Day31, Math151

Math 151 , Example of Two sample procedure, by hand.

10 Polyester cloth samples buried in landfill (BPS p. 343 ff). A random 5 dug up after 2 weeks, the others after 16
To compare the mean breaking strength (Does the cloth decay?)
Handout shows preliminary analysis (stemplot & boxplot) and SPSS results.

To do it by hand:

Group

Population

Sample size

Sample
mean

Sample
standard dev.

Standard Error
of the mean = SE

Sample
variance

(SE)²

2 weeks

1

n₁ = 5

xbar₁= 123.8

s₁= 4.6043

s₁/sqrt(n) =2.0591

s₁² =21.1996

s₁²/n =4.2399

16 wks

2

n₂ = 5

xbar₂= 116.4

s₂= 16.0873

s₂/sqrt(n) =7.1944

s₂² =258.8012

s₂²/n =51.7602

.

diff = 7.4

sum = 56.0001

SE _diff= sqrt(56.0001) = 7.4833

For hand work, d.f. = smaller of (5-1) and (5-1) = 4
t = diff/SE _diff= 7.4 / 7.483 = .989
     H₀: µ₁ - µ₂ = 0 same as µ₁= µ₂ , "no difference"
     H_a: µ₁ - µ₂ > 0 same as µ₁> µ₂ , shorter burial gives stronger cloth.
            Find t(4) in the t-table.   .989 is between .941 and 1.190, P-value is between .15 and .20
CI: (xbar₁ - xbar₂) + t*^. SE_diff     For 90% CI, t*(4) = 2.132
               7.4 + 2.132^. 7.4833
                7.4 + 15.96:   -8.56   to 23.36

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Group	Population	Sample size	Sample mean	Sample standard dev.	Standard Error of the mean = SE	Sample variance	(SE)²
2 weeks	1	n₁ = 5	xbar₁= 123.8	s₁= 4.6043	s₁/sqrt(n) =2.0591	s₁² =21.1996	s₁²/n =4.2399
16 wks	2	n₂ = 5	xbar₂= 116.4	s₂= 16.0873	s₂/sqrt(n) =7.1944	s₂² =258.8012	s₂²/n =51.7602
.			diff = 7.4				sum = 56.0001