STP 31 Example EXAM # Instructor: Ela Jackiewicz Honor Statement: I have neither given nor received information regarding this exam, and I will not do so until all exams have been graded and returned. PRINTED NAME: CLASS TIME: Signed Date: DIRECTIONS: This is a closed book examination. You may use a graphing calculator and an 8X11 page with hand written notes, no completely solved problems are allowed (one side only) Turn in the notes with your exam. Formulas and tables are provided on the last pages of the test. There are 16 problems. In problems #1=#4 you need to show all work by hand, do not use a calculator t test procedure or t interval procedure. Problems #5 #11 are multiple choice. Place letter answers for these problems in the table below. Questions #1 #16 are for 5 points extra credit, circle T or F in the table below. Relax and good luck! PLACE ANSWERS TO THE MULTIPLE CHOICE QUESTIONS BELOW, USE Letters A E as appropriate QUESTION Question#5 Question#6 Question#7 Question#8 ANSWER A B B B QUESTION Question#9 Question#10 Question#11 ANSWER A B C Extra credit questions: QUESTION Question#1 Question#13 Question#14 Question#15 Question#16 ANSWER (circle one) F F T F T
For Question #1 #4 show all work by hand using only basic functions of you calculator. Use following in questions #1 and # Sampling distribution of x Ayrshire cows average 47 lbs of milk a day with a standard deviation of 6 lbs of milk/day. Assume that the milk production for this breed has slightly right skewed distribution. Use Central Limit Theorem to answer questions #1 and #: Question#1 (6 points) Suppose we take samples of size 3 from the population of Ayrshire cows what is the distribution of the mean milk production for these samples, give mean and SD of that distribution. Distribution of x is approximately normal with μ x =47 and σ x = 6 3 1.061 Question# (10 points) If we randomly select 3 cows, compute the probability that the mean production level in a sample is less than 45 lbs of milk/day. P( x<45)=p(z < 1.89)=0.094 σ x = 6 3 1.061 z= 45 47 1.061 1.89 ANSWER: 0.094 Question#3 (8 points) Confidence interval for µ A random sample of 16 Jersey cows have a mean daily production of 43 lbs of milk/day with a standard deviation of 5 lbs of milk/day. Assume that normal models describe milk production for this breed. From that random sample of 16 cows construct a 90% confidence level for µ the population mean amount of milk produced. 43±1.753 5 16 Df=15, t 0.05 =1.753 ANSWER: (40.81, 45.19)
Question #4 (0 points) Body temperature. Summary statistics for female versus male body temperatures in Fahrenheit degrees for two independent samples of adult Americans are given below: Gender Sample size Sample mean body temperature Sample standard deviations Females 0 98.45 0.74 Males 3 98.10 0.70 Is there evidence at 10% significance that mean body temperature for females ( 1 ) is greater than the mean body temperature for males ( )? Assume normal populations with different standard deviations. Follow the steps below to answer the question: (i) (4 points)formulate appropriate null and alternative hypothesis (alternative is directional), use both, proper symbolic notation and express both in words. H 0 : 1 = H A : 1 > (ii)(4 points) Compute the test statistics, round your answer to decimal places. t s = 1.59 (iii)(4 points) Compute the p value, software produced 39.4 df. You may compute it exactly using calculator or bracket it using a t table. Include appropriate sketch explaining what you are computing. P VALUE= 0.06 Sketch: represents t curve with area shaded right of 1.59 or 0.05<p value<0.10 (tables) marked as a p value (iv) (4 points)decide if null hypothesis should be rejected or not at 10 % significance level and answer question posed in the problem. H 0 REJECTED H 0 NOT REJECTED (circle one) Select H 0 REJECTED ANSWER QUESTION: There is evidence at 10% significance level that mean body temperature is higher for females than for males. (v) (4 points)if you decided next to test a nondirectional hypothesis (with H A ), what would be the p value? Would there be evidence for that alternative hypothesis at 10% significance level? P VALUE= *(0.06)=0.1>0.10, null will not be rejected NO EVIDENCE FOR H A YES, EVIDENCE FOR H A (circle one) Select NO EVIDENCE FOR H A
In Questions #5 #11 select appropriate letter answer from given choices A E, 8 points each Question#5 Suppose 90% CI for = the mean menstrual cycle of U.S women ages 0 37 is (8.5 days, 9. days). We used a sample of 350 women. Is this interval providing evidence that is 9.5 days, the length of the lunar month? Briefly explain why or why not. Select on of the answers below. A) No, because the CI is below 9.5 days B) Yes, because 9.5 days is outside of the CI. C) Yes, because upper endpoint of CI is only a fraction below 9.5 days. D) Yes, because the CI is below 9.5 days E) none of the above are correct. Question#6 A bio technician would like to estimate the mean number of genes that she will have to analyze in her research. Assume that the standard deviation is 10 genes. How many genes does she have to sample if she would like the Standard Error to be no more than.3? A) n 7 B) n 73 C) n 53 D) n 5 E)none of these Question #7 Estimating p value from the t table Suppose we tested hypothesis H 0 = versus H A :μ 1 <μ and our test statistics is t s =.54. Assuming we have 15 degrees of freedom, use t table to estimate the p value. A) 0.0<p value<0.05 B) 0.01<p value<0.05 C) 0.05<p value< 0.01 D) none of these Question #8 Suppose 95% CI for difference between population means 1 is ( 1.5, 6.8). If we use the same data to test the hypothesis H 0 :μ 1 =μ versus H A :μ 1 μ which of the conclusions about H 0 are correct at 5 % significance level? A) H 0 is rejected, difference is significant at 5% significance level B) H 0 is not rejected, difference is not significant at 5% significance level C) H 0 is rejected, difference is not significant at 5% significance level D) not enough information Question #9 Recruiting T cells A scientific study investigated the use of T cells engaging antibody, blinatumomab, in controlling growth of cancerous tumors. The data below are the T cell counts (1000 per microliter) for 6 individuals at the beginning of the study (baseline) and after 0 days on blinatumomab. After 0 days 0.8 0.47 1.30 0.5 1. 0.44 Baseline 0.04 0.0 0.00 0.0 0.38 0.33 Difference 0.4 0.45 1.30 0.3 0.84 0.11 Compute 95 % Confidence Interval for the mean difference in T cell counts (After 0 days Baseline), use appropriate procedure. Round final answer to decimal places. Assume normally distributed differences. A) ( 0.05, 1.01) B) (0.19, 0.86) C) (0.03, 1.03) D) (0.07, 0.99) E) none of these
Use following information for Questions #10 and #11 Testing for a Difference between Reported and Measured Male Heights. Twelve randomly selected male statistics students self reported their heights (in inches), then a researcher measured their height. Results are given in the table below. Student 1 3 4 5 6 7 8 9 10 11 1 Reported Height =Y1 68 74 8.5 66.5 69 68 71 70 70 67 68 70 Measured Height=Y 66.8 73.9 74.3 66.1 67. 67.9 69.4 69.9 68.6 67.3 68 68.8 sign(d=y1 Y) + + + + + + + + + 0 + Differences do not suggest normal distribution can be assumed, so the Sign Test was used to find out if there is evidence at 5% significance level that self reported high is larger than measured height for male students. Question #10 Formulate the hypotheses for your Sign Test: A) H 0 : P(d>0)=0.5 H a : P(d> 0) 0.5 B) H 0 : P(d>0)=0.5 H a : P(d>0)>0.5 C) H 0 : P(d>0)=0.5 H a : P(d >0)<0.5 D) H 0 : P(d>0)=0.5 H a : P(d >0) 0 Question #11 Given that the test statistics for our test is B s =10 compute p value for your sign test (notice that one of the differences is zero). Round answer to 4 decimal places. A) 0.0117 B) 0.0193 C) 0.0059 D) 0.0005 Extra Credit Questions, answer True or False for each (5 points total) Question #1 For samples of size 4 sampling distribution of population distribution of y is extremely right skewed. y is approximately normal, even if the Question #13 Suppose 95% CI for mean age of some population is (34, 37). Interval was based on the random sample of 00 people. We can say that we can be 95% confident that all people in that population are somewhere between 34 and 37 years old. Question #14 Standard Error (SE) tends to go down with increasing sample size. Question #15 Suppose we compute 95% and 90% CI, for mean weight of individuals in the certain population based on the same data. 90% CI will be wider than 95% CI. Question #16 Suppose difference between the population means is considered important if it exceeds. If 95% CI for 1 is (.6,.7) then we can say that difference is significant and important.
Sampling Distribution of y : Normal or approx. Normal, y =, y = n Estimating population mean: Standard Error of the sample mean: df=n 1 Guessed SD Selecting sample size: n= desiredse Comparing two populations means: 1) (independent samples, 1 ) standard error of y 1 y : SE y1 y = SE y = s n, 1 100 % CI for : y±t /SE y, s 1 s = SE n 1 n 1 SE, degrees of freedom will be given in df = [(s 1/n 1 )+ (s /n )] each problem, they are estimated by software, using: (s 1 /n 1 ) n 1 1 + (s /n ) n 1 If not given liberal estimate df =n 1 n or conservative estimate df =min n 1 1,n 1 is used. 1 100 % CI for 1 : y 1 y ±t / SE y1 y Hypothesis test: H 0 = vs H A or H A or H A test statistics: t s = y 1 y 0 SE y1 y ) Paired samples (assume normal differences): d= y 1 y, standard error of d : SE d = s d n d, 1 100 % CI for d = 1 : d ±t / SE d Hypothesis test: H 0 =0 vs H A 0 or H A 0 or H A 0 test statistics: t s = d SE d df= n d 1 3) Paired samples, nonparametric sign test : H 0 : P(d>0)=0.5, d= y 1 y, N =number of positive differences N = number of negative differences, ignore all d=0 test statistics : B s =max N, N ) has binomial distribution under null hypothesis.