Statistics and Probability for Business Management 1100
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. .1+.1+.2=.4
   2. 2.5, square root of 1.45 = 1.2
   3. half are below .3, so that is the median
6.  
   1. P(B given A) = P(A and B)/P(A) = .56/.7 = .8
   2. .8 = P(B given A) does not equal P(B) = .6, so not independent. Alternatively, P(A and B) = .56 does not equal P(A)P(B) = .7(.6) = .42, so not independent.
   3. Not mutually exclusive because P(A and B) = .56, so there IS overlap.
   4. P(A or B) = P(A) + P(B) + P(A and B) = .7 +.6 - .56 = .74
7.  
   1. (4/3)(1/2) - (1/3)(1/2)**4 = 2/3 - 1/48 = 31/48 = .646
   2. 4/3 - 4/3x**3
   3. Integrate (4/3)x - (4/3)x**4 from 0 to 1; answer is 2/5.
   4. Integrate (4/3)x**2 - (4/3)x**5 from 0 to 1 and subtract 2/5 squared.
8.  
   1. (vi) .9 because it is strong and positive
   2. no, it would not stray from the other scatter points
   3. -42.3 + 1.01(90) = 48.6; off by about s = 4.797
   4. 13 - 2 = 11
   5. 0.000
   6. yes, because the P-value is so small
9.  
   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 < .01
   8. (i) P-value is small, providing evidence of a relationship
   9. z = (.086-.031)/square root of .05(.95)(1/1510 + 1/2900)=7.96, P-value is approximately 0, reject the null hypothesis and conclude unequal proportions are laid off between African Americans and whites.
10.  
    1. 5 choose 2 times .2 squared times .8 cubed = .205
    2. Probability of X greater than or equal to 2 is the same as 1 minus probability of X less than or equal to 1 = 1-.737 = .262
    3. No: .263 is not a very low probability.
    4. mu is 50(.2) = 10; sigma is square root of 50(.2)(.8) = 2.83; z=(20-10)/2.83 = 3.53; probability of z above 3.53 is .0002.
    5. Yes: .0002 is a very low probability.
    6. z = (.4-.2)/square root of .2(.8)/50 = 3.53; again, the P-value is .0002, and we conclude the subject did significantly better than if he had been guessing.
11.  
    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)
    4. yes; 64 is a fairly large sample.
12.  
    1. df = 24, alpha = .005; 99% CI is 30 plus or minus 2.797 * 3.5/(square root of 25) = (28, 32)
    2. (i) and (vi) are correct; rule out (ii) and (iii) because the confidence interval is about the mean, not about individual values of the variable; rule out (iv) because population mean is fixed; it does NOT vary; rule out (v) because the interval is for population mean, not sample mean
    3. not really---25 isn't a very large sample
13.  
    1. .0004
    2. yes
    3. no
14.  
    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. DFT=2, DFE=15; MST=145.5, MSE=3.7; F=39.3; P-value<.05 so we conclude that the mean time is not the same for all 3 groups

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