Applied Statistical Methods 1000
Solutions to Practice Final
-
- (vi) side-by-side boxplots (comparing values of a quantitative variable
for several groups)
- (iv) scatterplot (looking at the relationship between two quantitative
variables)
- (ii) two-way table (looking at the relationship between two categorical
variables)
-
(ii) is matched pairs
-
- (i) z test about a proportion
- (iv) t test about a mean with one-sided alternative
- (vii) two-sample t test with two-sided alternative
- (ii) z test about a mean with one-sided alternative
- (x) inference for regression
- (ix) ANOVA
- (viii) chi square test
- (ii) a 99% confidence interval
-
- (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 p-value 0.01
- (ii) prediction interval
- (ii) prediction interval
-
- (i) observational study; race and layoffs are not treatments that
would be imposed by researchers
- null hypothesis states no relationship between race and layoffs;
alternative states there IS a relationship
- (i) African Americans (.086 vs. .031 for whites)
- 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
- 39.7+20.7+2.1+1.1 = 63.6
- (2-1)*(2-1) = 1
- P-value < .0005
- (i) P-value is small, providing evidence of a relationship
-
- 30 plus or minus 2.576 * 3.5/(square root of 25) = (28.2, 31.8)
- (1.96*3.5/2)squared = 11.76; round up to 12
-
- mean = 15; standard deviation = square root of 60*.25*(1-.25) = 3.354
- P(Z>(18-15)/3.354) = .1867
- mean = .25; standard deviation = square root of .25*(1-.25)/60 = .056
- P(Z>.89) = .1867
- no, because the p-value is large
-
- 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.
- .10 and .20
- (iii)
- (ii)
-
- null hypothesis: means equal for SF and LA; alternative: means not equal
- plots of 14 SF values and 16 LA values should be roughly symmetric
(combined sample size is medium, 30)
- df is smaller of 14 and 16, or 13; p-value between .20 and .30 since
alternative is two-sided
- no because the p-value is large
- circle the StDev's 1.02 and 1.09
-
- A has highest mean
- B has largest standard deviation
- yes; Q3+1.5(IQR)= 74+1.5(74-58)=98 is the cutoff for high outliers
-
- (ii)
- yes; circle s1, s2, and s3 or s1 (largest) and s2 (smallest)
- null hypothesis: the three 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; P-value<.001
so we conclude that the mean time is not the same for all 3 groups
- 14,3,10,6,11,16
-
- mu for X1+X2+X3 is 10+10+10=30, sigma is square root of 4+4+4=3.46
- mu for 3X is 3(10)=30, sigma is 3(2)=6
- binomial with n=20, p=.40, k=7 so probability is .1659
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