Question

Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 5.341a 3 .148 Likelihood Ratio 5.123 3...

Chi-Square Tests

Value

df

Asymp. Sig. (2-sided)

Pearson Chi-Square

5.341a

3

.148

Likelihood Ratio

5.123

3

.163

Linear-by-Linear Association

5.237

1

.022

N of Valid Cases

787

a. 1 cells (12.5%) have expected count less than 5. The minimum expected count is 4.84.

  1. Identify the independent and dependent variable.
  2. What is the null and alternate hypothesis?
  3. What is the test statistic value?
  4. Interpret the results in terms of the original research hypothesis. Assume alpha = .05.

Homework Answers

Answer #1

Chi-Square Tests

Value

df

Asymp. Sig. (2-sided)

Pearson Chi-Square

5.341a

3

.148

Likelihood Ratio

5.123

3

.163

Linear-by-Linear Association

5.237

1

.022

N of Valid Cases

787

a. 1 cells (12.5%) have expected count less than 5. The minimum expected count is 4.84.

a ) the null Hypothesis : there is no relationship between two varibales

alternate hypothesis : there is relationship between two varibales

b )  the test statistic value is 5.341

c ) since p value of test stat is 0.148 which is greater then 0.05 so we do not reject Ho and conclude that there is no relationship between two varibales

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Researchers were interested in examining whether a history of diabetes influenced where patients recovering from Coronary...
Researchers were interested in examining whether a history of diabetes influenced where patients recovering from Coronary Artery Bypass Graft surgery completed their cardiac rehab Chi-Square Tests Value df Asymp. Sig. (2-sided) Pearson Chi-Square 5.341a 3 .148 Likelihood Ratio 5.123 3 .163 Linear-by-Linear Association 5.237 1 .022 N of Valid Cases 787 a. 1 cells (12.5%) have expected count less than 5. The minimum expected count is 4.84. Identify the independent and dependent variable. What is the null and alternate hypothesis?...
What do the following results show from the chi-square test regarding relationship happiness and the financial...
What do the following results show from the chi-square test regarding relationship happiness and the financial comfort? Crosstab Count Financial Comfort Total Comfortable Struggling Relationship happiness Very unhappy 7 9 16 Unhappy 5 11 16 Mixed 32 59 91 Happy 79 69 148 Very happy 93 36 129 Total 216 184 400 Chi-Square Tests Value df Asymptotic Significance (2-sided) Pearson Chi-Square 34.031a 4 .000 Likelihood Ratio 34.880 4 .000 Linear-by-Linear Association 26.309 1 .000 N of Valid Cases 400 a....
Look at the relationship between marital status (MSTAT) and college graduation using a chi-square test. What...
Look at the relationship between marital status (MSTAT) and college graduation using a chi-square test. What would you conclude? Married people are more often college graduates than singles College graduates are more often married than non-graduates There is not a significant relationship between marital status and college graduation Both “a” and “b” are true Case Processing Summary Cases Valid Missing Total N Percent N Percent N Percent COLLEGE * Marital Status 400 100.0% 0 0.0% 400 100.0% COLLEGE * Marital...
A researcher is conducting a test to assess the relationship between parents’ level of education and...
A researcher is conducting a test to assess the relationship between parents’ level of education and whether choose private or public school. The researcher got the following result. What does it mean from the below result? value df Asymp. Sig. (2-sided) Pearson Chi-Square Likelihood Ratio Linear-by-Linear Association N of Valid Cases 17.45 17.651 1.951 2099 6 6 1 .008 .007 .162 0 cells (.0%) have expected count less than 5. The minimum expected count is 14.82.
Chapter 18 Chi-square – Test of Independence Certain editors at Sage like to think they’re a...
Chapter 18 Chi-square – Test of Independence Certain editors at Sage like to think they’re a bit of a whiz at soccer. To see how well they play compared with Sussex lecturers and postgraduates, we invited various employees of Sage to join in our soccer matches. Every player was allowed only to play in one match. Over many matches, we counted the number of players that scored goals. After running a chi-square test of independence, determine whether or not the...
Researchers were interested in whether patients with a history of angina are more likely to take...
Researchers were interested in whether patients with a history of angina are more likely to take anticoagulant drugs. (5 points) angina2 * Taking anticoagulant drugs Crosstabulation Taking anticoagulant drugs Total Yes No Angina Yes Count 311 25 336 Expected Count 177.3 158.7 336.0 Std. Residual 10.0 -10.6 No Count 242 470 712 Expected Count 375.7 336.3 712.0 Std. Residual -6.9 7.3 Total Count 553 495 1048 Expected Count 553.0 495.0 1048.0 Value df Asymp. Sig. (2-sided) Exact Sig. (2-sided) Exact...
Researchers were interested in examining the impact poverty may have on self-perceptions of health. Participants were...
Researchers were interested in examining the impact poverty may have on self-perceptions of health. Participants were asked to rate their health as in good or poor. Their income level was categorized as above or below the poverty line. Use the two tables below to answer the following questions. Identify the null and alternate hypothesis. Interpret test statistic and state conclusion based on original hypothesis. Assume alpha = .05 Interpret the odds ratio based on original research hypothesis. Interpret the relative...
A political scientist wanted to learn whether there is an association between the education level of...
A political scientist wanted to learn whether there is an association between the education level of registered voters and his or her political party affiliation. He randomly selected 46 registered voters and ran a Chi-square test of Independence and Homogeneity in SPSS. The following is the test result from SPSS: Education * Party Crosstabulation Party Total Democrat Republican Education College Count 9 12 21 Expected Count 11.9 9.1 21.0 Grade School Count 7 2 9 Expected Count 5.1 3.9 9.0...
A political scientist wanted to learn whether there is an association between the education level of...
A political scientist wanted to learn whether there is an association between the education level of registered voters and his or her political party affiliation. He randomly selected 46 registered voters and ran a Chi-square test of Independence and Homogeneity in SPSS. The following is the test result from SPSS: Education * Party Crosstabulation Party Total Democrat Republican Education College Count 9 12 21 Expected Count 11.9 9.1 21.0 Grade School Count 7 2 9 Expected Count 5.1 3.9 9.0...
Researchers were interested in examining the impact poverty may have on self-perceptions of health. Participants were...
Researchers were interested in examining the impact poverty may have on self-perceptions of health. Participants were asked to rate their health as in good or poor. Their income level was categorized as above or below the poverty line. Use the two tables below to answer the following questions. Identify the null and alternate hypothesis. Interpret test statistic and state conclusion based on original hypothesis. Assume alpha = .05 Interpret the odds ratio based on original research hypothesis. Interpret the relative...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT