Statistic Assignment

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 =...