Question

1. for the data in the following sample: 1, 1, 9, 1

a. find the mean, ss, variance, and standard deviation.
b. now change the score of x = 9 to x = 3, and find the new values for ss, variance, and standard deviation.
c. describe how one extreme score influences the mean and standard deviation.

Explanation:

Step1: Calculate the mean for part a

The mean $\bar{X}=\frac{\sum X}{n}$, where $\sum X = 1 + 1+9 + 1=12$ and $n = 4$. So $\bar{X}=\frac{12}{4}=3$.

Step2: Calculate $SS$ for part a

$SS=\sum(X - \bar{X})^2$.
$(1 - 3)^2+(1 - 3)^2+(9 - 3)^2+(1 - 3)^2=(-2)^2+(-2)^2+6^2+(-2)^2=4 + 4+36 + 4 = 48$.

Step3: Calculate variance for part a

Variance $s^2=\frac{SS}{n - 1}=\frac{48}{4 - 1}=\frac{48}{3}=16$.

Step4: Calculate standard - deviation for part a

Standard deviation $s=\sqrt{s^2}=\sqrt{16}=4$.

Step5: Calculate the new mean for part b

When $X = 9$ is changed to $X = 3$, $\sum X=1 + 1+3 + 1=6$. The new mean $\bar{X}=\frac{6}{4}=1.5$.

Step6: Calculate new $SS$ for part b

$(1 - 1.5)^2+(1 - 1.5)^2+(3 - 1.5)^2+(1 - 1.5)^2=(-0.5)^2+(-0.5)^2+(1.5)^2+(-0.5)^2=0.25+0.25 + 2.25+0.25 = 3$.

Step7: Calculate new variance for part b

New variance $s^2=\frac{SS}{n - 1}=\frac{3}{4 - 1}=1$.

Step8: Calculate new standard - deviation for part b

New standard deviation $s=\sqrt{s^2}=\sqrt{1}=1$.

Step9: Describe the influence for part c

The extreme score of $X = 9$ in the original data increased the mean from $1.5$ (the mean without the extreme value) to $3$. It also increased the standard deviation from $1$ (without the extreme value) to $4$. An extreme score can greatly increase the mean and standard deviation, making them less representative of the typical values in the data set.

Answer:

a. Mean: 3, $SS$: 48, Variance: 16, Standard deviation: 4
b. New $SS$: 3, New variance: 1, New standard deviation: 1
c. An extreme score can increase the mean and standard deviation, making them less representative of typical values.