Basic Applied Statistics 200
Solutions to Practice Final

 (vi) sidebyside 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 responseear lengthincreases for every unit increase of
the explanatory variableyear)
 ear length

 (i) z test about a proportion
 (v) t test about a mean with twosided alternative
 (vi) twosample t test with onesided alternative
 (ii) z test about a mean with onesided 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 pvalue 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; drug or alcohol addiction is not a treatment that
would be imposed by researchers
 null hypothesis states no relationship between addiction and gray hair;
alternative states there IS a relationship
 addicts (0.53 vs. 0.1625 for nonaddicts)
 addicted and gray 47.2; addicted and not gray 152.8; not addicted and gray 188.8; not addicted and not gray 611.2
 73.25+22.63+18.31+5.66=119.85
 (21)*(21) = 1
 (i) Pvalue very small, since 119.83 is much more than 3.84
 (i) and (iv) because of small pvalue

 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;
pvalue 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
 pvalue is not small at all because t=1.18 is not large at all
 no because the pvalue 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=(1815)/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
 pvalue is P(Z>.89); (i) not small
 no, because the pvalue 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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