STAT 302 Practice Final Exam Solutions (Q28-Q50)

Quiz: Final - Practice Exam Question 28 2 pts Use the following information for questions 28-29: Suppose you conduct a hypothesis test to determine whether or not the average height of first-grade students is less than 46 inches. You conduct this test at the 0.05 significance level and come to the conclusion that 0.05 < p-value < 0.10. 28. What is the correct decision? o Accept the Null Hypothesis. o Accept the Alternative Hypothesis. O Reject the Null Hypothesis. � Fail to Reject the Null Hypothesis. p-value > 0.05 = alpha, we fail to reject the null Question 29 2 pts Use the following information for questions 28-29: Suppose you conduct a hypothesis test to determine whether or not the average height of first-grade students is less than 46 inches. You conduct this test at the 0.05 significance level and come to the conclusion that 0.05 < p-value < 0.10. 29. What is the appropriate conclusion? H0:mu = 46,HA:mu<46 we fail to reject the null, that means, data was rot .. ...J sufficiently in favor of HA a The data does not provide statistically significant evidence that the average height of first-grade students is less than 46 inches. o The data does provide statistically significant evidence that the average height of first-grade students is less than 46 inches. o The data does not provide statistically significant evidence that the average height of first-grade students is 46 inches. O The data does provide statistically significant evidence that the average height of first-grade students is 46 inches. Question 30 2 pts https://canvas.tamu .edu/courses/210724/quizzes/390987/take?preview = 1 14/33

Quiz: Final - Practice Exam Suppose a researcher wants to test the hypothesis that 1,1 = 25 versus the alternative that 1,1 #: 25 using data from a random sample of 95 people. We calculate the standardized test statistic to be 2.087. Which of the following best describes the p- value? O 0.05 < p-value < 0.04. 'SJ 0.04 < p-value < 0.05. o 0.02 < p-value < 0.04. O 0.02 < p-value < 0.025. O 0.01 < p-value < 0.02. Question 31 HO: mu = 25, HA: mu not equal to 25 n = 95, under the null, xbar ~ t (mu0 = 25, s/sqrt(n)), df = n-1 = 94 p-value = P(t94 >2.087)*2: between 0.04 and 0.05 Note. df 94 is not on t table. read df 80 instead (the next low r df) 2 pts Use the following information for questions 31-33: One treatment for strep throat is a tonsillectomy (having the tonsils removed). A research study wanted to determine wnetner or not having a tonsillectomy reduced the occurrence of strep throat in pediatric patients. They recruited 40 pediatric patients who had a tonsillectomy for their study and for each patient measured both the number of occurrences of strep throat in the year prior to the tonsillectomy (untreated) and the number of occurrences of strep throat in the year following the tonsillectomy (treated). The summary statistics for the study is shown in the table below (note: not all information in the table is required to solve the problem; you must decide what pieces of information are necessary). Assuming all conditions are met, conduct the appropriate hypothesis test at the 0.01 significance level to determine if patients have less cases of strep throat after treatment [difference = untreated (before tonsillectomy) - treated (after tonsillectomy)]. 1 G r oup "' i s � n t r e ated (Before TonSlfeclomy) �i 7 · t - � T reate d (After To n s i ll e c tom y } 40 2.3 4.9 Dinerence 1 40 4 . ' 9 1 3 .. 2 31. What are the hypotheses? paired t-test: two measurements are not independent (re eated measurements at pre- and post- treatment from the sam individuals) https://canvas.tamu .edu/courses/210724/quizzes/390987/take?preview = 1 15/33

� Ho: µd = 0 Quiz: Final - Practice Exam H A : µd > 0 since we claim that the treatment reduce it (untreated> treated) 0 Ho: µuntreated - µTreated = 0 0 Ho: µuntreated - µTreated = 0 0 Ho: µUntreated - µTreated = 0 H A : At least one group is different. Question 32 2 pts Use the following information for questions 31-33: One treatment for strep throat is a tonsillectomy (having the tonsils removed). A research study wanted to determine whether or not having a tonsillectomy reduced the occurrence of strep throat in pediatric patients. They recruited 40 pediatric patients who had a tonsillectomy for their study and for each patient measured both the number of occurrences of strep throat in the year prior to the tonsillectomy (untreated) and the number of occurrences of strep throat in the year following the tonsillectomy (treated). The summary statistics for the study is shown in the table below (note : not all information in the table is required to solve the problem; you must decide what pieces of information are necessary). Assuming all conditions are met, conduct the appropriate hypothesis test at the 0.01 significance level to determine if patients have less cases of strep throat after treatment [difference = untreated (before tonsillectomy) - treated (after tonsillectomy)]. 1 G r oup n , i s U n t r eat , e d ( B , e f or e T o n s · nectom y ) �� 7.2 6.3 � T r e at ed (After T on s i l l ect o m y } 4 0 2.3 4.9 o·tt , ere n ce • 1 4 0 4.'9 3 .. 2 https://canvas.tamu .edu/courses/210724/quizzes/390987/take?preview = 1 16/33

Quiz: Final - Practice Exam 32. What is the value of the test statistic? xbar_diff = 4.9, s_diff = 3.2 SE = s_diff/sqrt(n) = 3.2/sqrt(40) ) df = n-1 = 39 0 1.5313 0 4.2933 0 8.5919 ?i 9.6838 0 13.6948 t = (xbar_diff - mu_0) / (s_diff/sqrt(n)) = (4.9-0)/(3.2/sqrt(40)) = 9. 838 Question 33 2 pts Use the following information for questions 31-33: One treatment for strep throat is a tonsillectomy (having the tonsils removed). A research study wanted to determine whether or not having a tonsillectomy reduced the occurrence of strep throat in pediatric patients. They recruited 40 pediatric patients who had a tonsillectomy for their study and for each patient measured both the number of occurrences of strep throat in the year prior to the tonsillectomy (untreated) and the number of occurrences of strep throat in the year following the tonsillectomy (treated). The summary statistics for the study is shown in the table below (note: not all information in the table is required to solve the problem; you must decide what pieces of information are necessary). Assuming all conditions are met, conduct the appropriate hypothesis test at the 0.01 significance level to determine if patients have less cases of strep throat after treatment [difference = untreated (before tonsillectomy) - treated (after tonsillectomy)]. 1 G r oup "' i s � n t r e ated (Before T o n S lf ecl om y ) �i 7 · t -3 T r e at ed (After T on s i l l e c tom y } 40 2.3 4.9 o·nerence 1 40 4. ' 9 1 3 .. 2 33. What is the following best describes the p-value? t = 9.6838, df = 39 Right tailed HT, thus p-value = P(t_39 > = 9.6838 ), df 39 is Flot located on t table, instead read the next lower df 30 : O p-value = 0 � p-value < 0.0005 P(t_30 > = 9.6838 ) <0.0005, since the cutoff t_30 for upper p 0.0005 is .646 https://canvas.tamu .edu/courses/210724/quizzes/390987/take?preview = 1 17/33

O p-v a lue < 0.001 O p-v a lue > 0.0005 O p-v a lue = 1 Question 34 Quiz: Final - Practice Exam 2 pts A recent study compared tooth health and periodontal damage in a group of 46 young aclult males wearing a tongue piercing ancl a control group of 46 young adult males inde pende without a tongue piercing. One question of interest was whether individuals with nt tvo tongue piercings had more enamel cracks on average. The summary statistics for th e8 arr p i es study are shown in the table below (note: not all information in the table is required to solve the problem; you must decide what pieces of information are necessary). Assuming all conditions are met, what is the 99.5% confidence interval for the difference between the number of enamel cracks in the two groups (difference = Tongue Piercing - No Tongue Piercing). 99.5% confidence level corresponds to 0.005 two sided prob Group n i s T o n g u e Piercing 46, 4. 0 1 3. 5 No T o n g ue , P i e rc i n g 46 , 1.2 1.3 Difference 46 2. 8 1 0.9 - Cl for two s a mple difference (mu1-mu2) = (xb a r1 - xb a r2) + - t*_df x SE, 0 (-4.4355, -1.1645) � .1645, 4.4355) 0 (1.2547, 4.3453) 0 (1.8401, 3.7599) 0 (2.4057, 3.1943) where SE = sqrt( s 1"2/n 1 + s2"2/s2 ) = sqrt( 3.5"2/46 + 1.3"2/41 ) = 0.55 xb a r1 - xb a r2 = 4.0-1.2 =2.8 Question 35 df = min(46-1, 46-1) = 45 (not on the t a ble, re a d df=40 inste a d) t*_ 40 = 2.971 99.5% Cl = 2.8+ - 2.971*0.55 = (2.8-1.6355, 2.8+1.6355) 2 pts We are testing the following hypotheses: https://canvas.tamu .edu/courses/210724/quizzes/390987/take?preview = 1 18/33