(a)Report the t-test statistic value, degrees of freedom, and p-value assuming e

Question

(a)Report the t-test statistic value, degrees of freedom, and p-value assuming equal population variances from the SPSS output.

(b)Summarize your results – use the .05 level of significance. Use the p-value approach to make your statistical decision.

T-Test [DataSet0] Paired Samples Statistics Std. Error Mean Std. Deviation 7.03325 4.97996 Mean Pair 1 Commte 9.3333 2.87131 Solitary 7.0000 2.03306 Paired Samples Correlations Correlation Sig Pair 1 Committee &Solitary; 1.000 Paired Samples Test Paired Differences 95% Confidence Interval ofthe Difference Std. Error Mean Lower Upper df Sig. (2-tailed) 537 Mean Std. Deviation Pair 1 Committee - Solitary 2.33333 8.61781 3.51821 6.71050 11.37717 663

Explanation / Answer

From given SPSS output, we get,

a)

t-test statistic = 0.663

degrees of freedom = 5

P-value = 0.537

b) level of significance = 0.05

Decision rule:

If p-value is less than level of significance then we reject the null hypothesis(H0), otherwise do not reject null hypothesis(H0).

Here, p-value = 0.537 > 0.05

that is , p-value =0.537 is greater than level of significance =0.05

Therefore, we do not reject/fail to reject the null hypothesis(H0).

Conclusion:

There is no sufficient evidence that the two population means are different.

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