Download I need help with hypothesis testing and excel

I need help with hypothesis testing and excel.
I have attached the sample database for use.
Can you please explain how this is done in excel and what the statistical formulas are.
I do know that for a hypothesis test with .05 or 95% must compute the z-value which is
actual - predit/som var - how do I do this.
1. Test a hypothesis to see whether the average overall job satisfaction (in the population
of all workers in the USA) is equal to 4.5 with a = .05.
a. State the null hypothesis, the alternative hypothesis, and the significance level.
Null hypothesis H0: the average overall job satisfaction (in the population of all workers
in the USA) is equal to 4.5
Alternative hypothesis H1: the average overall job satisfaction (in the population of all
workers in the USA) is NOT equal to 4.5
The significance level is alpha=0.05
b. Using the data in our database, calculate the test statistic.
Denote the sample mean by x and the sample standard deviation by s, and sample size n
Use excel, we get
Column1
Mean
Standard Error
Median
Mode
Standard
Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
4.330208
0.080387
4.4
4.6
1.364211
1.86107
-0.46875
-0.22289
6
1
7
1247.1
288
So, x =4.330208, s=1.364211 and n=288.
So, we can compute the t-statistic
t
x 4.5
s
n
4.3302084.5
1.364211
288  2.112
c. What is the critical level for the significance level?
Degree of freedom is n-1=288-1=287. Use a website
http://duke.usask.ca/~rbaker/Tables.html , we get the critical level for the significance
level is 1.968.
d. What is your conclusion? Do we accept or reject the null hypothesis?
Since |t|=2.112>1.968, we should reject the null hypothesis H0. So, we conclude that the
average overall job satisfaction (in the population of all workers in the USA) is NOT
equal to 4.5
You may use Excel for the calculations, but you need to answer all four parts of this
question.
2. Propose a hypothesis test for the mean intrinsic job satisfaction, similar to the test from
problem 1, and answer parts a, b, c, and d of problem 1 for this hypothesis test. You may
use Excel for the calculations, but you need to answer the four questions.
a. State the null hypothesis, the alternative hypothesis, and the significance level.
Null hypothesis H0: the average intrinsic job satisfaction (in the population of all
workers in the USA) is equal to 4.5
Alternative hypothesis H1: the average intrinsic job satisfaction (in the population of all
workers in the USA) is NOT equal to 4.5
The significance level is alpha=0.05
b. Using the data in our database, calculate the test statistic.
Denote the sample mean by x and the sample standard deviation by s, and sample size n
Use excel, we get
Column1
Mean
Standard Error
Median
Mode
Standard
Deviation
Sample
Variance
Kurtosis
Skewness
Range
Minimum
Maximum
5.219722
0.058936
5.17
5
1.000183
1.000366
0.870918
-0.4306
6
1
7
Sum
Count
1503.28
288
So, x =5.219722, s=1.000183 and n=288.
So, we can compute the t-statistic
t
x 4.5
s
n
5.2197224.5
1.000183
288  12.21
c. What is the critical level for the significance level?
Degree of freedom is n-1=288-1=287. Use a website
http://duke.usask.ca/~rbaker/Tables.html , we get the critical level for the significance
level is 1.968.
d. What is your conclusion? Do we accept or reject the null hypothesis?
Since |t|=12.21>1.968, we should reject the null hypothesis H0. So, we conclude that the
average intrinsic job satisfaction (in the population of all workers in the USA) is NOT
equal to 4.5.
3. We believe that half of the population would have an extrinsic job satisfaction of 5.0 or
greater. Answer parts a, b, c, and d of problem 1 for this hypothesis test of a proportion.
You may use Excel for the calculations, but you need to answer the four questions.
a. State the null hypothesis, the alternative hypothesis, and the significance level.
Null hypothesis H0: the average extrinsic job satisfaction (in the population of all
workers in the USA) is equal to 5.0
Alternative hypothesis H1: the average extrinsic job satisfaction (in the population of all
workers in the USA) is greater than 5.0
The significance level is alpha=0.05
b. Using the data in our database, calculate the test statistic.
Denote the sample mean by x and the sample standard deviation by s, and sample size n
Use excel, we get
Column1
Mean
Standard Error
Median
Mode
Standard
Deviation
Sample
Variance
4.911111
0.096975
5.5
6
1.645715
2.708378
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
-0.16404
-0.9033
6
1
7
1414.4
288
So, x =4.911111, s=1.645715 and n=288.
So, we can compute the t-statistic
t
x 4.5
s
n
4.9111115
1.645715
288  1.031
c. What is the critical level for the significance level?
Degree of freedom is n-1=288-1=287. Use a website
http://duke.usask.ca/~rbaker/Tables.html , we get the critical level for the significance
level is 1.65.
d. What is your conclusion? Do we accept or reject the null hypothesis?
Since |t|=1.031<1.65, we fail to reject the null hypothesis H0. So, we conclude that the
average extrinsic job satisfaction (in the population of all workers in the USA) is greater
than 5.0.
4. We believe that the variance of the overall job satisfaction is equal to 1.0 Answer parts
a, b, c, and d of problem 1 for this hypothesis test of a variance. You may use Excel for
the calculations, but you need to answer the four questions.
a. State the null hypothesis, the alternative hypothesis, and the significance level.
Null hypothesis H0: The variance of the overall job satisfaction is equal to 1.0
Alternative hypothesis H1: The variance of the overall job satisfaction is NOT equal to
1.0
The significance level is alpha=0.05
b. Using the data in our database, calculate the test statistic.
Denote the sample mean by x and the sample standard deviation by s, and sample size n
Use excel, we get
Column1
Mean
Standard Error
Median
Mode
Standard
Deviation
4.330208
0.080387
4.4
4.6
1.364211
Sample
Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
1.86107
-0.46875
-0.22289
6
1
7
1247.1
288
So, s=1.364211 and n=288.
So, we can compute the chi-squared statistic
 2  ( n1) s  287 *1.3642112  534.12
2
2
c. What is the critical level for the significance level?
Degree of freedom is n-1=288-1=287. Use a website
http://duke.usask.ca/~rbaker/Tables.html , we get the critical level for the significance
level is 327.512
d. What is your conclusion? Do we accept or reject the null hypothesis?
Since  2  534.12  327.512 , we should reject the null hypothesis H0. So, we conclude
that the variance of the overall job satisfaction is NOT equal to 1.0.
5. We will call a “deskbody” a person whose intrinsic job satisfaction level is higher than
their extrinsic job satisfaction level (i.e. happy with their job more than their office). We
will call a “socialbody” a person whose extrinsic job satisfaction level is higher than their
intrinsic job satisfaction level (i.e. happy with the office more than their job). We believe
that there are equal deskbodies and socialbodies in the work force.
a. State an appropriate null hypothesis and its alternative hypothesis.
Null hypothesis H0: there are equal deskbodies and socialbodies in the work force
Alternative hypothesis H1: there are not equal deskbodies and socialbodies in the work
force.
b. In our database, what percent of the employees are deskbodies? Are socialbodies?
From given information, we count that there are 141 deskbodies and 140 socialbodies. So,
141/288=48.96% of the employees are deskbodies, and
140/288=48.61% of the employees are socialbodies.
c. What did you do with the employees who had equal intrinsics and extrinsics?
When we test the null hypothesis H0, we can remove those people who have equal
intrinsics and extrinsics.
6. Determine the required sample size if you need to estimate the number of workers in
the United States who are highly satisfied with their job and you want the estimate to be
within 2 percentage points with a 96% confidence interval.
If we choose the overall job satisfaction, then we use Excel to get
Column1
Mean
Standard Error
Median
Mode
Standard
Deviation
Sample
Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
4.330208
0.080387
4.4
4.6
1.364211
1.86107
-0.46875
-0.22289
6
1
7
1247.1
288
The sample standard deviation is s=1.364211. So, t-value=2.05 if we use 96% significance
level. So, the standard error E  s t . By hypothesis, E<=2%. So, the required sample size n
n
can be determined by
E
s
n
t  2%
So,
1.364211
n
* 2.05  2%
So, the standard error
n  ( 1.364211
* 2.05) 2  19552.89
2%
So, we choose the minimum sample size n=19553
So, the
required sample size is 19553 if you need to estimate the number of workers in
the United States who are highly satisfied with their job and you want the estimate to be
within 2 percentage points with a 96% confidence interval.
hypothesis testing
------------------------------------------------------------------------------I need help with hypothesis testing and excel.
I have attached the sample database for use.
Can you please explain how this is done in excel and what the
statistical formulas are.
When we do hypothesis testing, we need to know what the sample
mean and sample standard deviation are. We can get them by using
Excel. For example, in question 1), we use Excel to get
Column1
Mean
Standard Error
Median
Mode
Standard
Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
4.330208
0.080387
4.4
4.6
1.364211
1.86107
-0.46875
-0.22289
6
1
7
1247.1
288
How can you get it by Excel? Open your excel file, choose “Tools” on
the top menu, choose “Data Analysis” and click it, you will have a
window from which you choose “Descriptive Statistics” and click “OK”.
You will have another window in which you need to input
$F$2:$F$289 if you want to get something about the Overall
satisfaction—column F; if you want to get something about “Intrinsic”column G, then you need to input $G$2:$G$289; then click”
Descriptive Statistics” and “OK”, you will get a table above.
I do know that for a hypothesis test with .05 or 95% must compute
the z-value which is actual - predit/som var - how do I do this.
In many times, you have to compute the t-value since we don’t use Ztest but t-test. If we know the standard deviation of population, then
we can use Z-test. In this case, we need to compute z-value. If we
don’t know the standard deviation of population, then we can use ttest. In this case, we need to compute t-value. Either case, we can use
a website http://duke.usask.ca/~rbaker/Tables.html (maybe you
can look up some distribution tables or critical values tables) to get zvalue or t-value.
If you use this website, you need to choose “critical values for z-test”
or “critical values for t-test” and click ”GO”.
For example, if you choose “critical values for z-test”, then you need to
select “1-tailed” or “two-tailed” and “significance level”, for instance,
alpha=0.05—you can select 0.05 from a list. If alpha=0.01, then you
select 0.01 from a list. Then you will get the critical value, namely, zvalue. If alpha=0.05, we can get z-value=1.645 if it is a “1-tailed” test;
z-value=1.96 if it is a “two-tailed” test.
Hope it makes sense. If you have further questions, feel free to ask
me.
Good luck!
Changping