1Final Practice Exam B – Fall 2016 - Solutions 1) b) Reasoning: Statement III isn’t true because hypothesis tests are only designed to determine whether the data provides strong enough evidence for the HA that we can conclude it is true to a certain significance level. 2) d) 3) c) Test the relationship between 2 categorical variables 4) c) 5) b) - µ=22 is not a plausible population mean in the 90% CI and we would reject H0 at significance level α = 0.10 (10%). The P-value must be less than 0.10. - µ=22 is a plausible population mean in the 95% CI and we would fail to reject H0 at significance level α = 0.05 (5%). The P-value must be larger than α = 0.05. - µ=22 is a plausible population mean in the 99% CI and we would fail to reject H0 at significance level α = 0.01 (1%) The P-value must be larger than significance level α = 0.01. Hence we conclude 0.05 < P-value < 0.10 6) b) Identify this is a one sample t-test. Compute the test statistic ..√⁄2, use the t-table with df=15, estimate the one-sided P-value: 0.025 < P-value < 0.05 Alternatively, you could use calculator T-Test and find the P-value = 0.03197; or compute tcdf(14, 999, 15)=0.03197 7) e) 8) a) Because the study design is for 2 independent samples quantitative variable, we use the t test output (not the matched pairs output) and the p-value for  which is Prob > |t| 9) a) c) is wrong because it is a statement about only the subjects in the sample. Based on the results of hypothesis tests we make inferences about the entire population, not just the people who took part in the study. The conclusion for a hypothesis test is always about a parameter and never about the sample. 10) c) 11) b) 12) d) The estimated average wholesale price is -3.50+2.50x but this equation is only good for x values between 20 and 50 (see the scatterplot) because this is the range of our explanatory variable values in the data set. 13) a) Identify this is a 2 independent samples t-test (assume unequal variance) and apply the formula to compute the test statistic.  0.78 (or, use calculator procedure “2-SampTTest…”) 14) c) 15) b) They are pairing by year – the man and woman were hired the same year (salaries vary year by year so my pairing people hired the same year, differences due to year are eliminated). 16) b) This is a randomized comparative experiment and the null hypothesis was rejected. So the best conclusion is one which concludes some sort of causation rather than just association.2 17) a) Because it asks for the relative risk of “placebo gum” compared to Xylitol gum” we calculate relative risk as placebo/Xylitol = 70.1179/29178/49 18) a) Reasoning: The explanatory variable values (from an Aggie family and Not from an Aggie family) describe the groups whose proportions are being compared. So, the proportions of interest are: pA and pNA. These proportions must be proportion of “successes” = “maroon car” in each group. (read similar arguments in #10 from the solutions for the 1st set of practice final exam) 19) b) The conditions are about the expected counts on the bottom row of each cell. The expected cell count for the cell corresponds to “From Aggie family” and “Has Maroon car” is equal to (12)·(36)/120 = 3.6. 20) a) Reasoning: The explanatory variable consists of the treatment(s) being given to the patients. In this case that is OC versus placebo. The age of a subject isn’t of interest but is just used for blocking purposes. 21) b) Reasoning: The subjects are first divided up by age group. The age groups form the “blocks”. Then within each block, the subjects are randomly divided up into treatment groups. 22) a) 23) a) 24) b) Compute the two sample proportions: 45/60 = 0.75, 40/50=0.8 and  󰇛.󰇜∙󰇛.󰇜󰇛.󰇜∙󰇛.󰇜 0.0795 25) b) Under the H0: p1 = p2, which is equal to a common value of population proportion. We shall pool all the sample data together to estimate this common population proportion. For the SE of () in the test statistic, use the “pooled” sample proportion   0.7727, and  󰇛1󰇜󰇡󰇢󰇛0.7727󰇜∙󰇛0.2273󰇜󰇡󰇢0.0802 26) d) 27) e) Statement I is wrong because sample mean is not part of the margin of error. Statement II is wrong because decreasing the sample size increases the variability hence increases the size of margin of error. Statement III is wrong because increasing the confidence level increases the size of the margin of error 28) b) is not true c) is true because the box plots don’t overlap so there is statistical evidence the means differ (at least when alpha is .05 or larger). 29) c) df1 = k – 1 ; df2 = N – k where k is the number of groups and N is the total sample size3 30) b) 31) d) 32) b) Hypotheses are about population parameters, not sample statistics 33) e) 34) c) From the ANOVA table, MSE = pooled sample variance 35)