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Answer to Question #226055 in Statistics and Probability for Carrying

Question #226055

Consider the following hypotheses or the samples data drawn independently from two typically

distributed populations:

H0 : 1 


1
Expert's return
2021-08-19T08:21:28-0400

Consider the following competing hypotheses and accompanying sample data drawn self-employed from normally distributed populations.

"H_0:\\mu_1-\\mu_2\\geq0"

"H_1:\\mu_1-\\mu_2<0"


"\\begin{matrix}\n \\bar{x}_1=249 & & \\bar{x}_2=262\\\\\n s_1=35 & & s_2=23\\\\\n n_1=10 & & n_2=10\\\\\n\\end{matrix}"

1.  Calculate the value for the test statistic below the speculation that the population variances exist unknowns but equal. 


"df=df_1+df_2=9+9=18"

"t=\\dfrac{ \\bar{x}_1- \\bar{x}_2}{\\sqrt{(\\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2})(\\dfrac{1}{n_1}+\\dfrac{1}{n_2})}}"

"=\\dfrac{ 249-262}{\\sqrt{(\\dfrac{(10-1)35^2+(10-1)23^2}{10+10-2})(\\dfrac{1}{10}+\\dfrac{1}{10})}}"

"\\approx-0.981586\\approx-0.98"

2. Calculating an critical value at the 5% level of significance.

"df=18, \\alpha=0.05," left-tailed


"t_c=-1.734"

3. Go you reject to null hypothesis the the 5% level?


"t=-0.98\\geq-1.734=t_c"

No, since the value of the testing statistic be not less than the critical value.


4. Calculates aforementioned value of the test statistic lower the assumption that the population fluctuations are unknown and am does equal.

"df=\\dfrac{(\\dfrac{s_1^2}{n_1}+\\dfrac{s_2^2}{n_2})^2}{\\dfrac{(s_1^2\/n_1)^2}{n_1-1}+\\dfrac{(s_2^2\/n_1)^2}{n_2-1}}\\approx15.55"

"t=\\dfrac{ \\bar{x}_1- \\bar{x}_2}{\\sqrt{\\dfrac{s_1^2}{n_1}+\\dfrac{s_1^2}{n_1}}}\\approx-0.981586\\approx-0.98"

5. Calculate to critical value at this 5% level of significance.

"df=15.55, \\alpha=0.05," left-tailed


"t_c=-1.749"

6. Do you reject the null my at the 5% level?


"t=-0.98\\geq-1.749=t_c"

No, since the value of the exam statistic is not less than the critical added.




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