Basic Applied Statistics 200
Solutions to Practice Final
-
- (vi) side-by-side boxplots (comparing values of a quantitative variable
for several groups)
- (iv) scatterplot (looking at the relationship between two quantitative
variables)
- (ii) bar graph (looking at the relationship between two categorical
variables)
- (b) is matched pairs
- (b) a 99% confidence interval
-
- observational study
- age; quantitative
- ear length; quantitative
- b1 (the observed slope of the regression line which tells how
much the response---ear length---increases for every unit increase of
the explanatory variable---year)
- ear length
-
- (i) z test about a proportion
- (iv) t test about a mean with one-sided alternative
- (vii) two-sample t test with two-sided alternative
- (ii) z test about a mean with one-sided alternative
- (x) inference for regression
- (ix) ANOVA
- (viii) chi square test
-
- (v) +.6 because it is positive and moderate
- -113+36.5(9)=215.5; you should underline s = 75.93
- yes; you should circle the p-value 0.01
- no; 300 is not in the confidence interval
- (i) because lower x goes with lower y; the width of the PI only
widens slightly as x gets further from its mean [the width of the CI
widens more]
-
- (i) observational study; race and layoffs are not treatments that
would be imposed by researchers
- null hypothesis states no relationship between race and layoffs;
alternative states there IS a relationship
- (i) African Americans (.086 vs. .031 for whites)
- AA and Laid Off: 75.3; AA and Not Laid Off: 1434.7
White and Laid Off: 144.7; White and Not Laid Off: 2755.3
- 39.7+20.7+2.1+1.1 = 63.6
- (2-1)*(2-1) = 1
- (i) P-value very small, since 63.6 much more than 3.84
- (i) and (iv) because of small p-value
-
- 30.5 plus or minus 2 * 4.9/(square root of 81) = (29.4,31.6)
- (1) because multiplier is larger for 99% confidence (2) due to small
n, multiplier comes from t distribution and so it is larger (3) because of
dividing by square root of 11 instead of 81 in the margin of error
-
- null hypothesis: mu=50; alternative hypothesis: mu>50; t=2.67;
p-value between .05 and .025; yes there is evidence at the .05 level
- .10 and .05
- (iii)
- (i) administrators; (ii) students
-
- null hypothesis: population means equal for SF and LA; alternative:
population means not equal
- p-value is not small at all because t=1.18 is not large at all
- no because the p-value is not small
- yes
- no because sample sizes are medium
- circle the StDev's 1.02 and 1.09 (yes, the Rule of Thumb is satisfied)
-
- (i) more
- (ii) less
- (i) more
-
- (ii)
- large
- yes; circle s1, s2, and s3 or s1 (largest) and s2 (smallest)
- null hypothesis: the three population means are equal; alternative hypothesis:
not all three population means are equal
- DFG=2, DFE=15; MSG=145.5, MSE=3.7; F=39.3
- (ii) and (iii)
- (ii)
- (ii)
-
- mean = 15; standard deviation = square root of 60*.25*(1-.25) = 3.354
- z=(18-15)/3.354=.89 which is not large, so probability is (i) not small
- mean = .25; standard deviation = square root of .25*(1-.25)/60 = .056
- p-value is P(Z>.89); (i) not small
- no, because the p-value is not small
- .3 plus or minus 1.645 times square root of (.3)(1-.3)/60
=.3 plus or minus .1 = (.2, .4); yes, it contains .25
-
- (i) confounding variables
- observational study; (iv) ANOVA
- .44 plus or minus 1.96 times square root of (.44)(.56)/400
=.44 plus or minus .05 = (.39, .49)(i)
- reject Ho, conclude population proportion with increased desire to
quit is less than .5 (a minority)
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