Question

maintaining your balance may get harder as you grow older. a study was conducted to see how steady the elderly are on their feet. they had a random sample of subjects stand on a force platform and have them react to a noise. the force platform then measured how much they swayed forward and backward, and the data are in the table below. do the data show that the mean elderly sway measurement is higher than the mean forward sway of younger people, which is 18.125 mm? test at the 10% level.
forward sway in mm
43
43
23
7
30
25
11
21
39
28
38
24
30
30
20
30
17
45
20
20
26
24
12
16
32
14
19
12
29
34
14
17
27
24
39

Explanation:

Step1: Calculate sample mean

The sample mean $\bar{x}$ is calculated by summing all the data points and dividing by the number of data points $n$.
$$\bar{x}=\frac{43 + 43+23+7+30+25+11+21+39+28+38+24+30+30+20+30+17+45+20+20+26+24+12+16+32+14+19+12+29+34+14+17+27+24+39}{35}$$
$$\bar{x}=\frac{875}{35}=25$$

Step2: Calculate sample standard deviation

First, calculate the sum of squared deviations from the mean:
$$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(43 - 25)^{2}+(43 - 25)^{2}+(23 - 25)^{2}+(7 - 25)^{2}+(30 - 25)^{2}+(25 - 25)^{2}+(11 - 25)^{2}+(21 - 25)^{2}+(39 - 25)^{2}+(28 - 25)^{2}+(38 - 25)^{2}+(24 - 25)^{2}+(30 - 25)^{2}+(30 - 25)^{2}+(20 - 25)^{2}+(30 - 25)^{2}+(17 - 25)^{2}+(45 - 25)^{2}+(20 - 25)^{2}+(20 - 25)^{2}+(26 - 25)^{2}+(24 - 25)^{2}+(12 - 25)^{2}+(16 - 25)^{2}+(32 - 25)^{2}+(14 - 25)^{2}+(19 - 25)^{2}+(12 - 25)^{2}+(29 - 25)^{2}+(34 - 25)^{2}+(14 - 25)^{2}+(17 - 25)^{2}+(27 - 25)^{2}+(24 - 25)^{2}+(39 - 25)^{2}$$
$$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=18^{2}+18^{2}+(- 2)^{2}+(-18)^{2}+5^{2}+0^{2}+(-14)^{2}+(-4)^{2}+14^{2}+3^{2}+13^{2}+(-1)^{2}+5^{2}+5^{2}+(-5)^{2}+5^{2}+(-8)^{2}+20^{2}+(-5)^{2}+(-5)^{2}+1^{2}+(-1)^{2}+(-13)^{2}+(-9)^{2}+7^{2}+(-11)^{2}+(-6)^{2}+(-13)^{2}+4^{2}+9^{2}+(-11)^{2}+(-8)^{2}+2^{2}+(-1)^{2}+14^{2}$$
$$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=324+324 + 4+324+25+0+196+16+196+9+169+1+25+25+25+25+64+400+25+25+1+1+169+81+49+121+36+169+16+81+121+64+4+1+196$$
$$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=3360$$
The sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}=\sqrt{\frac{3360}{34}}\approx9.97$

Step3: Set up hypothesis test

The null hypothesis $H_{0}:\mu\leq18.125$ and the alternative hypothesis $H_{1}:\mu>18.125$
The test statistic for a one - sample $t$ - test is $t=\frac{\bar{x}-\mu_{0}}{s/\sqrt{n}}$
Here, $\mu_{0} = 18.125$, $\bar{x}=25$, $s\approx9.97$, $n = 35$
$$t=\frac{25 - 18.125}{9.97/\sqrt{35}}\approx\frac{6.875}{1.69}\approx4.07$$

Step4: Determine critical value and make decision

The degrees of freedom $df=n - 1=35 - 1 = 34$. For a one - tailed test with $\alpha = 0.10$, using a $t$ - table or calculator, the critical value $t_{\alpha,df}=t_{0.10,34}\approx1.307$
Since $t = 4.07>t_{0.10,34}=1.307$, we reject the null hypothesis.

Answer:

Yes, the data show that the mean elderly sway measurement is higher than the mean forward sway of younger people at the 10% level.