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ID: 3365448 • Letter: H
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Here . lear" ×My Grades-M × Homenork × y G Search. Testboo xVE nBnault.. X(@in OY LEWIS ·Socure | https://www 45631 MAT-240-R2441-17EW2 Homework: 5-2 MyStatLab: Module Five Problem Set Score: 0 of 5 pts 11.3.21 Save 130114 (10 complete) Hw Score: 38 72%, 25 17 of 65 pts Question Help Atandom sample of 40 adults with no children under the age of 18 years results n a mean daly lesure e of 5 97 hours wth a standard devation of 2 34 hours A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.23 hours, with a standard deviation of 1.81 hours Construct and interpret a 96% confdence interval for the mean dference in lesure tree between adults with no children and adults with children ( Let , regrese t the mean lesure hours of adults with no chairen under the age of 18 aan represent the mean lesse hours of ada wm eden aide he age or 18 The 96% cortoence rterval for (n-P2)"the range fromhours to hors Round to two decimal places as needed) Enter your answver in the edit feis and then cick Check Answer F10 F11 F12 insertP.ScExplanation / Answer
TRADITIONAL METHOD
given that,
mean(x)=5.97
standard deviation , s.d1=2.34
number(n1)=40
y(mean)=4.23
standard deviation, s.d2 =1.81
number(n2)=40
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((5.48/40)+(3.28/40))
= 0.47
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 39 d.f is 2.02
margin of error = 2.023 * 0.47
= 0.95
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (5.97-4.23) ± 0.95 ]
= [0.79 , 2.69]
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DIRECT METHOD
given that,
mean(x)=5.97
standard deviation , s.d1=2.34
sample size, n1=40
y(mean)=4.23
standard deviation, s.d2 =1.81
sample size,n2 =40
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 = [( 5.97-4.23) ± t a/2 * sqrt((5.48/40)+(3.28/40)]
= [ (1.74) ± t a/2 * 0.47]
= [0.79 , 2.69]
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interpretations:
1. we are 95% sure that the interval [0.79 , 2.69] 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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