Question

In a pilot study, a company’s new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing evidence that the mean cholesterol reduction with the new drug is more than $10$milligrams per deciliter of blood (mg/dl) greater than with the current drug, the company will begin the expensive process of mass-producing the new drug. For the $14$subjects who were assigned at random to the current drug, the mean cholesterol reduction was $54.1$mg/dl with a standard deviation of $11.93$mg/dl. For the $15$subjects who were randomly assigned to the new drug, the mean cholesterol reduction was $68.7$mg/dl with a standard deviation of $13.3$mg/dl. Graphs of the data reveal no outliers or strong skewness.

a. Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not $H_0:{\mu }_1-{\mu }_2=0$.)

b. Based on your conclusion in part (a), could you have made a Type I error or a Type II error? Justify your answer.

Short answer

Part a. 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.

Part b. Type II error.

Step-by-step solution

Part a. Step 1. Given information

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

Part a. Step 2. Explanation

The appropriate hypotheses for this is:

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

Now, find the test statistics:

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

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,

$P>0.5\Rightarrow FailtoReject{H}_0$

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.

Part b. Step 1. Explanation

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.