Statistics in a Modern World 800
Solutions to Exam 2
-
- (i) 5 is the typical distance of those values from their mean of 86;
all the other numbers are way too high
- (ii) fairly symmetric
- (iii) about equal to the median
-
- (iv) multiple boxplots (for values of one measurement variable compared
for two categorical groups)
- (i) stemplot (for one measurement variables
- (iii) scatterplot (for two measurement variables)
- (ii) piechart (for one measurement variable)
- (i) negative; price almost always tends to be lower for older cars
(I even did this example in lecture)
-
- z=(9-7.6)/1.1=1.27; the proportion OVER 1.27 would be the same as the
proportion UNDER -1.27, or about .10, according to Table A
- lightest 5% have z=-1.64, observed value = 7.6 + (-1.64)(1.1)
= 5.8 lbs.
-
- r is about halfway between 0 and 1, so we'd call the relationship
moderate
- predicted weight = -106 + 3.7(70) = 153
- (i) 60 is outside the range of x-values used to produce the regression
line
- (iii) stay the same: r is unaffected by assignment of roles for
explanatory and response variables
-
- $2490(172/82.4)=$5,198
- (iii) These universities became relatively more expensive (almost
double what they would have been in terms of cost of living)
-
- (i) long term trend
- (ii) seasonal components: each year it changes from summer to winter
- A lot of people got (b) 5 and (g) 4 mixed up; I only took off 5 points instead
of 10 because the difference is subtle.
- (7)
- 5 (the common cause is slower metabolism)
- 1
- 3
- 2
- 6
- 4 (income is a confounding variable)
-
- (i) males: 18/34 as opposed to 17/50 for the females
- calculate (column total * row total)/table total: 20,29,14,21
- .8, .6, 1.1, .8
- .8+ .6+ 1.1+ .8=3.3
- 3.3 < 3.84, so we do not have convincing statistical evidence of a relationship
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