Consider the following sample data drawn independently from normally distributed

Question

Consider the following sample data drawn independently from normally distributed populations with equal population variances. Use Table 2.

11.8

1 is the population mean for individuals with a CFA designation and 2 is the population mean of individuals with MBAs.

a. Construct the relevant hypotheses to test if the mean of the second population is greater than the mean of the first population.

H0: 1 2 0; HA: 1 2 > 0

b-1. Calculate the value of the test statistic. (Negative values should be indicated by a minus sign. Round all intermediate calculations to at least 2 decimal places and final answer to 2 decimal places.)

b-2. Calculate the critical value at the 5% level of significance. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

b-3. Using the critical value approach, can we reject the null hypothesis at the 5% significance level?

No, since the value of the test statistic is not less than the critical value of -1.761.

c. Using the critical value approach, can we reject the null hypothesis at the 10% level?

Sample 1 Sample 2 12.7 8.7 11.7 10.8 7.8 13.5 11.6 11.8 10.8 11.5 10.4 9.5 9.4 10.8 10.7

11.8

Explanation / Answer

Given that,
mean(x)=10.6375
standard deviation , s.d1=1.5109
number(n1)=8
y(mean)=11.05
standard deviation, s.d2 =1.4823
number(n2)=8
null, Ho: u1 - u2 >= 0
alternate,mean of the second population is greater than the mean of the first population H1: u1 - u2 < 0
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.761
since our test is left-tailed
reject Ho, if to < -1.761
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (7*2.282819 + 7*2.197213) / (16- 2 )
s^2 = 2.240016
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=10.6375-11.05/sqrt((2.240016( 1 /8+ 1/8 ))
to=-0.4125/0.748334
to=-0.551224
| to | =0.551224
critical value
the value of |t | with (n1+n2-2) i.e 14 d.f is 1.761
we got |to| = 0.551224 & | t | = 1.761
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: left tail - ha : ( p < -0.5512 ) = 0.29508
hence value of p0.05 < 0.29508,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 - u2 >= 0
alternate, H1: u1 - u2 < 0
test statistic: -0.55
critical value: -1.761
decision: do not reject Ho
b.
no, since the value of the test statistic is not less than the critical value of -2.145.
p-value: 0.29508
c.
No, since the value of the test statistic is not less than the critical value of -1.761.

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