N6208 Calculated Values of Chi Square

Running head: N6208 ASSIGNMENT 1
N6208 Assignment
Student’s Name
Institutional Affiliation

N6208 ASSIGNMENT 2
Chapter 8: Chi-Square
I. Given each of the following situations, determine whether the calculated values of chi-square
are statistically significant:
a) Χ2 = 3.72, df = 1, α = .05
The Chi-square tabulated value is 3.84.(from the chi-square table)
Chi-square Computed (3.72)< Chi-square Tabulated (3.84)
This implies that the chi-square value is not statistically significant at α = .05
b) Χ2 = 9.59, df = 4, α = .05
The Chi-square tabulated value is 9.49.(from the chi-square table)
Chi-square Computed (9.59)> Chi-square Tabulated (9.49)
This implies that the chi-square value is statistically significant at α = .05
c) Χ2 = 10.67, df = 3, α = .01
The Chi-square tabulated value is 11.34(from the chi-square table)
Chi-square Computed (10.67)< Chi-square Tabulated (11.34)
This implies that the chi-square value is not statistically significant at α = .01
d) Χ2 = 9.88, df = 2, α = .01
The Chi-square tabulated value is 9.21.(from the chi-square table)
Chi-square Computed (9.88)> Chi-square Tabulated (9.21)
According to Berenson, Timothy & David (2005),this implies that the chi-square value is
statistically significant at α = .01
II. Using the information provided, indicate which test you think should be used for each of the
following situations:
1) Independent variable: Infertility treatment A vs. infertility treatment B vs. control
condition; dependent variable: did vs. did not become pregnant; sample size: 180.
Chi-square test for independence
2) Independent variable: Drug A vs. Drug B vs. placebo; dependent variable: pain
measured on a 0 to 10 scale; sample size: 50 per group.
Analysis of Variance
3) Independent variable: gender (male versus female); dependent variable: preference
for nursing home placement versus home care for parent.
Chi-square test

N6208 ASSIGNMENT 3
III. Use the SPSS dataset Polit2SetB. Begin by running a crosstab (Analyze → Descriptive
Statistics → Crosstabs) for the variables bmicat and hlthlimit. The first is a variable that uses
the women’s body mass index (BMI), computed from their height and weight, to form four
groups: those classified in the normal BMI range (values under 25.0), overweight (values
from 25.00 to 29.99), obese (values from 30.0 to 40.0), and morbidly obese (values over
40.0). The second variable is dichotomous responses to whether or not the woman had a
health condition that limited her ability to work (yes is coded 1, no is coded 0). In the first
dialog box for the Crosstabs, enter hlthlimit as the row variable and bmicat as the column
variable. Click the pushbutton for Cells, and select Observed, Expected, and Row and
Column percentages. Click on Continue, then open the dialog box for Statistics. Select Chi-
square and also Phi and Cramer’s V. Then click Continue and OK to initiate the analysis.
Answer the following questions about the output:
1) How many normal-weight women were expected to have a health limitation, if
BMI and health limitations were unrelated? How many actually did have a
limitation? What are the two values for women who were morbidly obese?
For normal weight, 96.9 (approximately 97) were expected to have health
limitation. In this category,85 had actual health limitations.
In the category of morbidly obese women, 28 were expected to have health
limitations while 47 had actual limitations.
2) What percent of all women had a health condition that limited their ability to
work?
31.1% which represent 301 out of 969 women.
3) What is the null hypothesis being tested in this analysis?
Null hypothesis (Ho): The woman’s Body mass index and work limiting health
condition are independent.
4) What is the value of the chi-square statistic?
Chi-square value is 23.153
5) What is the degree of freedom?
Degrees of freedom is 3
6) Are the results statistically significant for α = .05? At what level (actual p value)?
The p=.000, this implies that p< .05 which suggest that this result is statistically
significant.( Ott & Longnecker,2015).

N6208 ASSIGNMENT 4
7) Comment on the nature of the relationship. Provide a short paragraph summarizing
your results.
The results of the study was .The p-value is
less than the alpha=.05 hence there is sufficient evidence to reject the null
hypothesis and accept the alternative hypothesis accepted (King'oriah, 2012).
8) What is the value of Cramer’s V?
Cramer’s V=0.155
9) Attach the SPSS output.
Table 1: Case Summary
Table 2: Health Condition Limit versus BMI
Table 3: Chi-square Tests

N6208 ASSIGNMENT 5
Table 4: Symmetric Measures

N6208 ASSIGNMENT 6
References.
Berenson, M. L., Timothy, C. K., David M. L. (2005). Basic business statistics: concepts and
applications. 10th ed. New York: Prentice Hall.
Grace-Martin, S. A. S. K. (2010). Data analysis with SPSS: A first course in applied statistics.
Statistics, 4, 27.
Johnson, R. A., & Bhattacharyya, G. K. (2019). Statistics: principles and methods. John Wiley & Sons.
King'oriah, G. K. ( 2012) "Fundamentals of applied statistics." Nairobi: The Jomo Kenyatta Foundation.
Ott, R. L., & Longnecker, M. (2015). An introduction to statistical methods and data analysis (7th ed.).
Pacific Grove, CA: Brooks Cole.
Pallant, J. (2013). SPSS survival manual. McGraw-Hill Education (UK).