1. The sum of all Z1 values is _____. a. 3396.730 b. 3923.730 c. 3936.730 d. 157
ID: 2927042 • Letter: 1
Question
1. The sum of all Z1 values is _____.
a.
3396.730
b.
3923.730
c.
3936.730
d.
157.469
1 points
QUESTION 2
[Use data from problem 1] Conduct a t test to test the hypothesis that mean value of Z1 equals 150 in the population. The p value for this test is:
a.- 0.5
b.- .005
c.- .05
d.- 5.0
1 points
QUESTION 3
[Use data from problem 1] Conduct a t test to test the hypothesis that mean value of Z1 equals mean value of Z2 in the population. The p value for this test is:
0.063
6.3
0.0063
0.63
1 points
QUESTION 4
[Use data from problem 1] Conduct a t test to test the hypothesis that mean value of Z1 equals mean value of Z2 in the population. The conclusion of this test is:
Fail to reject the null hypothesis
b.-
Reject the null hypothesis
1 points
QUESTION 5
[Use data from problem 1] Conduct a t test to test the hypothesis that mean value of Z1 equals mean value of Z2 in the population. The critical test statistic value is:
2.06
-1.96
1.96
0.05
1 points
QUESTION 6
[Use data from problem 1] Conduct a t test to test the hypothesis that mean value of Z1 for employees and managers is the same. The observed value of the test statistic is:
0.31
18
0.05
0.76
1 points
QUESTION 7
[Use data from problem 1] Conduct a t test to test the hypothesis that mean value of Z1 for employees and managers is the same. The conclusion is:
Reject the null hypothesis
Fail to reject the null hypothesis
a.
3396.730
b.
3923.730
c.
3936.730
d.
157.469
Person ID Group 10 157.90 156.64 160.45 153.13 170.14 150.09 163.74 134.47 174.17 150.05 141.47 156.59 161.88 150.99 174.95 166.90 128.43 169.11 153.73 151.45 172.85 146.05 171.48 153.55 166.52 157.469 12.159 163.90 Manager 148.64 Manager 155.45 Manager 160.13 Manager 168.14 Manager 149.09 Supervisor 161.74 Superviso 142.47 Supervisor 177.17 Supervisor 151.05 Supervisor 148.47 Employee 148.59 Employee 162.88 Employee 158.99 Employee 174.95 Employee 165.90 Employee 124.43 Employee 168.11 Employee 150.73 Employee 146.45 Employee 164.85 Employee 137.05 Employee 178.48 Employee 151.55 Employee 164.52 Employee 156.949 12.776 2 10 10 5 10 12 13 14 15 16 17 18 19 20 21 5 5 10 2 4 23 24 25 Arithmetic mean Standard deviation 2.784 2.880 5.200 5.280Explanation / Answer
Q1.
Descriptive statistics Z1
count 25
mean 157.4692
sample variance 147.8366
sample standard deviation 12.1588
total 3936.730
Descriptive statistics Z2
count 25
mean 156.9492
sample variance 163.2366
sample standard deviation 12.7764
Q2.
Given that,
population mean(u)=150
sample mean, x =157.4692
standard deviation, s =12.1588
number (n)=25
null, Ho: =150
alternate, H1: !=150
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.064
since our test is two-tailed
reject Ho, if to < -2.064 OR if to > 2.064
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =157.4692-150/(12.1588/sqrt(25))
to =3.072
| to | =3.072
p-value = 0.005
Q3.
Given that,
mean(x)=157.4692
standard deviation , s.d1=12.1588
number(n1)=25
y(mean)=156.9492
standard deviation, s.d2 =12.7764
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.064
since our test is two-tailed
reject Ho, if to < -2.064 OR if to > 2.064
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =157.4692-156.9492/sqrt((147.83642/25)+(163.2364/25))
to =0.147
| to | =0.147
critical value
the value of |t | with min (n1-1, n2-1) i.e 24 d.f is 2.064
we got |to| = 0.14742 & | t | = 2.064
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.1474 ) = 0.884
hence value of p0.05 < 0.884,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.147
critical value: -2.064 , 2.064
decision: do not reject Ho
p-value: 0.884
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