Statistics in a Modern World 800
Solutions to Practice Exam 2
-
- (iv) multiple boxplots (for values of one measurement variable compared
for two categorical groups)
- (ii) piechart (for one categorical variable)
- (iii) scatterplot (for two measurement variables)
-
- (iii) skewed right/high outliers
- (iv) 23: mean greater than median (outliers pull mean up) but 32 would
be way too high
- (iii) 6 is the typical distance of those values from their mean;
0.06 and 0.6 would be way too low and 60 would be way too high
- (a) +1 because it is a strong positive relationship
- (d) [(a) not appropriate because heights level off before 20, and the
relationship is not at all linear; (b) not appropriate because living situation
is categorical]
-
- 0.60(230)/82=$1.68
- (ii)$2.50 is too high because $1.68 would have kept pace with inflation
- -92+.0468(2013)=2.21
- extrapolation
-
- z=(7-12)/2=-2.5; the proportion below is 0.005, according to Table
- z=(14.8-12)/2=1.4; the proportion below is 0.92, so proportion above
is 0.08.
- shortest 1% have z=-2.33, observed value = 12-2.33(2)=7.34
- longest 10% have z=+1.28, observed value = 12+1.28(2)=14.56
- (ii) Mean equals median because normal curves are symmetric.
- The curve is marked off with the following seven points: 6, 8, 10, 12,
14, 16, 18.
-
- (ii) because sales overall are increasing
- (iii)
- (iii)
-
- (7)
- 5 (slower metaolism is a common cause)
- 1 (TV is a direct cause of obesity
- 3 (TV is a contributing cause to obesity)
- 2 (causation in the opposite direction, with obesity causing children
to watch TV)
- 6 (TV and weight both increasing over time)
- 4 (socio-economic status is a confounding variable)
-
- (ii)males: 61/120=0.51 as opposed to 89/240=0.37 for females
- calculate (column total * row total)/table total: 100, 50, 140, 70
- 1.2, 2.4, 0.9, 1.7
- 1.2+2.4+0.9+1.7=6.2
- (ii) because 6.2 > 3.84
- (ii): we have convincing statistical evidence of a relationship, but we
can't draw definite conclusions about the reason for the relationship.
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