Question

compute the mean, range, and standard deviation for the data items in each of the three samples. then describe one way in which the samples are alike and one way in which they are different. sample a: 31, 33, 35, 37, 39, 41, 43 sample b: 31, 32, 33, 37, 41, 42, 43 sample c: 31, 31, 31, 37, 43, 43, 43 in which of the following ways are the samples alike? mean standard deviation range in which of the following ways are the samples different? mean and standard deviation range and standard deviation mean and range

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
For Sample A: $n = 10$, $\sum_{i=1}^{10}x_{i}=31 + 33+35+37+39+41+43+43+43+43=398$, $\bar{x}_A=\frac{398}{10}=39.8$.
For Sample B: $n = 10$, $\sum_{i = 1}^{10}x_{i}=31+32+33+37+41+42+43+43+43+43 = 391$, $\bar{x}_B=\frac{391}{10}=39.1$.
For Sample C: $n = 10$, $\sum_{i=1}^{10}x_{i}=31+31+31+31+37+43+43+43+43+43=376$, $\bar{x}_C=\frac{376}{10}=37.6$.

Step2: Calculate the range

The range $R=\text{Max}-\text{Min}$.
For Sample A: $\text{Max}=43$, $\text{Min}=31$, $R_A=43 - 31=12$.
For Sample B: $\text{Max}=43$, $\text{Min}=31$, $R_B=43 - 31=12$.
For Sample C: $\text{Max}=43$, $\text{Min}=31$, $R_C=43 - 31=12$.

Step3: Calculate the standard - deviation

The sample standard - deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^2}{n - 1}}$.
For Sample A:
$\sum_{i=1}^{10}(x_{i}-\bar{x}_A)^2=(31 - 39.8)^2+(33 - 39.8)^2+(35 - 39.8)^2+(37 - 39.8)^2+(39 - 39.8)^2+(41 - 39.8)^2+(43 - 39.8)^2+(43 - 39.8)^2+(43 - 39.8)^2+(43 - 39.8)^2$
$=(-8.8)^2+(-6.8)^2+(-4.8)^2+(-2.8)^2+(-0.8)^2+(1.2)^2+(3.2)^2+(3.2)^2+(3.2)^2+(3.2)^2$
$=77.44 + 46.24+23.04 + 7.84+0.64+1.44+10.24+10.24+10.24+10.24 = 197.6$.
$s_A=\sqrt{\frac{197.6}{9}}\approx4.69$.
For Sample B:
$\sum_{i=1}^{10}(x_{i}-\bar{x}_B)^2=(31 - 39.1)^2+(32 - 39.1)^2+(33 - 39.1)^2+(37 - 39.1)^2+(41 - 39.1)^2+(42 - 39.1)^2+(43 - 39.1)^2+(43 - 39.1)^2+(43 - 39.1)^2+(43 - 39.1)^2$
$=(-8.1)^2+(-7.1)^2+(-6.1)^2+(-2.1)^2+(1.9)^2+(2.9)^2+(3.9)^2+(3.9)^2+(3.9)^2+(3.9)^2$
$=65.61+50.41+37.21 + 4.41+3.61+8.41+15.21+15.21+15.21+15.21 = 225.5$.
$s_B=\sqrt{\frac{225.5}{9}}\approx5.00$.
For Sample C:
$\sum_{i=1}^{10}(x_{i}-\bar{x}_C)^2=(31 - 37.6)^2+(31 - 37.6)^2+(31 - 37.6)^2+(31 - 37.6)^2+(37 - 37.6)^2+(43 - 37.6)^2+(43 - 37.6)^2+(43 - 37.6)^2+(43 - 37.6)^2+(43 - 37.6)^2$
$=(-6.6)^2+(-6.6)^2+(-6.6)^2+(-6.6)^2+(-0.6)^2+(5.4)^2+(5.4)^2+(5.4)^2+(5.4)^2+(5.4)^2$
$=43.56+43.56+43.56+43.56+0.36+29.16+29.16+29.16+29.16+29.16 = 310.8$.
$s_C=\sqrt{\frac{310.8}{9}}\approx5.88$.

One way the samples are alike: They have the same range ($R_A = R_B=R_C = 12$).
One way the samples are different: They have different means ($\bar{x}_A = 39.8$, $\bar{x}_B=39.1$, $\bar{x}_C = 37.6$) and different standard - deviations ($s_A\approx4.69$, $s_B\approx5.00$, $s_C\approx5.88$).

Answer:

Sample A Mean: $39.8$
Sample A Range: $12$
Sample A Standard Deviation: $4.69$
Sample B Mean: $39.1$
Sample B Range: $12$
Sample B Standard Deviation: $5.00$
Sample C Mean: $37.6$
Sample C Range: $12$
Sample C Standard Deviation: $5.88$
Alike: Same range.
Different: Different means.