Question

1. everyone experiences \ups and downs\ in life satisfaction. boehm, winning, segerstrom, and kubzansky (2015) studied whether such variability in life satisfaction is correlated with mortality rate. over a nine - year period, 4,458 australian older adult participants answered surveys about life satisfaction, and the researchers recorded whether the participants were deceased at the time of planned follow - up interviews. participants rated their life satisfaction on a scale of 0 (dissatisfied) to 10 (satisfied). the researchers observed data similar to the following:

low mortality sample	high mortality sample
6	10
6	10
7	4
6	9
5	3
5	0
3	10
8	2
3	3
5	4

a. compute m, ss, s², and s for these samples.
b. report the mean and standard deviation in a format that is appropriate for a scientific journal.

Explanation:

Step1: Calculate the mean ($M$) for Low Mortality Sample

The formula for the mean of a sample is $M=\frac{\sum X}{n}$. For the Low Mortality Sample, $n = 11$, and $\sum X=1 + 6+6+7+6+5+5+3+8+3+5=55$. So, $M=\frac{55}{11}=5$.

Step2: Calculate the sum - of - squares ($SS$) for Low Mortality Sample

The formula for $SS=\sum(X - M)^2$.
$(1 - 5)^2+(6 - 5)^2+(6 - 5)^2+(7 - 5)^2+(6 - 5)^2+(5 - 5)^2+(5 - 5)^2+(3 - 5)^2+(8 - 5)^2+(3 - 5)^2+(5 - 5)^2$
$=(-4)^2+1^2+1^2+2^2+1^2+0^2+0^2+(-2)^2+3^2+(-2)^2+0^2$
$=16 + 1+1+4+1+0+0+4+9+4+0=40$.

Step3: Calculate the variance ($s^{2}$) for Low Mortality Sample

The formula for the sample variance is $s^{2}=\frac{SS}{n - 1}$. Since $SS = 40$ and $n=11$, $s^{2}=\frac{40}{11 - 1}=\frac{40}{10}=4$.

Step4: Calculate the standard deviation ($s$) for Low Mortality Sample

The formula for the sample standard deviation is $s=\sqrt{s^{2}}$. Since $s^{2}=4$, $s = 2$.

Step5: Calculate the mean ($M$) for High Mortality Sample

For the High Mortality Sample, $n = 11$, and $\sum X=0+10+10+4+9+3+0+10+2+3+4=55$. So, $M=\frac{55}{11}=5$.

Step6: Calculate the sum - of - squares ($SS$) for High Mortality Sample

$(0 - 5)^2+(10 - 5)^2+(10 - 5)^2+(4 - 5)^2+(9 - 5)^2+(3 - 5)^2+(0 - 5)^2+(10 - 5)^2+(2 - 5)^2+(3 - 5)^2+(4 - 5)^2$
$=(-5)^2+5^2+5^2+(-1)^2+4^2+(-2)^2+(-5)^2+5^2+(-3)^2+(-2)^2+(-1)^2$
$=25+25+25 + 1+16+4+25+25+9+4+1=180$.

Step7: Calculate the variance ($s^{2}$) for High Mortality Sample

Since $SS = 180$ and $n = 11$, $s^{2}=\frac{180}{11 - 1}=\frac{180}{10}=18$.

Step8: Calculate the standard deviation ($s$) for High Mortality Sample

Since $s^{2}=18$, $s=\sqrt{18}\approx4.24$.

Answer:

a.
Low Mortality Sample: $M = 5$, $SS = 40$, $s^{2}=4$, $s = 2$.
High Mortality Sample: $M = 5$, $SS = 180$, $s^{2}=18$, $s\approx4.24$.
b.
Low Mortality Sample: $M = 5\pm2$
High Mortality Sample: $M = 5\pm4.24$