Basic Applied Statistics 200
Solutions to Midterm 2

 .2*.1=.02 (AND means MULTIPLY)
 .35+.35=.70 (OR means ADD)
 (iii) skewed right (sketch a histogram; right tail is longer)
 1.76/(square root of 36)=.2933

 (.90)(50)=45
 (.10)(50)=5

 150/500=.3
 125/315=.4
 (iii)
 (ii)
 (v)
 (b)

 30
 (ii)
 (i)
 mean is 30(.65)=19.5; sd is square root of 30(.65)(.35)=2.61
 P(X>25)=P(Z>(2519.5)/2.61)=P(Z>2.11)=.0174
 (i) very unusual because Pvalue is very small

 not disjoint, because there is some overlap (males with earrings)
 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.

 Ho:mu=571; Ha: mu not equal to 571 ("Is this significantly different
from..." suggests the general, twosided alternative.)
 z=(587571)/(112/square root of 322)=2.56
 Pvalue=2P(Z>2.56)=2(.0052)=.0104 [I also gave credit if you
produced the interval .02 to .01 using Table D.]
 (ii) .0104 is small
 (ii) 322 is large
 .0052 (or the interval from .01 to .005)

 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)
 (i) yes, since 7 is in the interval
 (v)
 (i) reducing C reduces t* which makes the interval narrower;
(iii) larger n leads to a narrower interval [larger sigma means s would
tend to be larger, too, which results in a wider interval]

 (i) yes
 (i) yes definitely, because the Pvalue .013 is quite small
 (i) flipping a coin lets order be random
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