - Prof. Raoul Lepage, Study notes of Data Analysis & Statistical Methods

Download Statistical Analysis: Confidence Intervals and Hypothesis Testing - Prof. Raoul Lepage and more Study notes Data Analysis & Statistical Methods in PDF only on Docsity! STT 200 Review 2-2-09 1. A random sample of 400 hospital admis- sions from a week's total of 5400 finds 88 were emergency contacts. Give a 98% con- fidence interval for p = rate of emergency contacts among admissions. p̀ = 88400= 22 100= 0.22 DF ¶ 1.96 2.326 p̀ ± z p̀ H1- p̀L n N-nN-1 Conf 95% 98% 2. A random sample of 36 elk selected from the Jackon, Wy. Elk Refuge in win- ter are scored for x = lead exposure finding sample mean x = 27.6 sample standard deviation s = 11.4 It is believed that x scores in this winter herd are normal distributed. Give the 80% confidence interval for population mean lead exposure m. DF 35 1.306 x ± t s n H1L ¶ Conf 80% 2. A random sample of 36 elk selected from the Jackon, Wy. Elk Refuge in win- ter are scored for x = lead exposure finding sample mean x = 27.6 sample standard deviation s = 11.4 It is believed that x scores in this winter herd are normal distributed. Give the 80% confidence interval for population mean lead exposure m. DF 35 1.306 x ± t s n H1L ¶ Conf 80% 3. What does estimated margin of error of x actually estimate? population sd s sd of the list of all possible x 1.96 s 1.96 sd of the list of all possible x 2 Lecture 2-2-09.nb 5. We have obtained estimated standard errors for sample means of concrete hard- ness 0.037 for xmixes with latex 0.042 for xmixes without latex Give the estimated margin of error for x latex-xno latex. 1.96 0.0372 - 0.0422 6. Estimate the mean and sd by eye. 80 100 120 140 0.005 0.010 0.015 0.020 0.025 Lecture 2-2-09.nb 5 6. Estimate the mean and sd by eye. 80 100 120 140 0.005 0.010 0.015 0.020 0.025 7. Amount of genetic material in a given plot is normal distributed with m = 9 s = 3 Determine the standard score z of a plot with score x = 10.5. Determine the amount x of genetic mate- rial of a plot with standard score z = 2.5. 6 Lecture 2-2-09.nb 7. Amount of genetic material in a given plot is normal distributed with m = 9 s = 3 Determine the standard score z of a plot with score x = 10.5. Determine the amount x of genetic mate- rial of a plot with standard score z = 2.5. 8. What is the exact chance that a 95% confidence interval for m will in fact cover m if the population is normal distributed and the t-CI is used? Lecture 2-2-09.nb 7 10. Determine the 86th percentile of Z. z 0.08 1.0 0.8599 IQ is normal distributed and has mean 100 and sd 15. Determine the 86th percentile of IQ. IQ = 100 + z 15 11. Determine the 86th percentile of Z. Calculate the sample standard deviation s for the list x = {0, 0, 4, 8}. avg = 12/4 = 3 sx = H0-3L 2+H0-3L2+H4-3L2+H8-3L2 4-1 = 3.82971 s4 x+9 = † 4 † sx = 4 (3.82971) 10 Lecture 2-2-09.nb 11. Determine the 86th percentile of Z. Calculate the sample standard deviation s for the list x = {0, 0, 4, 8}. avg = 12/4 = 3 sx = H0-3L 2+H0-3L2+H4-3L2+H8-3L2 4-1 = 3.82971 s4 x+9 = † 4 † sx = 4 (3.82971) 12. We've selected random samples of peo- ple with or without medication, the score being x = blood pressure decrease over a 5 minute period. Assume large populations. xwith med = 12.3 swith med = 3.2 n = 60 xwithout med = 3.7 swith med = 1.2 n = 90 Give the 95% CI for mwith med - mwithout med. (12.3 - 3.7) ± 1.96 3.2260 + 1.22 90 Lecture 2-2-09.nb 11 12. We've selected random samples of peo- ple with or without medication, the score being x = blood pressure decrease over a 5 minute period. Assume large populations. xwith med = 12.3 swith med = 3.2 n = 60 xwithout med = 3.7 swith med = 1.2 n = 90 Give the 95% CI for mwith med - mwithout med. (12.3 - 3.7) ± 1.96 3.2260 + 1.22 90 12 Lecture 2-2-09.nb