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.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
