1. (2 points) To evaluate the efficacy of two different smoking cessation programs, smokers are randomized to receive either Program A or Program B. Of the six smokers in Program A, one stopped smoking; of the six smokers in Program B, four stopped smoking. Which statistical procedure would you use to test the null hypotheses that the programs are equally effective and to obtain a p-value? a. Paired t-test b. Chi-square test of independence c. Two-sample t-test d. Fisher's exact test e. Wilcoxon rank-sum test 2. (2 points) In a cross sectional study of bicycle helmet use, authors were interested in assessing neighborhood socio-economic status (as measured by percentage of school children receiving free or reduced-fee lunches at school) and percentage of bicycle rider wearing a helmet. If the assumptions of the test were met, what is the appropriate statistical test to perform: a. two independent sample t-test b. Wilcoxon sign-rank test c. correlation analysis d. chi-square one-sample test of the variance e. none of the above 3. (2 points) A study was performed on 200 elementary school students to investigate whether taking regular Vitamin A supplements during the month of March would prevent colds. One hundred students were randomized to receive a daily Vitamin A supplement of 3,000 IU and 100 students to receive a placebo. If you were interested in testing for a statistical relationship between taking Vitamin A and cold status in the population of elementary school students, the correct statistical test is: a. two independent sample t-test b. chi-square test of the variance c. chi-square test of independence (homogeneity) d. paired t-test 4. (6 points) A researcher wants to determine if an experimental drug significantly affects the body temperature. Seven test subjects are randomly selected and the body temperature of each subject is measured (initial temperature). After that, the subjects are given the drug and after 60 minutes the body temperature of each subject is measured again (follow-up temperature). The results are given in the StatCrunch dataset 'body_temperature_study'. We know that the body temperature is a normally distributed variable. Test if the drug significantly affects the body temperature. - What test will you use? - State H0 and H1 for this test. - Provide the p-value and make a final conclusion about the data using the significance level of 0.01. 5. (6 points) A sports psychologist believes that listening to music affects the length of athletes' workout sessions. To test this assumption a random sample of ten athletes was selected. Each athlete had two workouts, one while listening to music and the other without music. For each athlete, each of the two workouts was on a separate day and the order of \"with music\" and \"without music\" workouts was random. The length of each workout (in minutes) was recorded for each athlete. The results are given in the StatCrunch dataset 'length_of_workout_epi551'. Assume that the data exhibit significant deviations from normality and the t-test is not applicable. Test if the psychologist's claim is justified. - What test will you use? - State H0 and H1 for this test. - Provide the p-value and make a final conclusion about the data using the significance level of 0.05. 6. (6 points) Suppose, you wish to compare salaries in the manufacturing and construction industries. You obtained a random sample of 10 manufacturing workers and another random sample of 10 construction workers. The annual salaries for each individual (in thousands of dollars) are given in the StatCrunch dataset 'comparing_industry_salaries'. Assume that the data exhibit significant deviations from normality and the t-test is not applicable. Test if the salaries in these two industries are significantly different. - What test will you use? - State H0 and H1 for this test. - Provide the p-value and make a final conclusion about the data using the significance level of 0.01. \f\f\f1. (2) A certain drug is tested for its effect on blood pressure. Seven male patients have their systolic blood pressure measured before and after receiving the drug. What is the outcome variable? Answer: systolic blood pressure 2. (1) The following data are the numbers of people who have called the New York State Center for Environmental Health's hotline for the past two weeks: 9, 15, 3, 8, 8, 11, 7, 12, 11, 15 What is the mean number of calls for the two weeks? Write your answer to the nearest tenths. Answer: 9.9 3.(1) The following data are the numbers of people who have called the New York State Center for Environmental Health's hotline for the past two weeks: 9, 15, 3, 8, 8, 11, 7, 12, 11, 15 What is the median number of calls? Answer: 10 4. Des Jarles et al. examined the failure to maintain AIDS risk reduction in a study of intravenous drug users from NYC. The following table shows the study subjects cross-classified according to risk reduction status (practices risk reduction) and number of sexual partners in an average month: Number of sexual partners/month None 1 >1 None 20 37 20 Risk reduction status Not maintained Maintained (occasionally) (always) 17 43 45 95 54 67 a. (2) What is the measurement scale of risk reduction status? qualitiative, ordinal b. (2) What is the measurement scale of number of sexual partners/month? qualitative, ordinal c. (2) What is the probability of a randomly selected person having one sexual partner in an average month? 177/398 = 0.4447 d. (2) What is the probability of a person who has one sexual partner in an average month having a risk reduction status of none? 37/177 = 0.2090 e. (2) What is the probability of a person whose risk reduction status is none having one sexual partner in an average month? 37/77 = 0.4805 5. The stress scores (on a scale 1 to 10) of students before a statistics test are found to be approximately normally distributed with a mean of 7.08 and a s.d. of .63. Find the probability that the stress score before a statistics test for a randomly selected student will be: a. (3) more than 6.25 P(X > 6.25) = P(Z > 6.257.08 ) = P(Z > -1.32) = 0.9066 0.63 b. (3) between 7.4 and 8.9 P(7.4 < X < 8.9) = P( 7.47.08 0.63 Proportion stats>One sample>with summary>enter #successes and sample size, and null proportion of 0.15 and select >Compute: Proportion Count Total p 10 100 Sample Prop. 0.1 Std. Err. Z-Stat P-value 0.035707142 -1.4002801 0.0807 Pvalue > 0.05. Accept H0. The researcher's claim is NOT justified. 12. (4) The Consumer Reports magazine tested two types of winter tires, SnowFlake and SnowMan, to determine how well each brand performs on ice when stopping from 15 mph. 12 sets of the SnowFlake tires and 12 sets of the SnowMan tires were tested. The results showed that for the SnowFlake tires the sample mean stopping distance was 55 feet with the sample standard deviation of 7 feet, while for the SnowMan tires the sample mean stopping distance was 57 feet with the sample standard deviation of 4 feet. Assume that the data are normally distributed. Can you conclude that there is a significant difference in the stopping distance between these two types of tires? - What test(s) will you use? - State H0 and H1 for these tests. - Provide the p-value and make a final conclusion about the data using the significance level of 0.05. - What test will you use? The Ftest for the equality of two variances, followed by an appropriate version of the ttest for two independent samples. - State H0 and H1 for this test. Ftest: H0: 2 1 = 2 2 H1: 2 1 2 2 ttest: H0: 1 = 2 H1: - Provide the p-value and make a final conclusion about the data using the significance level of 0.05. >Compute: Ratio 1 2 /2 2 Num. DF 11 Den. DF 11 Sample Ratio 3.0625 F Stat 3.0625 Pvalue 0.0765 pvalue > 0.05, accept H0. The variances are equal. t-test for two independent samples, equal variances case (use 'pooled variances' in StatCrunch) [Stat]->T stats->Two sample>with summary->enter sample parameters Difference 1 - 2 Sample Diff. Std. Err. 2.3273733 DF 22 T-Stat -0.85933785 Pvalue 0.3994 pvalue > 0.05, accept H0. There is NO significant difference in the stopping distance between these two types of tires 1. (2) A certain drug is tested for its effect on blood pressure. Seven male patients have their systolic blood pressure measured before and after receiving the drug. What is the outcome variable? Systolic blood pressure 2. (1) The following data are the numbers of people who have called the New York State Center for Environmental Health's hotline for the past two weeks: 9, 15, 3, 8, 8, 11, 7, 12, 11, 15 What is the mean number of calls for the two weeks? Write your answer to the nearest tenths. xx = 9.9 3.(1) The following data are the numbers of people who have called the New York State Center for Environmental Health's hotline for the past two weeks: 9, 15, 3, 8, 8, 11, 7, 12, 11, 15 What is the median number of calls? median = 10 4. Des Jarles et al. examined the failure to maintain AIDS risk reduction in a study of intravenous drug users from NYC. The following table shows the study subjects cross-classified according to risk reduction status (practices risk reduction) and number of sexual partners in an average month: Number of sexual partners/month None 1 >1 None 20 37 20 Risk reduction status Not maintained Maintained (occasionally) (always) 17 43 45 95 54 67 a. (2) What is the measurement scale of risk reduction status? No risk reduction (none), occasionally risk reduction not maintained, , always maintain risk reduction and therefore qualitative and ordinal (ordered) b. (2) What is the measurement scale of number of sexual partners/month? 0, 1, and more than 1 and therefore qualitative, and ordinal c. (2) What is the probability of a randomly selected person having one sexual partner in an average month? P(Number of sexual partners/month is 1) = 177/398 = 0.444723618 or 44.47% d. (2) What is the probability of a person who has one sexual partner in an average month having a risk reduction status of none? P(risk reduction is none | number of sexual partners/month is 1) = 37/177 = 0.209039548 or 20.90% e. (2) What is the probability of a person whose risk reduction status is none having one sexual partner in an average month? P(number of sexual partners is 1 | risk reduction status is none) = 37/77 = 0.480519481 or 48.05% 5. The stress scores (on a scale 1 to 10) of students before a statistics test are found to be approximately normally distributed with a mean of 7.08 and a s.d. of .63. Find the probability that the stress score before a statistics test for a randomly selected student will be: Mean = 7.08 s.d. = 0.63 a. (3) more than 6.25, formally: P(X > 6.25) Calculate z-score: (6.25 - 7.08) / 0.63 = -1.3175 I am using tables at: http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf The area under the curve from the left up to -1.3175 is 0.0934. Subtract this from 1 to get the area from -1.3175 to the right (greater than) P(X > 6.25) = 1 - 0.0934 = 0.9066 or 90.66% b. (3) between 7.4 and 8.9, formally: P(7.4 < X < 8.9) Calculate both z-scores: First, 7.4: (7.4 - 7.08) / 0.63 = 0.5079 Second, 8.9: (8.9 - 7.08)/ 0.63 = 2.8889 Using the table, the area under the curve from the left to 0.5079 is: 0.6950 The area under the curve from the left to 2.8889 is: 0.9981 To get the area under the curve between 7.4 and 8.9, subtract the result for 7.4 from 8.9 P(7.4 < X < 8.9) = 0.9981 - 0.6950 = 0.3031 or 30.31% c. (3) between 3.4 and 6.2, formally P(3.4 < X < 6.2) Calculate both z-scores: First, 3.4: (3.4 - 7.08) / 0.63 = -5.8412 Second, 6.2: (6.2 - 7.08)/ 0.63 = -1.3968 Using the table, the area under the curve from the left to -5.84 is: 0.0000 The area under the curve from the left to -1.40 is: 0.0808 To get the area under the curve between 3.4 and 6.2, subtract the result for 3.4 from 6.2 P(3.4 < X < 6.2) = 0.0808 - 0.0000 = 0.0808 or 8.08% 6. (1)Nine men with a genetic condition that causes obesity entered a weight reduction program. After four months the statistics of weight loss were XBAR = 11.2, S = 9.0. Provide a 95% confidence interval for the mean of the population of which this is a sample (assuming normality and randomness) . xx = 11.2 s = 9.0 n=9 confidence level is 95% Compute : = 1 - (confidence level / 100) = 1- 0.95 = 0.05 Critical value for d.f. = 8 at the two-tailed 95% confidence interval t-test is: 2.306 (See t-table at: http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf) Sample standard error of the mean: xx = s / sqrt( n ) = 9/(9) = 3 Confidence intervals are: = xx - 2.306 * (s/ n) < < xx + 2.306 * (s/ n) = 11.2 - 2.306 * (3) < < 11.2 + 2.306 * (3) = 11.2 - 6.918 < < 11.2 + 6.918 = 4.282 < < 18.118 7. (1) In a survey study on the incidence of depression in a population of psychiatric hospital administrators, scores on a depression measure were obtained from 86 respondents. The observed mean score was 62, with a standard deviation of 16. What kind of table would you use in constructing a 99% confidence interval for the mean depression score? Use t-statistic and t table because the population standard deviation is not known. 8. (1) State True or False. As the sample size (n) increases, the mean of a random sample is less likely to be near the mean of the population. False 9. (1) Which of the following is a true statement regarding the comparison of t-distributions to the standard normal distribution? a. The normal distribution is symmetrical whereas the t-distributions are slightly skewed. False b. The proportion of area beyond a specific value of t is less than the proportion of area beyond the corresponding value of z. False c. The greater the df, the more the t-distributions resemble the standard normal distribution. True d. All of the above. False e. None of the above False Answer is (c.) 10. (2) A sample of size 26 is taken from a normal population. Then a 95% confidence interval is set up with XBAR = 30 and s = 20. The value you would find in the table to complete the information necessary to obtain the interval would be: Select one: a. 1.711 b. 1.708 c. 2.060 d. 2.064 e. 1.960 Need to use t-table to estimate the population mean from a sample. The degrees of freedom is n26 = 25, and the alpha level is 0.05. This is a two-tailed test, so use two-tail for alpha=0.05 and d.f. = 25 to find the critical value. Answer is c: 2.060 11. (3 ) A researcher claims that less than 15% of adults in the general US population are allergic to a medication. The results from his study showed that in a random sample of 100 adults only 10 said that they had such an allergy. Test if the researcher's claim is justified (use a one-sided test because you are given the anticipated direction of the difference). - What test will you use? The one-sample test of a binomial proportion - State H0 and H1 for this test. H0: p = 0.15 HA: p > 0.15 - Provide the p-value and make a final conclusion about the data using the significance level of 0.05. (Hint: this question is about comparing a proportion to some given value) n=100 The sample proportion pp = 10 adults/100 sampled = 0.1 The test statistics Z is calculated: Z= p ^ p0 p 0 (1 p0 ) n = 0.10.15 0.15(10.15) = 100 0.05 0.05 = =1.4006 0.0357 0.001275 Using the binomial calculator at: http://stattrek.com/onlinecalculator/binomial.aspx, with probability of success in a single trial as 0.1, number of trials as 100, and P(X15) = 0.072. The P(X15) > 0.05, so we accept (fail to reject) H0. There is sufficient evidence at the = 0.05 level to conclude that the proportion of allergic adults is not less than 15%. 12. (4) The Consumer Reports magazine tested two types of winter tires, SnowFlake and SnowMan, to determine how well each brand performs on ice when stopping from 15 mph. 12 sets of the SnowFlake tires and 12 sets of the SnowMan tires were tested. The results showed that for the SnowFlake tires the sample mean stopping distance was 55 feet with the sample standard deviation of 7 feet, while for the SnowMan tires the sample mean stopping distance was 57 feet with the sample standard deviation of 4 feet. Assume that the data are normally distributed. Can you conclude that there is a significant difference in the stopping distance between these two types of tires? - What test(s) will you use? Two sample t test for equality of means, assuming equal population variances Table: SnowMan (SM) n = 12 xx = 57 sx = 4 s2x = 16 - State H0 and H1 for these tests. SnowFlake (SF) m = 12 yx = 55 sy = 7 s2y = 49 H0:SF = SM HA:SF SM - Provide the p-value and make a final conclusion about the data using the significance level of 0.05. Calculate the T statistic: T= x y s 2 s2y x + n m = 5755 2 2 = = =0.8593 16 49 5.4167 2.3274 + 12 12 Degrees of freedom d.f. = = n + m - 2 = 12 + 12 - 2 = 22 dfSM = nSM - 1 = 12-1 = 11 dfSF = nSF - 1 = 12-1 = 11 d.f.(pooled) = (n1 - 1) + (n2 - 1) = 9+9 = 18 Critical value is two-tailed cumulative probability from the t-table for = 0.05 and /2 = 0.025 and 22 d.f. is 1.717 critical value Decision rule: if t statistic is less than -1.717 or t statistic is greater than +1.717, reject the null hypothesis Calculate sum of squares (SS) for each group: SSSM = s2SM(dfSM) = 16 * 11 = 176 SSSF = s2SF(dfSF) = 49 * 11 = 539 where s2 is just the variance for each group (the standard deviation squared). We calculate the pooled variance, sp: 2 s p= SS SM + SSSF 176+539 715 = = =32.5 df SM +df SF 11+11 22 Then, we calculate the t-test statistic: t= x x ( 1 2) 2 p 2 p s s + n1 n2 = (5755) 2 2 = = =0.3692 32.5 32.5 5 .4167 5.4167 + 12 12 Since t = 0.3692 > 1.717 we do not reject the null hypothesis. There is sufficient evidence at the = 0.05 level to conclude that there is no significant difference in the stopping distance between these two types of tires