Applied Statistical Methods 1000
Solutions to Midterm 2

1.  
   1. .2*.1=.02 (AND means MULTIPLY)
   2. .35+.35=.70 (OR means ADD)
   3. (iii) skewed right (sketch a histogram; right tail is longer)
   4. 1(.35)+2(.35)+3(.20)+4(.10)=2.05
   5. 1.76/(square root of 36)=.2933
2.  
   1. (.90)(50)=45
   2. (.10(50)=5
3.  
   1. 150/500=.3
   2. 125/315=.4
   3. (iii)
   4. (ii)
   5. (v)
4. (b)
5.  
   1. 30
   2. (ii)
   3. (i)
   4. mean is 30(.65)=19.5; sd is square root of 30(.65)(.35)=2.61
   5. P(X>25)=P(Z>(25-19.5)/2.61)=P(Z>2.11)=.0174
   6. (i) very unusual because P-value is very small
   7. mean is p=.65, sd is square root of (.65)(.35)/30=.087
6.  
   1. not disjoint, because there is some overlap (males with earrings)
   2. not independent, because knowing whether or not one event occurred gives us information about the probability of the other. For example, if I picked a student at random and told you the student had pierced ears, you would know the student is more likely to be a female.
7.  
   1. Ho:mu=571; Ha: mu not equal to 571 ("Is this significantly different from..." suggests the general, two-sided alternative.)
   2. z=(587-571)/(112/square root of 322)=2.56
   3. P-value=2P(Z>|2.56|)=2(.0052)=.0104 [I also gave credit if you produced the interval .02 to .01 using Table D.]
   4. (ii) .0104 is small
   5. (ii) 322 is large
   6. .0052 (or the interval from .01 to .005)
8.  
   1. Note that standard deviation comes from the sample, so this should be a t confidence interval, not z. There are 15 degrees of freedom. CI is 7.1 plus or minus 2.947(1.56)/square root of 16=7.1 plus or minus 1.15=(5.95,8.25)
   2. (i) yes, since 7 is in the interval
   3. (v)
9.  
   1. (i) yes
   2. (i) yes definitely, because the P-value .013 is quite small
   3. (i) flipping a coin lets order be random
10. Probability of 8 successes in 10 trials, when probability of success is .6, is the same as probability of 2 failures in 10 trials, when probability of failure is .4: .1209 is the correct entry in Table D.

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