QUESTION 1 Please use sample data provided in the following table in order to an

Question

QUESTION 1

Please use sample data provided in the following table in order to answer questions on this assignment. Round answers to three places after the decimal point. Unless noted otherwise use 5% level of significance for tests of hypothesis. For between-group mean comparisons you should assume equal variances across groups. After you enter data into the computer verify that there was no data entry error by computing and sample mean and sample standard deviation of each variable in your worksheet. These should be identical to statistics provided in the last two rows of the data table.

The sum of all Z1 values is _____.

a) 3396.730

b) 3936.730

c) 3923.730

d) 157.469

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) .005

.05

0.5

5.0

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

0.0063

6.3

0.63

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

Reject the null hypothesis

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

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:

18

0.31

0.05

0.76

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

QUESTION 8

[Use data from problem 1] If we subtract the mean value of Y for managers from the mean value of Y for supervisors, then the answer is _____.

0.220

-2.200

2.200

1.000

QUESTION 9

[Use data from problem 1] Median value of Z2 for all individuals in the sample is _____.

158.990

158.909

156.640

156.949

QUESTION 10

[Use data from problem 1] The coefficient of variation of Y is _____ times the the coefficient of variation of X.

10.19

1.019

0.982

101.879

QUESTION 11

[Use data from problem 1] The mean value of Z1 for all individuals designated as employees in the sample is _____.

2365.950

157.730

13.096

157.469

QUESTION 12

[Use data from problem 1] Assume that a new (26th) value of X becomes available. As a result the arithmetic mean of all 26 X values decreases to 5. The new X value must be _____.

5

10

5.2

0

QUESTION 13

[Use data from problem 1] Construct a new variable J such that J = 1 + 2*X. Mean and standard deviation of J are _____ and _____ respectively.

7.200, 2.784

11.400, 5.568

5.568, 11.400

2.784, 7.200

QUESTION 14

115.204

3731.015

151.204

None of the above

Person ID X Y Z1 Z2 Group 1 9 10 157.90 163.90 Manager 2 3 6 156.64 148.64 Manager 3 2 7 160.45 155.45 Manager 4 8 8 153.13 160.13 Manager 5 2 7 170.14 168.14 Manager 6 8 4 150.09 149.09 Supervisor 7 10 5 163.74 161.74 Supervisor 8 4 3 134.47 142.47 Supervisor 9 1 6 174.17 177.17 Supervisor 10 6 9 150.05 151.05 Supervisor 11 5 10 141.47 148.47 Employee 12 3 5 156.59 148.59 Employee 13 5 4 161.88 162.88 Employee 14 4 3 150.99 158.99 Employee 15 8 1 174.95 174.95 Employee 16 5 1 166.90 165.90 Employee 17 8 10 128.43 124.43 Employee 18 1 7 169.11 168.11 Employee 19 1 2 153.73 150.73 Employee 20 5 4 151.45 146.45 Employee 21 3 2 172.85 164.85 Employee 22 8 5 146.05 137.05 Employee 23 9 8 171.48 178.48 Employee 24 7 4 153.55 151.55 Employee 25 5 1 166.52 164.52 Employee Arithmetic mean 5.200 5.280 157.469 156.949

Explanation / Answer

Solution 1: Mean= 157.469

n= 25 and Mean= Sum/n= 157.469= Sum/25

Sum= 25*157.469= 3936.730 (option B)

Solution 2: t= xbar-mu/S/sqrt(n)

= 157.469-150/2.880/sqrt(25)

= 7.469/2.880/5

= 7.469/0.576

t = 12.967

The P-Value is < .00001. The result is significant at p < .05.

So option A will be most appropriate

solution 3) P value for t test 0.63

t stat= xbar-ybar/sqrt(S^2p(1/n1+1/n2))

where s^2p ispooled variance

d.f= n1+n2-2= 48

S^2p= 155.54

t stat= 157.469-156.949/sqrt(155.54*(1/25+1/25))

t stat= 0.1474

Solution 4) Conclusion of this test is Fail to Reject Null Hypothesis.

Solution 5) The critical test statistic value is: 2.06 (option A)

Note: I have done the first five. Please Repost rest question along with data. Thank you!

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