Statistical Reasoning 90-707
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) +0.6 because it is positive and moderate
   2. -113+36.5(9)=215.5; s = 75.93 estimates about how far off your prediction should be. If the actual receipts are 295 million, the residual is observed minus predicted: 295-215.5=79.5 million
   3. yes; the p-value 0.01 is what you should use to decide
   4. r=0.6 suggests a moderately strong relationship, while p-value=0.01 suggests strong evidence that there is a relationship in general between average viewer rating and gross receipts.
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
   3. addicts (0.53 vs. 0.1625 for non-addicts)
   4. addicted and gray 47.2; addicted and not gray 152.8; not addicted and gray 188.8; not addicted and not gray 611.2
   5. 73.25+22.63+18.31+5.66=119.85
   6. (2-1)*(2-1) = 1
   7. (i) P-value very small, since 119.85 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 0.05 and 0.025; yes there is evidence at the 0.05 level
   2. between 2(0.05) and 2(0.025); that is, between 0.10 and 0.05
   3. (iii)
   4. administrators would want to avoid Type I Error, which would cause needless extra work for them; students would want to avoid Type II Error, which would cause them to continue to take a test that's too long.
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. We note that the StDev's 1.02 and 1.09 satisfy the Rule of Thumb: the larger isn't more than twice the smaller. (They're actually very close.)
11.  
    1. (i) more
    2. (ii) less
    3. (i) more
12.  
    1. (ii)
    2. large
    3. yes, because the largest standard deviation s1=2.168 is not more than twice the smallest, s2=1.414.
    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 = 0.25; standard deviation = square root of 0.25*(1-0.25)/60 = 0.056
    2. z=(0.30-0.25)/0.056 = 0.89
    3. p-value is P(Z>0.89); (i) not small
    4. no, because the p-value is not small
    5. 0.3 plus or minus 1.645 times square root of (0.3)(1-0.3)/60 =0.3 plus or minus 0.1 = (0.2, 0.4); yes, it contains 0.25
14.  
    1. (i) confounding variables
    2. observational study; (iv) ANOVA
    3. 0.44 plus or minus 2 times square root of (0.44)(1-0.44)/400 =0.44 plus or minus 0.05 = (0.39, 0.49)(i)
    4. Just barely reject Ho, conclude population proportion with increased desire to quit is less than 0.5, so it is a minority.

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