Solutions to Practice Test 1, Section I
- (B) Using the outlier guideline, any value above Q3 + 1.5 IQR = 28.5 + 1.5 10.5 = 44.25, or below
Q1 − 1.5 IQR = 18 − 1.5 10.5 = 2.25, is considered an outlier. Because the maximum is 33, there
are no values above 44.25. The minimum of 2 is an outlier. There may be more than one value below
2.25, so all we can say is that there is at least one outlier on the lower end.
- (B) You should recognize that this is the interpretation of r 2 . According to the printout, that value is
0.542313, or about 54.2%.
- (C) A Type II error means that the null hypothesis is false, but we fail to reject it. In this case, a false
null hypothesis means that Heartaid actually does work better. Failing to reject that null hypothesis
means they don’t have convincing evidence that Heartaid is better, so they won’t use it.
- (B) This is asking for a conditional probability. P (Morafka|Female). Out of the 48 female tortoises,
15 are Morafka. 15/48 = 0.3125, or about 31.3%.
- (C) Randomization is what fights against confounding by making the groups as alike as possible. With
blocking, similar units are grouped together so the effect of being in that group can be accounted for
in the analysis.
- (E) On the midterm, the z -score is So she scored 1.5 standard deviations above the
mean. To achieve the same standardized score on the final exam, she must score 1.5 standard
deviations, or 1.5 10 = 15 points, above the mean on that exam. 112 + 15 = 127 points.
- (A) A simple random sample is already unbiased. A stratified random sample cannot improve on that.
- (E) The correlation coefficient remains unchanged when converting units.
- (E) The median of the HFCS group is the same as Q3 of the No HFCS group.
- (D) . 0.06 = 0.09. 1 – 0.09 = 0.91.
- (D) Because there is a curve in residual plot I, which is for weight vs. length, the relationship
between those two variables is not linear. Because residual plot II shows a no curve and plot III
does, plot II represents a better model.
- (D) The null hypothesis always contains equality. Because they are interested in whether the
candidate’s support has improved, the alternative hypothesis is one-sided.
- (A) The correlation, r , is the square root of r 2 , which is given in the table. So r could be ±0.1887.
Since the slope is negative, r is negative.
- (E) Because this is a properly conducted randomized experiment, a causal relationship can be
determined. E is the choice that is not true.
- (B) The population is divided into homogenous subgroups, and a random sample is selected from
each.
- (E)
- (B) The formula for the confidence interval for the slope is b 1 ± t * · sb 1 · From the printout b 1 =
4.3441 and sb 1 = 1.811. t * for 18 – 2 = 16 degrees of freedom and 99% confidence is 2.921.
