Question

Determine the decision criterion for rejecting the null hypothesis in the given hypothesis​ test; i.e., describe the values of the test statistic that would result in rejection of the null hypothesis. We wish to compare the means of two populations using paired observations. Suppose that d overbar equals​3.125, subscript dequals​2.911, and nequals​8, and that you wish to test the following hypothesis at the​ 10% level of significance. H0​: mu Subscript d equals 0 against H1​: muSubscript dgreater than0 What decision rule would you​ use?

Answer 1

Answer:

$t = \frac{3.125-0}{\frac{2.911}{\sqrt{8}}}= 3.036$

We can use the p value as a decision rule is $p_v$ we reject the null hypothesis.

Now we can find the degrees of freedom given by:

$df = n-1=8-1=7$

And the p value would be:

$p_v = P(t_{7} >3.036) = 0.0094$

And since the $p_v$ we have enough evidence to reject the null hypothesis in favor to the alternative hypothesis.

Step-by-step explanation:

For this case we have the following statistics for the difference between the paired observations:

$\bar d = 3.125$ the sample mean for the paired difference

$s_d = 2.911$ the sample deviation for the paired difference data

$n =8$ the sample size

The system of hypothesis that we want to check is:

Null hypothesis: $\mu_d =0$

Alternative hypothesis: $\mu_d > 0$

And the statistic is given by:

$t = \frac{\bar d -\mu_d}{\frac{s_d}{\sqrt{n}}}$

And replacing we got:

$t = \frac{3.125-0}{\frac{2.911}{\sqrt{8}}}= 3.036$

We can use the p value as a decision rule is $p_v$ we reject the null hypothesis.

Now we can find the degrees of freedom given by:

$df = n-1=8-1=7$

And the p value would be:

$p_v = P(t_{7} >3.036) = 0.0094$

And since the $p_v$ we have enough evidence to reject the null hypothesis in favor to the alternative hypothesis.