Question

Refer to Exercise $37$.

a. Find the probability of getting a difference in sample means $\overline{x}M-\overline{x}B$that’s less than $0$mg/dl.

b. Should we be surprised if the sample mean cholesterol level for the $14-$year-old boys exceeds the sample mean cholesterol level for the men? Explain your answer.

Short answer

Part a. The probability is $0.0307$

Part b. Yes

Step-by-step solution

Part a. Step 1. Given information

From the previous exercise, we have:

$$\begin{gathered} \mu =18 \\ \sigma =9.6042 \\ x=0 \end{gathered}$$

Part a. Step 2. Explanation

The $z$ score is the value decreased by the mean and divided by the standard deviation. Then we have,

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0-18}{9.6042} \\ =-1.87 \end{gathered}$$

Now we will determine the value of the probability, that is:

$$\begin{gathered} P({\overline{x}}_M-{\overline{x}}_B<0)=P(Z<-1.87) \\ =0.0307 \\ =3.07\% \end{gathered}$$

Part b. Step 1. Explanation

It is given that:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0-18}{9.6042} \\ =-1.87 \end{gathered}$$

And the probability is $0.3070$

Thus, a probability is considered to be small when the probability is less than $0.05$.

We then note that the probability of the mean difference${\overline{x}}_M-{\overline{x}}_B$ being less than zero is small which means that it is unlikely that the mean difference is negative and thus we would be surprised if the sample mean cholesterol level for the fourteen year old boys exceeds the sample exceeds the sample mean cholesterol level for the mean. Thus, yes we should be surprised.