Based on the analysis picture below please help me answer the following:

1. Explain why the t-test you chose is most appropriate for analyzing the data and whether the assumptions were met.
2. What are the null and alternative hypotheses for this scenario?
3. What is the critical value? What is the decision rule?
4. Per the output, what is the test statistic and p-value?
5. Interpretation of the t-test results (What was done? What was found? What does it mean? What suggestions are there for the creation of a health promotion intervention?).

Group Statistics 1=Female, 2=Male N Mean Sid. Deviation Sid. Error Mean Weight in kilograms Female 7507 61.7909 14.04763 16213 Male 7393 73.5642 17.06722 19850 Independent Samples Test Levene's Test for Equality of I-test for Equality of Variances Means F Sig. Weight in kilograms Equal variances assumed 244.339 4001 46.004 14898 Equal variances not 45.908 14284.173 assumed Page 1 Independent Samples Test I-test for Equality of Means Significance One-Sided p Two-Sided p Mean Difference Weight in kilograms Equal variances assumed 000 000 -11.77313 Equal variances not .000 1000 -11.77333 assumed Independent Samples Test I-est for Equality of Means 95% Confidence Interval of the Sid. Error Difference Difference Lower Upper Weight in kilograms Equal variances assumed 25592 12.27496 11.27170 Equal variances not .25630 -12.27571 -11.27096 assumed