Chi-Square Test; - ANOVA; (D) Thesis

Open, View, and/or
Download this Document

Total Length: 852 words ( 3 double-spaced pages)

Total Sources: 3

Page 1 of 3

75

The standard value of 27.75 represents the distance of each score or frequency of representation of each employment category to the average or mean score or frequency for the distribution (i.e., employment categories.

Looking at the gender-commuting relationship, a possible hypotheses that can be developed from these variables are the following:

Ho: There is no significant relationship between commuting and gender.

H1: There is a significant relationship between commuting and gender.

SPSS results showed that generally, respondents went to work by driving a car. Directionally, females are more likely to take the passenger train (73%), while more than the majority of male respondents work at home (62%). However, these findings are not significant, and the X2 asymp. sig value of 0.101 showed that p< 0.05, which means that Ho is retained -- that is, there is no significant relationship between commuting and gender.

To determine the significance of the relationship between the variables savings and loans and other financiers, an independent samples t-test will be conducted as the statistical analysis. The null hypothesis for this analysis is:

Ho: There is no significant difference between savings and loans and other financiers in the average payback period necessary to justify solar heating systems for residences.

Stuck Writing Your "Chi-Square Test; - ANOVA; (D)" Thesis?

View All Our Example Chi-Square Test; - ANOVA; (D) Thesis

Have A Custom Example Essay Written



To conduct the t-test, the following formula will be used:

observed difference between sample means / standard error of the difference between means

X1-X2 / S (x1-x2) where: S (x1-x2) = ?sx12 + sx22

Applying the given data to the formula above, we get:

Savings & Loans

Other Financiers

Sample mean

Standard error

S (x1-x2)

Sample size df (n1 + n2 -2)

(8.7-7.7) / 1.05 = 1 / 1.05 = 0.95

At t=0.95 and df=162, the difference between the sample of savings and loans and sample of other financiers represents no real difference between the larger population of these two variables/groups. Thus, null hypothesis is retained.

Testing for the significance of the difference between evaluation ratings of managers from Western or Eastern region, the following data is used:

West

East

Sample mean

Standard error

S (x1-x2)

Sample size df (n1 + n2 -2)

The same formula for t-test will be used:

X1-X2 / S (x1-x2) where: S (x1-x2) = ?sx12 + sx22

Applying the figures in the table, we get:

(91-71) / 4.9 = 20 / 4.9 = 0.41

At t=0.41 and df=12, the difference between t between managers from the West and East and their evaluation ratings represent no real difference between.....

Show More ⇣

     Open the full completed essay and source list


OR

     Order a one-of-a-kind custom essay on this topic


sample essay writing service

Cite This Resource:

Latest APA Format (6th edition)

Copy Reference
"Chi-Square Test - ANOVA D " (2008, December 13) Retrieved June 5, 2026, from
https://www.aceyourpaper.com/essays/chi-square-test-anova-d-25820

Latest MLA Format (8th edition)

Copy Reference
"Chi-Square Test - ANOVA D " 13 December 2008. Web.5 June. 2026. <
https://www.aceyourpaper.com/essays/chi-square-test-anova-d-25820>

Latest Chicago Format (16th edition)

Copy Reference
"Chi-Square Test - ANOVA D ", 13 December 2008, Accessed.5 June. 2026,
https://www.aceyourpaper.com/essays/chi-square-test-anova-d-25820
   

Open, View, and/or
Download this Document
order custom essay example