Applied Statistical Methods 1000
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) two-way table (looking at the relationship between two categorical variables)
2. (ii) is matched pairs
3. (ii) a 99% confidence interval
4.  
   1. (i) z test about a proportion
   2. (iv) t test about a mean with one-sided alternative
   3. (vii) two-sample t test with two-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
5.  
   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. (ii) prediction interval
   5. (ii) prediction interval
6.  
   1. (i) observational study; race and layoffs are not treatments that would be imposed by researchers
   2. null hypothesis states no relationship between race and layoffs; alternative states there IS a relationship
   3. (i) African Americans (.086 vs. .031 for whites)
   4. 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
   5. 39.7+20.7+2.1+1.1 = 63.6
   6. (2-1)*(2-1) = 1
   7. P-value < .001, since 63.6 >10.83
   8. (i) P-value is small, providing evidence of a relationship
7. 30.5 plus or minus 2.64 * 4.9/(square root of 81) = (29.1, 31.9)
8.  
   1. observational study
   2. age; quantitative
   3. ear length; quantitative
   4. $b_1$ (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
9.  
   1. null hypothesis: mu=50; alternative hypothesis: mu>50; t=2; df=15; p-value between .05 and .025, so YES, there is compelling evidence that mu>50.
   2. .10 and .05
   3. (iii)
   4. (i) administrators; (ii) students
10.  
    1. null hypothesis: means equal for SF and LA; alternative: means not equal
    2. df is smaller of 14-1 and 16-1, or 13; p-value > 2(.05)=.10 since alternative is two-sided
    3. no because the p-value is not small
    4. no, because samples are small
    5. circle the StDev's 1.02 and 1.09 (yes, the Rule of Thumb is satisfied)
11.  
    1. 1 has highest mean
    2. 2 has largest standard deviation
12.  
    1. (ii)
    2. yes; circle s1, s2, and s3 or s1 (largest) and s2 (smallest)
    3. null hypothesis: the three means are equal; alternative hypothesis: not all three population means are equal
    4. DFG=2, DFE=15; MSG=145.5, MSE=3.7; F=39.3>6.36; P-value<.01 so we conclude that the mean time is not the same for all 3 groups
    5. 14,3,10,6,11,16
    6. (i)
13.  
    1. mean = 15; standard deviation = square root of 60*.25*(1-.25) = 3.354
    2. P(Z>(18-15)/3.354) = .1867
    3. mean = .25; standard deviation = square root of .25*(1-.25)/60 = .056
    4. P(Z>.89) = .1867
    5. no, because the p-value is large
    6. .3 plus or minus 1.645 times square root of (.3)(1-.3)/60 =.3 plus or minus .1 = (.2, .4)
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)

[ Home | Calendar | Assignments | Handouts ]