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Nancy Pfenning
nancyp+@pitt.edu
  

Basic Applied Statistics 200
Solutions to Practice Midterm 2

  1.  
    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<5)=P(Z<(5-19.5)/2.61)=P(Z<-5.56)=0 (approximately)
    6. (i) virtually impossible because P-value is practically zero
    7. mean is p=.65, standard deviation is square root of (.65)(.35)/30=.087
  2.  
    1. mean plus or minus 2 sds: between 600 and 1440
    2. P(X>820)=P(Z>(820-1020)/210)=P(Z>-.95)=P(Z<+.95)=.8289
    3. top 3% have z=1.88, so x=1020+1.88(210)=1414.8, or 1415
  3.  
    1. mean is 3.5, sd is 1.7/5=.34
    2. (x) 90% of 50 is 45
    3. (ii) 10% of 50 is 5
  4.  
    1. 16/89=.18
    2. Ho:p=.20 vs. Ha:p not equal to .20
    3. z=(.18-.20)/sq rt of (.2)(.8)/89 = .47
    4. 2P(Z>|.47|)=2P(Z>.47)=2P(Z<-.47)=2(.3192)=.6384
    5. (ii) large P-value means no reason to reject Ho
    6. .18 plus or minus 2 times square root of .18(.82)/89 =.18 plus or minus .08 = (.10, .26)[You can also multiply by 1.96 instead of 2, but after rounding, the interval will be the same.]
    7. conservative margin of error is m=1/sq rt of n, so n=1/m squared= 1/(.10)(.10)=100
  5.  
    1. Note that sample size is small, so this should be a t confidence interval, not z. There are 15 degrees of freedom. CI is 7.1 plus or minus 2.95(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)
    4. (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]
  6. (b)
  7. Note: The problem statement should have presented 112 as sample standard deviation, not population standard deviation. Please fix this error on your practice exam.
    1. Ho:mu=571; Ha: mu not equal to 571 ("Is this significantly different from..." suggests the general, two-sided alternative.)
    2. t=(611-571)/(112/square root of 36)=2.14
    3. df = 36-1 = 35; use 30 df in Table A.2 (safer than 40)
    4. for 30 df, 2.14 is between 2.04 and 2.46, so the P-value is between 2(.025) and 2(.01); between .05 and .02
    5. (ii) from (d), P-value is smaller than alpha = .05 so we reject Ho and conclude Ha is true: mean math SAT differs from 571
    6. (iii)
    7. between .025 and .01
  8.  
    1. (i) parameter
    2. (iii) parameter
    3. (ii) statistic
    4. (iv) statistic
  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.  
    1. (ii)
    2. (i)
  11.  
    1. (i)
    2. (ii)
  12. (a)

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