Question

Suppose we want to estimate the difference between the average number of hours worked per week by all Americans with a college degree and those without a college degree. Summary information for each group is shown in the tables. 200 College degree 100 20 80 250 No college degree 50 80 Hours worked per week Statistic College Degree No College Degree Mean42.5 hours SD 39.1 hours 14.9 hours 14.7 hours 467 646 1. Create a 95% confidence interval for the difference in number of hours worked between the two groups. Round results to three decimal places. 2. Based on your confidence interval, is it reasonable to think that people with a college degree work more hours on average than people without a college degree? Why or why not? OA. No, because my confidence interval contains negative values. B. No, because my confidence interval does not contain any negative values. C. Yes, because my confidence interval does not contain any negative values. O D. Yes, because my conridence interval contains negative values.

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=42.5
standard deviation , σ1 =14.9
population size(n1)=467
y(mean)=39.1
standard deviation, σ2 =14.7
population size(n2)=646
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((222.01/467)+(216.09/646))
= 0.9
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 0.9
= 1.764
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (42.5-39.1) ± 1.764 ]
= [1.636 , 5.164]
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DIRECT METHOD
given that,
mean(x)=42.5
standard deviation , σ1 =14.9
number(n1)=467
y(mean)=39.1
standard deviation, σ2 =14.7
number(n2)=646
CI = x1 - x2 ± Z 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)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 42.5-39.1) ±Z a/2 * Sqrt( 222.01/467+216.09/646)]
= [ (3.4) ± Z a/2 * Sqrt( 0.81) ]
= [ (3.4) ± 1.96 * Sqrt( 0.81) ]
= [1.636 , 5.164]
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interpretations:
1. we are 95% sure that the interval [1.636 , 5.164] contains the difference between
true population mean U1 - U2
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 difference between
true population mean U1 - U2
3. Since this Cl does contain a zero we can conclude at 0.05 true mean
difference is zero
Answers:
1.
confidence interval = [1.636 , 5.164]
2.
option:B
no,
because my confidence interval does not contain negative values
and average hours of college degree work more hours on average than without a college degree,confidence
interval average not in the range.