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The Confidence Interval is [tex]763.885\leq \mu\leq 840.115[/tex] and the true average stance duration is not larger among elderly individuals than among younger individuals.
Given the sample size of the older, [tex]n_1[/tex] = 27
The sample mean of the older, [tex]\bar x_1[/tex] = 802
The sample Standard Deviation, [tex]s_1[/tex] = 116
Here the population mean [tex]\mu[/tex] is unknown.
So the confidence interval for true average (population mean) is given by [tex]\bar x_1-t_\frac{\alpha }{2} \frac{s_1}{\sqrt{n_1-1} } \leq \mu\leq \bar x_1+t_\frac{\alpha }{2} \frac{s_1}{\sqrt{n_1-1} }[/tex]
Here [tex]t_\frac{\alpha }{2}[/tex] can be found from the tables of Student's 't' distribution for [tex]\alpha =0.01[/tex] ,the level of significance and m-1 = 26 degrees of freedom.
[tex]\alpha[/tex] is taken as 0.01 since we will be finding a 99% confidence interval for the true average stance duration among elderly individuals.
So here the [tex]t_\frac{\alpha }{2}[/tex] = 1.706
Hence the confidence interval will be
[tex]802-1.706 \times \frac{116}{\sqrt{26} }\leq \mu\leq 802+1.706 \times \frac{116}{\sqrt{26} }[/tex]
That is, [tex]763.885\leq \mu\leq 840.115[/tex] is the 99% confidence interval for true average stance duration among elderly individuals.
We will 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.
Testing of hypothesis concerning the true average stance among elderly and younger individuals.
That is, [tex]H_0:\mu_1=\mu_2[/tex]
against [tex]H_1:\mu_1 > \mu_2[/tex] ,the alternative hypothesis.
where [tex]\mu_1[/tex] is the true average for the older and [tex]\mu_2[/tex] is the true average for the younger.
Here the significance level, [tex]\alpha =0.05[/tex]. Hence [tex]t_\alpha = 1.65[/tex].
The test statistic is,
[tex]t=\frac{\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }[/tex]
i.e., [tex]t=\frac{802 - 770}{\sqrt{\frac{116^2}{27} +\frac{73^2}{19} } }[/tex]
= 1.147
Here [tex]t < t_\alpha[/tex] . So we cannot reject the null hypothesis [tex]H_0[/tex].
Thus the true average stance duration is not larger among elderly individuals than among younger individuals.
Your question is incomplete. Here you can find the full question:
The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the knee. An article 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 27 802 116 Younger 19 770 73 Assume that both stance duration distributions are normal.
a) Calculate and interpret a 99% Confidence interval for true average stance duration among elderly individuals.
b) Carry out a test hypothesis at significance level 0.05 to decide whether true average stance duration is larger among elderly individuals than among younger individuals.
Find out more about testing of hypothesis at https://brainly.com/question/14042255
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