# Solved-The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the kn

#### ByDr. Raju Chaudhari

Oct 7, 2020

The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. The article "Evidence of Mechanical Load Redistribution at the Knee Joint in the Elderly When Ascending Stairs and Ramps" (Annals of Biomed. Engr., 2008: 467-476) presented the following summary data on stance duration (ms) for samples of both older and younger adults.

Age Sample Size Sample Mean Sample SD
Older 28 801 117
Younger 16 780 72

Assume that both stance duration distributions are normal.

a. Calculate and interpret a 99% CI for true average stance duration among elderly individuals.
b. Carry out a test of hypotheses at significance level 0.05 to decide whether true average stance duration is larger among elderly individuals than among younger individuals.

### Solution

(a) Given that sample size $n =28$, sample mean $\overline{X}=801$, sample standard deviation $s=117$.

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is
 \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned}
where $E = t{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}$, andand $t{\alpha/2, n-1}$ is the $t$ value providing an area of $\alpha/2$ in the upper tail of the students' $t$ distribution.

The critical value of $t$ for given level of significance and $n-1$ degrees of freedom is $t_{\alpha/2,n-1}$.

Thus $t{\alpha/2,n-1} = t{0.005,28-1}= 2.771$.

The margin of error for mean is

 \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}\\ & = 2.771 \frac{117}{\sqrt{28}} \\ & = 61.269. \end{aligned}

$99$% confidence interval estimate for population mean is

 \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 801 - 61.269 & \leq \mu \leq 801 + 61.269\\ 739.731 &\leq \mu \leq 862.269. \end{aligned}

Thus, $99$% CI for true average stance duration among elderly individuals is $(739.731,862.269)$.

(b) Given that the sample size $n_1 = 28$, $n_2 = 16$, sample mean $\overline{x}_1= 801$,
$\overline{x}_2= 780$, sample standard deviation $s_1 = 117$ and $s_2 = 72$.

State the hypothesis testing problem

The hypothesis testing problem is
$H_0 : \mu_1 = \mu_2$ against $H_1 : \mu_1 > \mu_2$ ($\textit{right-tailed}$)

Define test statistic

The test statistic for testing above hypothesis testing problem is
 \begin{aligned} t=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \end{aligned}
The test statistic $t$ follows Students' $t$ distribution with $\nu$ degrees of freedom, where

 \begin{aligned} \nu = \frac{\bigg(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\bigg)^2}{\frac{s_1^4}{n_1^2(n_1-1)}+\frac{s_2^4}{n_2^2(n_2-1)}}=42 \end{aligned}
rounded to nearest integer.

Level of significance

The significance level is $\alpha = 0.05$.

Determine the critical value

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $t$ using $\alpha = 0.05$ and degrees of freedom $=42$ $\text{is}$ $\text{1.682}$.

The rejection region (i.e. critical region) is $\text{t > 1.682}$.

Computation

The test statistic for testing above hypothesis testing problem under the null hypothesis is
 \begin{aligned} t&=\frac{(\overline{x}_1 -\overline{x}_1)-0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ &= \frac{(801-780)}{\sqrt{\frac{117^2}{28}+\frac{72^2}{16}}}\\ &= 0.7366 \end{aligned}

Decision

The rejection region (i.e. critical region) is $\text{t > 1.682}$. The test statistic is $t =0.7366$ which falls $\text{outside}$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

We conclude that true average stance duration is not larger among elderly individuals than among younger individuals.