3) Make a Decision and Interpret the Results: Identify your p value: Describe ho

Question

3) Make a Decision and Interpret the Results: Identify your p value: Describe how you determined the p value:

Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper 401k% Equal variances assumed 2.153 .143 -2.412 298 .016 -1.641 .680 -2.979 -.302 Equal variances not assumed -2.395 278.338 .017 -1.641 .685 -2.990 -.292 Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper 401k% Equal variances assumed 2.153 .143 -2.412 298 .016 -1.641 .680 -2.979 -.302 Equal variances not assumed -2.395 278.338 .017 -1.641 .685 -2.990 -.292

Independent Samples Test

Levene's Test for Equality of Variances t-test for Equality of Means

F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference

Lower Upper

401k% Equal variances assumed 2.153 .143 -2.412 298 .016 -1.641 .680 -2.979 -.302

Equal variances not assumed -2.395 278.338 .017 -1.641 .685 -2.990 -.292

Independent Samples Test

Explanation / Answer

Result:

3) Make a Decision and Interpret the Results: Identify your p value: Describe how you determined the p value:

Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper

Equal variances assumed 2.153 .143 -2.412 298 .016 -1.641 .680 -2.979 -.302

Equal variances not assumed             -2.395 278.338 .017 -1.641 .685 -2.990 -.292

Independent Samples Test used.

Levene's Test for Equality of Variances shows that F=2.153, P=0.143, Equality of Variances assumption is not violated.

We have to take the Equal variances assumed t test results.

The calculated t= -2.412, df= 298, P= 0.016.

Since P=0.016 < 0.05 level, The null hypothesis is rejected.

We concludes that there is significant difference between the two group means.

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