Question

The output below is for a t-test for the hypothesis: Individuals living in bad neighborhoods commit more crime than those living in good neighborhoods. The data in the Group Statistics section provides you with the average number of crimes committed by individuals living in good and bad neighborhoods and the standard deviation of this same variable. The findings of significance are located in the Independent Samples Test section. Pay particular attention to the "t" column and the "Sig (2-tailed)"column. The "t" column is the t value that you would have calculated by hand (as we learned in class). The "Sig (2-tailed)"column provides you with the p value (the level of significance of this relationship). In this column, any value below .05 indicates that you would have rejected your null hypothesis.

Can someone tell me about the relationship of all these data? I'm trying to determine if my hypothesis, listed below, can be proven or disproven? What is the independent variable and dependent variable? Identify the level of measurement of each variable. You must indicate whether each variable is: (A) Nominal, (B) Ordinal, or (C)Interval/Ratio? And Which four tests should I be using and to solve it (A) t-test, (B) Chi-Square, (C) Analysis of Variance, or (D) Correlation Coefficient?

Output 4 [Document4] - IBM SPSS Statistics Viewer 00 + Output T-TEST GROUPS=Dum_Hood (1) Log /MISSINGEANALYSIS T-Test /VARIABLES=crime Title /CRITERIA=CI(.95). Notes Group Statis Independen → T-Test Group Statistics Std. Error Mean N Dum_Hood Good Nhood Std. Deviation 2.31003 2.66743 crime Mean 1.6087 1.9293 391 184 .11682 .19665 Bad Nhood Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means Sig. (2- tailed) F Sig. .108 df 573 Mean Difference -.32065 Std. Error Difference .21723 95% Confidence interval of the Difference Lower Upper -.74732 .10602 crime 2.585 -1.476 .140 Equal variances assumed Equal variances not assumed -1.402 316.470 .162 -.32065 .22873 -.77067 .12937

Answer 1

Given a output for t tsst for testing the hypothesis:

1) Ho: There is no impact of neighbourhood either good or bad on individuals commiting crimes.

H1: Individuals living in bad neighborhoods commit more crime than those living in good neighborhoods.

so, the data gives the relationship between individuals commiting crimes and the neighbourhood they have.

2) From the table of the independent sample t - test the values of the test statistic for equal variance assumed     and unequal variance assumed are -14.476 and -1.402 and the p - values are 0.140 and 0.162 respectively.

Since, both the p-values are greater than 0.05 so, we fail to reject the null hypothesis in both the cases and conclude that there is no impact of neighbourhood either good or bad on individuals commiting crimes.

3) Here, the individuals commiting crimes are depend on either good or bad neighbourhoods so, the dependent      variables is crime and independent variable is neighbourhood.

4) Here, the independent variable Neighbourhood measured in Nominal scale.

And the dependent variable crime measured in Intervel/Ratio scale

5) Here, we are using (A) t-test to solve this problem.