91Ó°ÊÓ

Normal Probability Plot

Box Plot:

The mean of two samples can be calculated as:

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

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

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

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

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.

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.

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

\((-150.3370, -38.6166)\)

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.