6. Construct a confidence interval for the difference of two population means ex

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6. Construct a confidence interval for the difference of two population means ex: A biologist who studies spiders was interested in comparing the lengths of female and male green lynx spiders. Assume that the length X of the male spider is approx imately N(Ax, VO) and that the length Y of the female spider is approximately N(uy, V ' ). Following are n = 30 observations of X 5.20 4.70 5.70 5.65 5.75 4.70 4.80 6.20 5.50 5.95 5.75 5.95 5.40 5.65 5.90 7.50 5.20 6.20 5.85 7.00 6.45 6.35 5.85 5.75 6.10 6.55 6.95 6.80 6.35 5.80 Following are m = 30 observations of Y 8.25 9.80 6.10 9.00 9.50 9.95 10.80 9.30 6.30 8.30 5.90 6.60 8.75 8.35 7.05 7.05 7.55 7.00 8.70 8.30 8.45 8.10 7.80 8.00 7.95 7.55 9.10 8.00 7.50 9.60 The units of measurement for both sets of observations are millimeters. Find an approximate 95% confidence interval for ex: Consider the butterfat production (in pounds) of a cow during a 305-day milk production period following the birth of a calf. Let X and Y be the butterfat production for such cows on a farm in Wisconsin and a farm in Michigan. Twelvie observations of X are 649 567 657 849 714 721 877 791 975874 468 405 Sixteen observations of Y are 699 597 891 868 632 652 815 978 589 479 764 733 524 549 727 790 v 2) ex: Students are weighed (in kilograms) at the beginning and end of a semester-long ram weight). Assume that the distribution of D is approx Assuming that X is approximately N(ux v 2) and Y is approximately N( find a 95% confidence interval for health-fitness program. Let D be the weight change for a student (post-program weight minus p imately N(HD, VaB). A random sample of n 12 female students yielded the following observations of D: 2.0 3.7-0.5 -0.1 1.4 0.6 -2.2 0.2 0.3 0.9-0.8-0.1 Find a 95% confidence interval for

Explanation / Answer

PART 2:
TRADITIONAL METHOD
given that,
mean(x)=712.25
standard deviation , s.d1=173.0833
number(n1)=12
y(mean)=705.4375
standard deviation, s.d2 =141.7114
number(n2)=16
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((29957.829/12)+(20082.121/16))
= 61.25
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, =
from standard normal table, two tailedand
value of |t | with min (n1-1, n2-1) i.e 11 d.f is 2.201
margin of error = 2.201 * 61.25
= 134.812
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (712.25-705.4375) ± 134.812 ]
= [-128 , 141.625]
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DIRECT METHOD
given that,
mean(x)=712.25
standard deviation , s.d1=173.0833
sample size, n1=12
y(mean)=705.4375
standard deviation, s.d2 =141.7114
sample size,n2 =16
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 712.25-705.4375) ± t a/2 * sqrt((29957.829/12)+(20082.121/16)]
= [ (6.813) ± t a/2 * 61.25]
= [-128 , 141.625]
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interpretations:
1. we are 95% sure that the interval [-128 , 141.625] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

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