91影视

$$\begin{gathered} {\stackrel{炉}{x}}_1=54.1 \\ {\stackrel{炉}{x}}_2=68.7 \\ {n}_1=14 \\ {n}_2=15 \\ {s}_1=11.93 \\ {s}_2=13.3 \\ 伪=0.05 \end{gathered}$$

The appropriate hypotheses for this is:

$$\begin{gathered} H_0:{渭}_1-{渭}_2=-10 \\ {H}_a:{渭}_1-{渭}_2<-10 \end{gathered}$$

Now, find the test statistics:

$$\begin{gathered} t=\frac{({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2)-({渭}_1-{渭}_2)}{\sqrt{{\displaystyle \frac{s_1^2}{n_1}}+{\displaystyle \frac{s_2^2}{n_2}}}} \\ =\frac{54.1-68.7-(-10)}{\sqrt{{\displaystyle \frac{11.{93}^2}{14}}+{\displaystyle \frac{13.3^2}{15}}}} \\ =-0.982 \end{gathered}$$

Now, the degree of freedom will be:

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(14-1,15-1) \\ =13 \end{gathered}$$

So the P-value will be:

$0.15<P<0.20$

And we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, then,

$P>0.5鈬�FailtoReject{H}_0$

Thus, we conclude that there is no convincing evidence that the true mean cholesterol reduction with the new drug is more than ten milligrams per deciliter of blood greater than the mean cholesterol reduction with the current drug.

We conclude in part (b) that,

There is no convincing evidence that the true mean cholesterol reduction with the new drug is more than ten milligrams per deciliter of blood greater than the mean cholesterol reduction with the current drug.

A type I error occurs if we reject a null hypothesis when the null hypothesis is true. And the Type II error occurs if we fails to reject the null hypothesis when the null hypothesis is false.

Thus, in this case we fail to reject the null hypothesis then it is a Type II error.