Basic Applied Statistics 200
Solutions to Practice Final

1. (ii) is matched pairs
2.  
   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)
3.  
   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
4. (ii) a 99% confidence interval
5.  
   1. A has highest mean
   2. B has largest standard deviation
6.  
   1. (v) +.6 because it is positive and moderate
   2. - 67.3+23.8(9) = 146.9; you should underline s = 45.64
   3. yes; you should circle the p-value 0.036
   4. (ii) prediction interval
   5. (ii) prediction interval
7.  
   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 < .0005
   8. (i) P-value is small, providing evidence of a relationship
8.  
   1. 30 plus or minus 2.576 * 3.5/(square root of 25) = (28.2, 31.8)
   2. (1.96*3.5/2)squared = 11.76; round up to 12
9.  
   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
10.  
    1. null hypothesis: mu=28; alternative hypothesis: mu>28; t=1.311; df=63 (use 60); p-value between .05 and .10, so NO, there is not compelling evidence that mu>28.
    2. .10 and .20
    3. (iii)
11.  
    1. null hypothesis: means equal for SF and LA; alternative: means not equal
    2. plots of 14 SF values and 16 LA values should be roughly symmetric (combined sample size is medium, 30)
    3. df is smaller of 14 and 16, or 13; p-value between .20 and .30 since alternative is two-sided
    4. no because the p-value is large
    5. circle the StDev's 1.02 and 1.09
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, DFT=17; MSG=145.5, MSE=3.7; F=39.3; P-value<.001 so we conclude that the mean time is not the same for all 3 groups

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