Question

Ertugrul et al. conducted a study to determine the association between insulin growth factor 1 (IGF-1) and bone mineral density (BMD) in men over \(65\) years of age. The researchers published their results in the paper "Relationship Between Insulin-Like Growth Factor 1 and Bone Mineral Density in Men Aged over \(65\) Years". Forty-one men over \(65\) years old were enrolled in the study as was a control group consisting of \(20\) younger men, ages \(19-62\) years. On the WeissStats site, we provide data on IGF-1 levels, based on the results of the study. Use the technology of your choice to do the following.

a. Obtain normal probability plots, boxplots and the standard deviation for the two samples

b. Based on your results from part (a), which are preferable here, pooled or non-pooled \(t-\)procedure? Explain your reasoning.

c. At the \(1%\) significance level, do the data provide sufficient evidence to conclude that on average men over \(65\) have a lower IGF-1 level than younger men?

d. Find and interpret a \(99%\) confidence interval for the difference between the mean IGF-1 levels of men over \(65\) and younger men.

e. Are your procedures in part (c) and (d) justified? Explain your answer.

Short answer

Part a. \(36.6349\)

\(84.8438\)

Part b. non-pooled t-procedures.

Part c.There is sufficient proof at\(1%\) significance level that men over \(65\) have a lower IGF-1 level than younger men.

Part d. \((-150.3370, -38.6166)\)

Part e. No

Step-by-step solution

Part a. Step 1. Explanation

Normal Probability Plot

Box Plot:

The mean of two samples can be calculated as:

\(\bar{x_{1}}=\frac{1135+81+..+91+106}{41}=102.0732\)

\(\bar{x_{2}}=\frac{229+172+..+217+233}{20}=196.55\)

\(s_{1}=\sqrt{\frac{(135-102.0732)^{2}+(81-102.0732)^{2}+..+(106-102.0732)^{2}}{41-1}}\approx 36.6349\)

\(s_{2}=\sqrt{\frac{(229-196.55)^{2}+(1172.3-196.55)^{2}+..+(233-196.55)^{2}}{20-1}}\approx 84.8438\)

Part b. Step 1. Explanation

It can be noted from part (a) that the standard deviation of second sample is teice of the standard deviation of first sample. Therefore, the standard deviations of the two sample are approximately not equal. Therefore, the standard deviations of the the two populations are not approximately not equal either.

As, they are different, non-pooled t-procedures are preferred.

Part c. Step 1. Explanation

Calculate the test-statistic as per follows:

Calculate the degrees of freedom:

As the P-value is less than the significance level, therefore the null hypothesis is rejected.

Therefore, there is sufficient proof at\(1%\) significance level that men over \(65\) have a lower IGF-1 level than younger men.

Part d. Step 1. Explanation

Calculate the degrees of freedom:

Now, calculate the t-value for \(t_{\frac{(1-c}{2}}=t_{\frac{(1-0.99)}{2}}=t_{0.005}\)

\(t_{\frac{\alpha}{2}}=2.819\)

Therefore, the margin of error is

The endpoints for confidence interval is

\((-150.3370, -38.6166)\)

Part e. Step 1. Explanation

From Box-Plot in part (a), it can be seen that there is an outlier and they strongly influence \(t-\)procedure, therefore the results in part (c) and (d) can not be justified.