Q. 7.46 - Elementary Statistics

Question

Young Adults at Risk. Research by R. Pyhala et al. shows that young adults who were born prematurely with very low birth weights (below 1500 grams) have higher blood pressure than those born at term. The study can be found in the article. "Blood Pressure Responses to Physiological Stress in Young Adults with Very Low Birth Weight" (Pediatrics, Vol. 123, No, 2, pp. 731-734 ). The researchers found that systolic blood pressures, of young adults who were born prematurely with very low birth weights have mean 120.7 mm Hg and standard deviation 13.8 mm Hg.
a. Identify the population and variable.
b. For samples of 30 young adults who were born prematurely with very low birth weights, find the mean and standard deviation of all possible sample mean systolic blood pressures. Interpret your results in words.
c. Repeat part (b) for samples of size 90. 

Step-by-Step Solution

Answer

Part a. The population includes all interested people who were examined and the variable is systolic blood pressure.

Part b. The mean and standard deviation of all possible sample systolic blood pressures for samples of 30 young adults are 120.7 mm Hg and 2.52 mm Hg.

Part c. The mean and standard deviation of all possible sample systolic blood pressures for samples of 90 young adults are 120.7 mm Hg and 1.45 mm Hg.

Part (a) Step 1. Given Information

We are given that the researchers found that systolic blood pressures, of young adults who were born prematurely with very low birth weights have mean 120.7 mm Hg and standard deviation 13.8 mm Hg. 

Part (a) Step 2. Identify the population and the variable

The population, in this case, includes all young adults who were born prematurely with very low birth weights. 

In this case, the systolic blood pressure of the young adults was measured. And also the systolic blood pressure varies from person to person. So the variable, in this case, is systolic blood pressure.

Part (b) Step 1. Find the mean for the sample

We know that the sample mean of a sample is equal to the population mean irrespective of the sample size.

The population mean in this case is given as $\mu =120.7$ mm Hg.

So when the sample size includes 30 young adults then the sample mean would be the same as the population mean.

Thus the mean of all possible sample mean systolic blood pressure of sample size $30$ is $120.7$ mm Hg.

Part (b) Step 2. Find the standard deviation

We know that the sample standard deviation of a sample is equal to the standard deviation of the variable under consideration divided by the square root of the sample size.

It is given that the standard deviation of the systolic blood pressure is $\sigma =13.8$ mm Hg.

So when the sample size is of 30 young adults who were born prematurely with very low birth weights then the standard deviation is given as

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{30}} \\ {\sigma }_{\overline{x}}=\frac{13.8}{\sqrt{30}} \\ {\sigma }_{\overline{x}}\approx 2.52 \end{gathered}$$

Thus the standard deviation of all possible sample mean systolic blood pressure of sample size $30$ is ${\sigma }_{\overline{x}}=2.52$ mm Hg.

Part (c) Step 1. Find the mean for the sample

We know that the sample mean of a sample is equal to the population mean irrespective of the sample size.

The population mean in this case is given as $\mu =120.7$ mm Hg.

So when the sample size includes $90$ young adults then the sample mean would be the same as the population mean.

Thus the mean of all possible sample mean systolic blood pressure of sample size $90$ is $120.7$ mm Hg.

Part (c) Step 2. Find the standard deviation

We know that the sample standard deviation of a sample is equal to the standard deviation of the variable under consideration divided by the square root of the sample size.

It is given that the standard deviation of the systolic blood pressure is $\sigma =13.8$ mm Hg.

So when the sample size is of $90$ young adults who were born prematurely with very low birth weights then the standard deviation is given as

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{90}} \\ {\sigma }_{\overline{x}}=\frac{13.8}{\sqrt{90}} \\ {\sigma }_{\overline{x}}\approx 1.45 \end{gathered}$$

Thus the standard deviation of all possible sample mean systolic blood pressure of sample size $90$ is ${\sigma }_{\overline{x}}=1.45$ mm Hg.