Question

1.) Problem: You are applying for jobs after graduating and are looking at a particular company. You’ve heard that they don’t pay women as much as men starting out and want to test it at the p<.01 level before deciding to work there. Luckily you are able to get publicly available salary information and randomly select the starting salary of 7 women and 6 men to see if women are paid less than men at this company. The salaries of the 7 women are: 45k, 48k, 50k, 42k, 46k, 51k, 43k. The salaries of the 6 men are: 46k, 48k, 52k, 49k, 47k, 51k.

1. Restate the question:

   • Population 1:

   • Population 2 (comparison):

   • Research hypothesis:

   • Null hypothesis:

   • b. Determine the characteristics of the comparison distribution to compare to sample distribution:

     • M1(Mean, Sample 1):

     • SS (Sum of squares, Sample 1):

     • df1 (degrees of freedom, Sample 1):

     • S21(estimated variance, Population 1):

     • M2(Mean, Sample 2):

     • SS (Sum of Squares, Sample 2):

     • df2 (degrees of freedom, Sample 2):

     • S22(estimated variance, Population 2):

     • dfTotal (total degrees of freedom):

     • S2Pooled (variance, pooled from Population 1 and Population 2):

     • c. Determine the critical t-value (cutoff score) on the comparison distribution at which the null hypothesis should be rejected

     • d. Determine the t score of your sample

     • e Decide whether to reject the null hypothesis

     • f. Explain the findings to one of your classmates who has not taken statistics

     • S2Difference (variance, distribution of the difference of means):

     • SDifference (standard deviation, distribution of the difference of means):

Answer 1

1.

Given that,
mean(x)=46.4286
standard deviation , s.d1=3.4087
number(n1)=7
y(mean)=48.8333
standard deviation, s.d2 =2.3166
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.01
from standard normal table,left tailed t α/2 =3.365
since our test is left-tailed
reject Ho, if to < -3.365
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =46.4286-48.8333/sqrt((11.61924/7)+(5.36664/6))
to =-1.5046
| to | =1.5046
critical value
the value of |t α| with min (n1-1, n2-1) i.e 5 d.f is 3.365
we got |to| = 1.5046 & | t α | = 3.365
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:left tail - Ha : ( p < -1.5046 ) = 0.09638
hence value of p0.01 < 0.09638,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 < u2
b.
mean(x)=46.4286
standard deviation , s.d1=3.4087
number(n1)=7
y(mean)=48.8333
standard deviation, s.d2 =2.3166
number(n2)=6
c.
critical value: -3.365
d.
test statistic: -1.5046
e.
decision: do not reject Ho
p-value: 0.09638
f.
we do not have enough evidence to support the claim that if women are paid less than men at this company.