91Ó°ÊÓ

Given in the question that, 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.

For the two samples, we need normal probability plots, boxplots, and standard deviations.

The normal probability chart is a graphic technique that can be used to determine if a data set is nearly normally distributed. The dots form an approximate straight line when the data is plotted against a theoretical normal distribution.

The two normal probability plots are given below :

The box plot will be:

Let's use the following formula to find the mean of two samples:

$$\begin{gathered} {\stackrel{¯}{x}}_1=\frac{1135+81+…+91+106}{41} \\ =102.0732 \end{gathered}$$

$$\begin{gathered} {\stackrel{¯}{x}}_2=\frac{229+172+…+217+233}{20} \\ =196.55 \end{gathered}$$

$$\begin{gathered} s_1=\sqrt{\frac{(135-102.0732{)}^2+(81-102.0732{)}^2+…+(106-102.0732{)}^2}{41-1}} \\ ≈36.6349 \end{gathered}$$

$$\begin{gathered} s_2=\sqrt{\frac{(229-196.55{)}^2+(1172.3-196.55{)}^2+…+(233-196.55{)}^2}{20-1}} \\ ≈84.8438 \end{gathered}$$

Given in the question that,

Standard deviations for the two samples are: $s_1≈36.6349$and$s_2≈84.8438$

Which is preferable here, pooled or non pooled t methods, based on your results from portion (a)? Justify your decision.

The standard deviation of the second sample is teice of the standard deviation of the first sample, as shown in part (a). As a result, the two samples' standard deviations are roughly not comparable. As a result, the standard deviations of the two populations are also not quite identical.

Non-pooled t-procedures are favored since they are different.

To summaries, men over 65 have a lower IGF-1 level than younger men at a 1% significance level.

The control mean is higher than that of men over 65.

This claim could be either a null or alternative hypothesis.

$$\begin{gathered} H_0:{\mathrm{μ}}_d=0 \\ {H}_{\mathrm{Î\pm }}:{\mathrm{μ}}_d>0 \end{gathered}$$

The test statistic should be calculated as follows:

$$\begin{gathered} t=\frac{{\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2-\left({\mathrm{μ}}_1-{\mathrm{μ}}_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{102.0732-196.55-0}{\sqrt{\frac{36.{6349}^2}{41}+\frac{84.{8438}^2}{20}}} \\ ≈-4.768 \end{gathered}$$

Determine the degree of freedom:

$$\begin{gathered} \mathrm{Δ}=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}} \\ =\frac{{\left(\frac{36.{6349}^2}{41}+\frac{84.{8438}^2}{20}\right)}^2}{\frac{{\left(\frac{36.63492}{41}\right)}^2}{41-1}+\frac{{\left(\frac{84.843382}{20}\right)}^2}{20-1}} \\ ≈22 \end{gathered}$$

The null hypothesis is rejected since the P-value is less than the significance level.

We must determine the difference between the mean IGF-1 levels of men over 65 and younger men at a 95% confidence level.

Let's determine the degrees of freedom:

$$\begin{gathered} \mathrm{Δ}=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s^2}{n_2}\right)}^2}{n_2-1}} \\ =\frac{{\left(\frac{36.{6349}^2}{41}+\frac{84.{8438}^2}{20}\right)}^2}{\frac{{\left(\frac{36.63492}{41}\right)}^2}{41-1}+\frac{{\left(\frac{84.{84388}^2}{20}\right)}^2}{20-1}} \\ ≈22 \end{gathered}$$

Then, compute the $t$value for

$$\begin{gathered} t_{\frac{(1-c)}{2}}=t_{\frac{(1-0.99)}{2}} \\ =t_{0.005} \end{gathered}$$

$t_{\frac{a}{2}}=2.819$

So, the margin of error will be:

$$\begin{gathered} E=t_{\frac{a}{2}}×\sqrt{\frac{s_1^2}{n_1}+\frac{{\frac{s}{2}}_2^2}{n_2}} \\ =2.819×\sqrt{\frac{36.{6349}^2}{41}+\frac{84.{8438}^2}{20}} \\ ≈55.8602 \end{gathered}$$

Therefore, the endpoints for confidence interval will be as follow:

$$\begin{gathered} \left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)-E=(102.0732-196.55)-55.8602 \\ =-150.3370 \end{gathered}$$

$$\begin{gathered} \left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)+E=(102.0732-196.55)+55.8602 \\ =-38.6166 \end{gathered}$$

We must determine if the procedures in parts (c) and (d) are justified.

Part (a) of the Box-Plot shows that there is an outlier, which greatly influences the t-procedure, hence the results in parts (c) and (d) cannot be supported.