Question

Can someone send me a screen shot of how to input this into SPSS data and variable view to obtain the information provided below?

Question 1

The nominal scale of measurement is used to measure academic program in this example because data collected for the students is nothing but the names of their major subjects or classes and their status of mood whether they are nervous or excited.

Question 2

The nominal scale of measurement is used to measure feeling about PSY 3002 because the feeling is explained by two categories such as nervous and excited.

Question 3

This scenario required a Chi-square test of independence between two categorical variables. For this scenario, two categories are given as major subject and feeling. Here, we have to check the hypothesis whether the two categorical variables major subject of student and feeling of a student are independent of each other or not.

Question 4

For this scenario, the null and alternative hypotheses are given as below:

Null hypothesis: H₀: The two categorical variables major subject of student and the feeling of the student are independent of each other.

Alternative hypothesis: H_a: The two categorical variables major subject of student and feeling of student are not independent from each other.

We can also write these hypotheses as below:

Null hypothesis: H₀: There is no any relationship between two categorical variables such as major subject of student and feeling of student.

Alternative hypothesis: H_a: There is a relationship between the two categorical variables such as major subject of student and feeling of student.

Question 5

The Chi square test statistic is given as 16.94235589.

Question 6

Number of rows = r = 2

Number of columns = c = 2

Degrees of freedom = (r – 1)*(c – 1) = (2 – 1)*(2 – 1) = 1*1 = 1

Degrees of freedom = 1

Question 7

P-value = 0.0000385

(by using Chi square table or excel)

Question 8

Reject the null hypothesis because p-value is less than alpha value 0.05.

Question 9

Yes, results are statistically significant because p-value for this test is given as 0.0000385 which is less than alpha value 0.05.

Question 10

There is sufficient evidence to conclude that there is a relationship between the two categorical variables such as major subject of student and feeling of student.

There is sufficient evidence to conclude that two categorical variables major subject of student and feeling of student are not independent from each other.

Required output for Chi square test is given as below:

Observed Frequencies
	Column variable
Row variable	Nursing	Psychology	Total
Nervous	16	3	19
Excited	4	17	21
Total	20	20	40
Expected Frequencies
	Column variable
Row variable	Nursing	Psychology	Total
Nervous	9.5	9.5	19
Excited	10.5	10.5	21
Total	20	20	40

Data
Level of Significance	0.05
Number of Rows	2
Number of Columns	2
Degrees of Freedom	1
Results
Critical Value	3.841459149
Chi-Square Test Statistic	16.94235589
p-Value	0.0000385
Reject the null hypothesis

Answer 1

Can someone send me a screen shot of how to input this into SPSS data and variable view to obtain the information provided below?

Observed Frequencies
	Column variable
Row variable	Nursing	Psychology	Total
Nervous	16	3	19
Excited	4	17	21
Total	20	20	40

Enter data in SPSS in three variables. Row, col and frequency.

Now weight the data by frequency:

In menu DATA ---weight cases--- ( weight cases by frequency)

Or spss command is

WEIGHT BY frequency.

Now you do chi square analysis for the two variables row and column.

Analyze---descriptive statistics ---cross tabs ( select row in row variable and col in col variable, in sub menu statistics select chi square and ok

CROSSTABS

/TABLES=row BY col

/FORMAT=AVALUE TABLES

/STATISTICS=CHISQ

/CELLS=COUNT

/COUNT ROUND CELL.

row * col Crosstabulation
Count
		col		Total
		1	2
row	1	16	3	19
	2	4	17	21
Total		20	20	40

Chi-Square Tests
	Value	df	Asymptotic Significance (2-sided)	Exact Sig. (2-sided)	Exact Sig. (1-sided)
Pearson Chi-Square	16.942a	1	.000
Continuity Correctionb	14.436	1	.000
Likelihood Ratio	18.427	1	.000
Fisher's Exact Test				.000	.000
Linear-by-Linear Association	16.519	1	.000
N of Valid Cases	40
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 9.50.
b. Computed only for a 2x2 table