Answer to Question #327420 in Statistics and Probability for yyy

Question

The number of grams of carbohydrates contained in a 1-ounce servings of randomly selected chocolates and non-chocolate candy is listed below. Is there sufficient evidence to conclude that the difference in the carbohydrates content is significant?

Chocolate

29 25 17 36 41 25 32 29 38 34 24 27 29 31 23 42

Non-chocolate

41 41 37 29 30 38 39 10 29 55 29 34 32 38 53 36

 

 

 

 

 

 

 

 

Answer 1

$n_1=16\\n_2=16\\n_1=10\\n_2=9\\\bar{x}_1=30.125\\\bar{x}_2=35.6875\\{s_1}^2=46.7833\\{s_2}^2=106.3625\\\nu =\frac{\left( \frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2} \right) ^2}{\frac{1}{n_1-1}\left( \frac{{s_1}^2}{n_1} \right) ^2+\frac{1}{n_2-1}\left( \frac{{s_2}^2}{n_2} \right) ^2}=\frac{\left( \frac{46.7833}{16}+\frac{106.3625}{16} \right) ^2}{\frac{1}{15}\left( \frac{46.7833}{16} \right) ^2+\frac{1}{15}\left( \frac{106.3625}{16} \right) ^2}=26.0564\sim 26\\T=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{{s_1}^2}{n_1}+\frac{{s_2}^2}{n_2}}}=\frac{30.125-35.6875}{\sqrt{\frac{46.7833}{16}+\frac{106.3625}{16}}}=-1.79795\\P\left( \left| t \right|>1.79795 \right) =2F_{t,26}\left( -1.79795 \right) =2\cdot 0.0419=0.0838>0.05\Rightarrow \\\Rightarrow the\,\,difference\,\,is\,\,not\,\,significant$