1 Solutions –Practice Final Exam A– Fall 2016 Multiple Choice Answers: 1) a) The output gives P-value < 0.0001 and is less than α = 0.05. We reject the null hypothesis. The test is significant at α=0.05. b) is wrong because it was an observational study and we cannot establish cause and effect. c) is wrong because the chi-square test is two-sided where the alternative hypothesis only states “there is an association …” and doesn’t give a direction of more or less... 2) a) Calculation for relative risk: 50.1322584/2067211581/2026ˆˆCarTruckpp 3) c) Test the slope of linear regression line for linear relationship of two quantitative variables. Use one-sided alternative hypothesis: :0 to test “positively linear relationship”. e) is wrong because  is a notation used for estimated slope from sample data. Hypotheses are statements about the population, not sample. 4) d) Reasoning: the statistical software output gives 2-sided P-value for testing slope as a default. Since the test statistic is 8.45 (a positive value) and the alternative hypothesis is HA : 1 > 0. By the definition of P-value, you divide the P-value of 0.0052 (read in the Regression table) by 2. 5) c) The definition of the slope statistic (b1) is the estimated change in average response (increase if slope > 0; decrease if slope < 0) for each increase in 1 of the explanatory variable value. 6) a) A very small P-value provides strong evidence against the null hypothesis and hence support the alternative hypothesis. However, it cannot measure the strength of association. 7) e) There are 2 independent samples and it’s a one-sided test based on the description of the study. 8) c) Reasoning: Statistical output tells us they are Comparing B – A. This implies, the null hypotheses that software is calculating the p-values for is: H0: B - A = 0. With this order and the wheat farmer’s hypotheses, the hypotheses become: H0: B - A = 0 vs HA: B - A < 0. So the p-value is the number next to Prob < t. Note that 67.88 x Ax B 76.70 so it makes sense that the p-value is close to 1. 9) c) 10) b) There are 4 possible outcomes (i.e. four cells) so the conditions require each expected count be 5 or larger. 11) d) With “two very large independent samples”, the t-score is very close to the z-score. Use the t-Table and read the critical value z* with corresponding one-sided and two-sided tail probabilities at the very bottom of the table. Given α = 0.02, we can estimate the P-value:  one-sided P-value < 0.02 (hence reject the H0)  0.02 < two-sided P-value < 0.04 (hence fail to reject H0) 2.054 < critical value z* < 2.326 corresponds to 2-tail probability 0.02 < P-value < 0.04. Hence 2.22 could be the value of computed test statistic2 12) a) Based on the UT student’s belief, the hypotheses are H0: UT = TAMU vs. Ha: UT > TAMU . You are told the two population means (from census of all records) TAMU = 2.95 and UT = 2.81. Since the p-value = .04 <  they will reject a true null hypothesis and make a Type I error. The phrase “the average GPR of UT students is statistically significantly greater ” is equivalent to saying “the data provides statistical evidence the average UT GPR is greater ” 13) c) Apply the formula: 󰇛󰇜󰇛󰇜 where 0.175, 0.35 14) a) Reasoning: Note that P-value is a conditional probability with the condition being “given that Ho is true”. Hence, the P-value interpretation must always have at the statement, something telling the reader that Ho is true. So, we must include the phrase “ = 75” in the interpretation. 15) Both d) & e) are FALSE Review the “BEAN” regarding the power. 16) d) 17) b) The null hypothesis states all the population means are the same (hence, no association). By rejecting H0, the evidence supports the alternative hypothesis which states “not all the population means are equal”(i.e., there is some association). We should not make inference of which population mean is larger at this point. We could only do so after performing pairwise multiple comparisons. 18) b) statement III is wrong because we don’t usually randomly select subjects for experiments. III would be a correct statement if experimental studies was replaced with observational studies. 19) d) Apply 2-proportion-z test. Be careful how you set up the alternative hypothesis : where p1 and p2 represent the proportion of cyanobacteria in riparian and coastal populations, respectively. 20) d) a) is wrong because it if an inference about the sample mean and confidence intervals are inferences about population parameters. b) would be correct if the word “sample” was omitted. c) is wrong because again it has the sample mean in it. Confidence interval statements are about the population parameter, not the sample statistic. 21) d) a) is wrong because the sample mean have nothing to do with margin of error. The formula for the margin of error is *stnand this doesn’t have xin it. b) is wrong because increasing the confidence level increases the margin of error because t* is increased. c) is wrong because increasing the standard deviation causes s to increase. 22) c) The test was statistically significant because we rejected Ho but the effect size was only .21 on a scale of 0 – 10. So, this is a very small increase for a course that costs $5000.3 23) a) Reasoning: There are two independent samples. Use 2-sample t test because the response is numerical and the explanatory variable has 2 groups (yes or no). It isn’t paired because there is no obvious pairing described. Answer d) is wrong because it is a statement about proportions but the response is numerical, and with numerical responses, we summarize the data using the mean not a proportion. 24) b) The is paired because the subjects (plots of land) were measured twice (had # cacti measured twice). 25) a) Read the three boxplots and find the computed difference for each pair has smaller standard deviation as it has smaller spread. 26) c) The power is solely a function of the sample size, the