please show all work and round to four decimals. r the Question(s) by Using the
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please show all work and round to four decimals.
r the Question(s) by Using the TI-30XA (or other approved Make sure you carry all decimals until you reach your answer, then round to ote: Answe PORTANT: Consider the following data for two independent random samples taken from two normal populations with equ variances. Find the 95% confidence interval for 1-2. problem.) (See exercise 11 on page 449 of your textbook for a si Sample 17 Sample 2 10 10 12 13 teft Endpoint Right Endpoint Daily ProblemExplanation / Answer
TRADITIONAL METHOD
given that,
mean(x)=8
standard deviation , s.d1=2.7568
number(n1)=6
y(mean)=12
standard deviation, s.d2 =2.3664
number(n2)=6
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((7.5999/6)+(5.5998/6))
= 1.4832
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 5 d.f is 2.5706
margin of error = 2.571 * 1.4832
= 3.8134
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (8-12) ± 3.8134 ]
= [-7.8134 , -0.1866]
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DIRECT METHOD
given that,
mean(x)=8
standard deviation , s.d1=2.7568
sample size, n1=6
y(mean)=12
standard deviation, s.d2 =2.3664
sample size,n2 =6
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 = [( 8-12) ± t a/2 * sqrt((7.5999/6)+(5.5998/6)]
= [ (-4) ± t a/2 * 1.4832]
= [-7.8134 , -0.1866]
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interpretations:
1. we are 95% sure that the interval [-7.8134 , -0.1866] 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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