Answer to Question #327843 in Statistics and Probability for Jade

Question

A course in mathematics is taught to 15 students by explicit method. Another group of 17 students was given the same course by means of another method. At the end of the semester, the same test was administered to each group. The 15 students under method A made an average of 85 with a standard deviation of 4, while the 17 students under method B made an average of 81 with a standard deviation of 5. Test

the null hypothesis of no significant difference in the performance of the two groups of

students at 5% level of significance.

Solution:

1. H0:

H1:

2. Level of Significance:

3.Test:

4. Critical Region:

5. Solution:

Make a decision and draw a

conclusion based from the results.

1. Decision:

2. Conclusion:

Answer 1

$1:\\H_0:\mu _1=\mu _2\\H_1:\mu _1\ne \mu _2\\2:\\\alpha =0.05\\3.Test:\\t-test\\n_1=15\\n_2=17\\\bar{x}_1=85\\\bar{x}_2=81\\s_1=4\\s_2=5\\\nu =\frac{\left( \frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2} \right) ^2}{\frac{1}{n_1-1}\left( \frac{{s_1}^2}{n_1} \right) ^2+\frac{1}{n_2-1}\left( \frac{{s_2}^2}{n_2} \right) ^2}=\frac{\left( \frac{4^2}{15}+\frac{5^2}{17} \right) ^2}{\frac{1}{14}\left( \frac{4^2}{15} \right) ^2+\frac{1}{16}\left( \frac{5^2}{17} \right) ^2}=29.7442\approx 30\\T=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}}=\frac{15-17}{\sqrt{\frac{4^2}{15}+\frac{5^2}{17}}}=-1.25559\\4.CriticalRegion:\\\left| T \right|>t_{inv}\left( \frac{1+\gamma}{2},\nu \right) =t_{inv}\left( 0.975,30 \right) =2.0422\\5.Solution:\\\left| T \right|<2.0422\\Make\,\,a\,\,decision\,\,and\,\,draw\,\,a\,\,conclusion\,\,based\,\,from\,\,the\,\,results.\\1.Decision:\\Not\,\,reject\,\,H_0\\2.Conclusion:\\The\,\,results\,\,in\,\,two\,\,methods\,\,are\,\,not\,\,significally\,\,different$