Answer to Question #323246 in Statistics and Probability for ana

Question

In a study of inter spousal aggression and its possible effect on child behavior, the Behavior Problem Checklist (BPC) scores were recorded for 47 children and whose parents were classified as aggressive. The sample mean and standard deviation were 7.92 and 3.45, respectively. For a sample of 38 children whose parents were classified as non-aggressive, the mean and standard deviation of the BPC scores were 5.80 and 2.87, respectively. Do these observations substantiate the conjecture that the children of aggressive families have higher mean BPC than those of non- aggressive families? Use a significance level of 0.05.

Answer 1

$H_0:\mu _1\leqslant \mu _2\\H_1:\mu _1>\mu _2\\n_1=47\\n_2=38\\\bar{x}_1=7.92\\\bar{x}_2=5.80\\s_1=3.45\\s_2=2.87\\\nu =\frac{\left( \frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2} \right) ^2}{\frac{1}{n_1-1}\left( \frac{{s_1}^2}{n_1} \right) ^2+\frac{1}{n_2-1}\left( \frac{{s_2}^2}{n_2} \right) ^2}=\frac{\left( \frac{3.45^2}{47}+\frac{2.87^2}{38} \right) ^2}{\frac{1}{46}\left( \frac{3.45^2}{47} \right) ^2+\frac{1}{37}\left( \frac{2.87^2}{38} \right) ^2}=82.9204\approx 83\\T=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}}=\frac{47-38}{\sqrt{\frac{3.45^2}{47}+\frac{2.87^2}{38}}}=13.1278\\P-value:\\P\left( T>13.1278 \right) =1-F_{t,83}\left( 13.1278 \right) =F_{t,83}\left( -13.1278 \right) =2.95\cdot 10^{-22}<0.05\Rightarrow \\\Rightarrow \mu _1>\mu _2$