# Basic Applied Statistics 200 Solutions to Practice Final

1.
1. (vi) side-by-side boxplots (comparing values of a quantitative variable for several groups)
2. (iv) scatterplot (looking at the relationship between two quantitative variables)
3. (ii) bar graph (looking at the relationship between two categorical variables)
2. (b) is matched pairs
3. (b) a 99% confidence interval
4.
1. observational study
2. age; quantitative
3. ear length; quantitative
4. 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)
5. ear length
5.
1. (i) z test about a proportion
2. (v) t test about a mean with two-sided alternative
3. (vi) two-sample t test with one-sided alternative
4. (ii) z test about a mean with one-sided alternative
5. (x) inference for regression
6. (ix) ANOVA
7. (viii) chi square test
6.
1. (v) +.6 because it is positive and moderate
2. -113+36.5(9)=215.5; you should underline s = 75.93
3. yes; you should circle the p-value 0.01
4. no; 300 is not in the confidence interval
5. (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]
7.
1. (i) observational study; drug or alcohol addiction is not a treatment that would be imposed by researchers
2. null hypothesis states no relationship between addiction and gray hair; alternative states there IS a relationship
5. 73.25+22.63+18.31+5.66=119.85
6. (2-1)*(2-1) = 1
7. (i) P-value very small, since 119.83 is much more than 3.84
8. (i) and (iv) because of small p-value
8.
1. 30.5 plus or minus 2 * 4.9/(square root of 81) = (29.4,31.6)
2. (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
9.
1. 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
2. .10 and .05
3. (iii)
10.
1. null hypothesis: population means equal for SF and LA; alternative: population means not equal
2. p-value is not small at all because t=1.18 is not large at all
3. no because the p-value is not small
4. yes
5. no because sample sizes are medium
6. circle the StDev's 1.02 and 1.09 (yes, the Rule of Thumb is satisfied)
11.
1. (i) more
2. (ii) less
3. (i) more
12.
1. (ii)
2. large
3. yes; circle s1, s2, and s3 or s1 (largest) and s2 (smallest)
4. null hypothesis: the three population means are equal; alternative hypothesis: not all three population means are equal
5. DFG=2, DFE=15; MSG=145.5, MSE=3.7; F=39.3
6. (ii) and (iii)
7. (ii)
8. (ii)
13.
1. mean = 15; standard deviation = square root of 60*.25*(1-.25) = 3.354
2. z=(18-15)/3.354=.89 which is not large, so probability is (i) not small
3. mean = .25; standard deviation = square root of .25*(1-.25)/60 = .056
4. p-value is P(Z>.89); (i) not small
5. no, because the p-value is not small
6. .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
14.
1. (i) confounding variables
2. observational study; (iv) ANOVA
3. .44 plus or minus 1.96 times square root of (.44)(.56)/400 =.44 plus or minus .05 = (.39, .49)(i)
4. reject Ho, conclude population proportion with increased desire to quit is less than .5 (a minority)