Key Bits

Comprehensive Final Examination

Math 1107

Summer 2009

Case One

There are 4*4=16 Pairs:

(1,1), (1,2), (1,3), (1,4)

(2,1), (2,2), (2,3), (2,4)

(3,1), (3,2), (3,3), (3,4)

(4,1), (4,2), (4,3), (4,4)

Pr{(1 from 1^st, 1 from 2^nd)} = Pr{1 from 1^st)}* Pr{1 from 2^nd} = (1/10)*(4/10) = 4/100

Pr{(1 from 1^st, 2 from 2^nd)} = Pr{1 from 1^st)}* Pr{2 from 2^nd} = (1/10)*(3/10) = 3/100

Pr{(1 from 1^st, 3 from 2^nd)} = Pr{1 from 1^st)}* Pr{3 from 2^nd} = (1/10)*(2/10) = 2/100

Pr{(1 from 1^st, 4 from 2^nd)} = Pr{1 from 1^st)}* Pr{4 from 2^nd} = (1/10)*(1/10) = 1/100

Pr{(2 from 1^st, 1 from 2^nd)} = Pr{2 from 1^st)}* Pr{1 from 2^nd} = (2/10)*(4/10) = 8/100

Pr{(2 from 1^st, 2 from 2^nd)} = Pr{2 from 1^st)}* Pr{2 from 2^nd} = (2/10)*(3/10) = 6/100

Pr{(2 from 1^st, 3 from 2^nd)} = Pr{2 from 1^st)}* Pr{3 from 2^nd} = (2/10)*(2/10) = 4/100

Pr{(2 from 1^st, 4 from 2^nd)} = Pr{2 from 1^st)}* Pr{4 from 2^nd} = (2/10)*(1/10) = 2/100

Pr{(3 from 1^st, 1 from 2^nd)} = Pr{3 from 1^st)}* Pr{1 from 2^nd} = (3/10)*(4/10) = 12/100

Pr{(3 from 1^st, 2 from 2^nd)} = Pr{3 from 1^st)}* Pr{2 from 2^nd} = (3/10)*(3/10) = 9/100

Pr{(3 from 1^st, 3 from 2^nd)} = Pr{3 from 1^st)}* Pr{3 from 2^nd} = (3/10)*(2/10) = 6/100

Pr{(3 from 1^st, 4 from 2^nd)} = Pr{3 from 1^st)}* Pr{4 from 2^nd} = (3/10)*(1/10) = 3/100

Pr{(4 from 1^st, 1 from 2^nd)} = Pr{4 from 1^st)}* Pr{1 from 2^nd} = (4/10)*(4/10) = 16/100

Pr{(4 from 1^st, 2 from 2^nd)} = Pr{4 from 1^st)}* Pr{2 from 2^nd} = (4/10)*(3/10) = 12/100

Pr{(4 from 1^st, 3 from 2^nd)} = Pr{4 from 1^st)}* Pr{3 from 2^nd} = (4/10)*(2/10) = 8/100

Pr{(4 from 1^st, 4 from 2^nd)} = Pr{4 from 1^st)}* Pr{4 from 2^nd} = (4/10)*(1/10) = 4/100

High

Pair:High

(1,1):1, (1,2):2, (1,3):3, (1,4):4

(2,1):2, (2,2):2, (2,3):3, (2,4):4

(3,1):3, (3,2):3, (3,3):3, (3,4):4

(4,1):4, (4,2):4, (4,3):4, (4,4):4

Pr{High=1} = Pr{(1,1)} = 4/100 = .04 = 4%

Pr{High=2} = Pr{One of (1,2),(2,2),(2,1) Shows} = Pr{(1,2)}+Pr{(2,2)}+Pr{(2,1)} = (3/100)+(6/100)+(8/100) = 17/100 = .17 = 17%

Pr{High=3} = Pr{One of (1,3),(2,3),(3,3).(3,2),(3,1) Shows} = Pr{(1,3)}+Pr{(2,3)}+Pr{(3,3)} + Pr{(3,2)}+Pr{(3,1)} = (2/100)+(4/100)+(6/100) + (9/100)+(12/100) = 33/100 = .33 = 33%

Pr{High=4} = Pr{One of (1,4),(2,4),(3,4).(4,4),(4,1),(4,2),(4,3) Shows} = Pr{(1,4)}+Pr{(2,4)}+Pr{(3,4)} + Pr{(4,4)} + Pr{(4,3)} + Pr{(4,2)} + Pr{(4,1)} =

(1/100) + (2/100) + (3/100) + (4/100) + (8/100) + (12/100) + (16/100) = 46/100 = .46 = 46%

Case Two

Numbers

n=64

min=24

25^th pctl=37

50^th pctl=38

75^th pctl=40

max=45

upper quarter range = max – p75 = 45 – 40 = 5

lower middle quarter range = p50 – p25 = 38 – 37 = 1

lower quarter range = p25 – min = 37 – 24 = 13

middle half range = p75 – p25 = 40 – 37 = 3

Interpretation

There are 64 United Kingdom year 2007 resident live births in the sample.

The infant in the sample with the shortest gestation has a gestational age of 24 weeks.

Approximately 25% of the infants in the sample have gestational ages of 37 weeks or less.

Approximately 50% of the infants in the sample have gestational ages of 38 weeks or less.

Approximately 75% of the infants in the sample have gestational ages of 40 weeks or less.

The infant in the sample with the longest gestation has a gestational age of 45 weeks.

Approximately 25% of the infants in the sample gestated between 40 and 45 weeks. The largest difference in gestational age between any pair of infants in the upper quarter sample is 5 weeks.

Approximately 25% of the infants in the sample gestated between 37 and 38 weeks. The largest difference in gestational age between any pair of infants in the lower middle quarter sample is 1 week.

Approximately 25% of the infants in the sample gestated between 24 and 37 weeks. The largest difference in gestational age between any pair of infants in the lower quarter sample is 13 weeks.

Approximately 50% of the infants in the sample gestated between 37 and 40 weeks. The largest difference in gestational age between any pair of infants in the middle half sample is 3 weeks.

Case Three

Numbers

n=64

m=37.5937

sd=3.8861

Z=2.6

lower99 = m – Z*(sd/sqrt(n)) = 37.5937 – 2.6*(3.8861/sqrt(64)) = 36.33

upper99 = m + Z*(sd/sqrt(n)) = 37.5937 – 2.6*(3.8861/sqrt(64)) = 38.85

Interpretation

We estimate the population mean gestational age for year 2007 United Kingdom resident live births.

Each member of the Family of Samples is a single random sample of 64 live born infants. The FoS consists of all such possible samples.

From each member sample, compute: sample mean gestational age(m), sample standard deviation(sd) and the interval

[lower99 = m – 2.6*(sd/sqrt(n)), upper99 = m + 2.6*(sd/sqrt(n))]. Computing this interval for each member of the FoS yields a Family of Intervals, approximately 99% of which cover the true population mean.

If our interval resides in this supermajority, then the population mean gestational age for year 2007 UK resident liveborn infants is between 36.3 and 38.8 weeks.

Case Four

Numbers

n = 30

error form = “guess is too large/count strictly below”

error = 22

p = .8%

Interpretation

Each member of the Family of Samples is a single random sample of 30 Green Lynx spiders. The FoS consists of all such possible samples.

From each member sample, compute the number of spiders whose body length is strictly less than 13 millimeters. Computing this interval for each member of the FoS yields a Family of Errors.

If the population median body length for Green Lynx spiders is 13 millimeters, then approximately 0.8% of the member samples yield errors as bad as or more severe than our single error. Our sample appears to present highly significant evidence against the null median hypothesis.