Basic Applied Statistics 200
Solutions to Midterm 2 Spring 2002

 MORE THAN 3 means 4, 5, or 6: add the probabilities to get .2+.3+.1=.6
 Sketch a quick histogram with bars of height .1, .1, .2, .2, .3, and .1
to see that the distribution is skewed left (has a long left tail)

 distribution of sample mean has mean equal to population mean (3.5),
standard deviation equal to population standard deviation divided by square
root of sample size (.5/8=.0625) and shape approximately normal because (ii)
the sample is fairly large [according to the Central Limit Theorem]
 P(Xbar>3.2)=P(Z>(3.23.1)/.0625)=P(Z>1.6)=P(Z<1.6)=.0548

 P(T)=85/565=.15
 P(TP)=42/141=.30
 (iii) since the answer to (b) is twice as large as the answer to (a)

 .3 is a parameter because it describes the population (all students
at the university).
 The variable of interest is categorical (whether they commute or not)
 approx. binomial because (i) 200<1/10 of several thousand [this makes
sampling without replacement approximately independent]
 approx. normal because (iii) 200(.3)>10 and 200(.7)>10 [this makes the
sample size large enough relative to shape so that the Central Limit Theorem
holds.
 expect about 60 (mean is np=200(.3)), give or take about 6.48
(standard deviation is square root of np(1p), or square root of 200(.3)(.7))
 P(X<50)=P(Z<(5060)/6.48)=P(Z<1.54)=.0618

 285 plus or minus 2.576(94)/square root of 12= (215,355)
 YES, we anticipate the test to reject Ho:mu=200, because 200 is NOT
in the above interval.
 (iii) is the only correct choice
 (ii) smaller sample size makes the interval wider because we are
dividing by a larger number in the margin of error.
 (iv) 12 is a small sample size; xbar from a skewed population also
tends to be skewed.

 Ho:mu=7 vs. Ha: mu > 7
 (7.57)/(3/square root of 18)=.707
 For 181=17 df, .707 is between two critical values that have
p between .20 and .25
 (iii) The Pvalue is NOT small; NO evidence against Ho.
 If 3 had been sigma instead of s, we would have had a z distribution
instead of t, and could look up the probability in Table A: .2389

 (ii) NO, this was not matched pairs; it couldn't have been. It's
called Twosample, and has data for 23 females and 16 males.
 (iii) With a Pvalue of .07, the difference is not quite significant.
 twosided Pvalue is twice the onesided Pvalue: 2(.07)=.14
 P(FW)=146/(146+27)=.84
 Pvalue is smaller if the test statistic is larger;
 (ii) larger because it's in the numerator of the test statistic
 (ii) larger because it's in the numerator of the test statistic
 (i) smaller because it's in the denominator of the test statistic


 (v)
 (iv)
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