Submitted by: Submitted by Lauraba
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Category: Other Topics
Date Submitted: 10/30/2014 08:25 AM
Q1.
a)
Mean hourly wage rate for the population of men = 6.313021159
Mean hourly wage rate for the population of women = 5.146923868
b)
Confidence interval = ( x1-x2) ± tdf*s12n1+s22n2
( x1-x2)= (6.313021159 - 5.146923868) = 1.166097291
df = n1-1=1725-1=1724
= n2-1=1569-1=1568
Taking the smaller n = 1568
t1568,0.025*= 1.962
Margin of error = tdf* s12n1+s22n2= 1.962(3.4988613117)21725+(2.876237138)21569 = 1.962 x 0.1112180238 = 0.2182097627
Confidence interval =1.166097291 ± 0.2182097627 = [1.166097291 – 0.2182097627, 1.166097291 + 0.2182097627] = [0.9478875283, 1.384307054]
We are 95% confident that the true mean lies between 0.9478875283 and 1.384307054.
c)
Variance in the mean hourly wage rate in the population of men = 12.24203
Variance in the mean hourly wage rate in the population of women =8.27274
d)
Testing the null hypothesis:
H0 :σ12= σ22
Against the alternative hypothesis:
Ha: σ12 ≠ σ22
At significance level α= 0.05
The value of the test statistic is:
F=larger s2smaller s2= 3.49886122.8762372 = 1.47980347
The degree of freedom is: n1 – 1 = 1725 – 1 = 1724 (numerator)
n2 – 1= 1569 – 1 = 1568 (denominator)
Critical value of F1724,1568,0.05= 1.11
We reject HO if: the value of the test statistic > critical value = 1.47980347 > 1.11
This is true and so we reject HO. Now we can conclude that the two populations have the same variance.
e)
Testing the null hypothesis:
HO: µ1 = µ2 HO: (µ1 - µ2) = 0
Against the alternative hypothesis:
Ha: µ1 < µ2 Ha: (µ1 - µ2) < 0
At significance level α= 0.05
Sp2= n1-1s12+n2-1s22n1+n2-2 =1725-13.4988613172+1569-12.87623713821725+1569-2=21105.26061+12971.656443292=34076.917053292=10.35143288
Sp=3.217364275
t= x1-x2-(µ1-µ2)Sp1n1+1n2 =...