Basic Applied Statistics 200
Solutions to Practice Midterm 2

1.  
   1. no: np = 7 < 10
   2. yes: np = 28 > 10, n(1-p) = 12 > 10
   3. no: n(1-p) = 4 < 10
2.  
   1. n = 10
   2. mean is np = 10(.7) = 7; standard deviation is square root of 10(.7)(1-.7) = 1.45
   3.  
      1. circled, because being in the first row may change the probability of being female from .7
      2. not circled
      3. circled, because population size (200) not more than 10 times sample size (50) results in too much dependence when sampling without replacement
3.  
   1. probability of female but NOT off campus is .28, probability of female AND off campus is .42, probability of NOT female and off campus is .18, probability of neither is .12
   2. .7 + .6 - .42 = .88
   3. .42/.70 = .6
   4. independent because P(F and O) = .42 = P(F) P(O) = .7(.6) [not disjoint because there IS overlap]
4.  
   1. 38
   2. 4.5 divided by square root of 900, or .15
   3. (iii) approximately normal because sample size is large (this is the Central Limit Theorem in action)
   4. (iii) not unusual (just 1 standard deviation above the mean)
5. 95%, by definition of the confidence interval
6.  
   1. population size 5000, sample size 81, parameter 7, statistic 53
   2. 53 plus or minus 2.576(7)/9 = (51, 55)
   3. (i) narrower (ii) wider (iii) narrower
   4. yes, because 50 is outside the interval (51, 55)
   5. 1.645(7)/3 squared is 14.7; round up to 15
7.  
   1. null hypothesis is that mu = 2.0; alternative that mu > 2.0
   2. t = (2.2-2.0)/(1.3/5) = .77
   3. for 24 df, .77 is between .685 and .857, so P-value is between .25 and .20
   4. (iv)
   5.  
      1. not circled, because population is still more than 10 times sample size
      2. circled; for a medium sample size of 25, severe right skewness is a problem
      3. circled, because the sample must be random
8.  
   1. not true
   2. true
   3. true
9. (ii) large alpha (because if there are any doubts about the null hypothesis, it should be rejected)
10.  
    1. B (matched pairs)
    2. yes (P-value is .019)
    3. A (two-sample)
    4. no (P-value is .14)
    5. (1) matched pairs study is better because it controls for differences among individuals studied, and better pins down the effects of the explanatory variable of interest (dieting or not)

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