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dxu46 wrote:

Jacobi wrote:

dxu46 wrote:

Click to Show Answer

1. at least 3
2. at least 4

Actually, only 3 CVs are needed.

Oh, then they changed it, it was 4 last year.

dxu46 wrote:You test the affect of running speed on perspiration. You results are, for 4 mph, 67 mL, 69 mL, and 81 mL. For 6 mph, your results are 87, 82, and 91 mL. For 8 mph, your results are 101, 101, and 109 mL. List and calculate at least 6 useful statistics.

(disclaimer: this is a FICTIONAL experiment. The results are highly improbable, and the experiment is unlikely to be done. But the question is about statistics, not ideas.)

1. Mean Perspiration (mL): 4 mph -> 72 mL (= (67 + 69 + 81)/3), 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2)}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is the second-highest value:[/b] [math]67, \underline{69}, 81[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = max - min = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]MAD = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = (|67 - 72.33| + |69 - 72.33| + |81 - 72.33|) / 3 = 5.8 mL[/math], 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y})}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

Jacobi wrote:

dxu46 wrote:You test the affect of running speed on perspiration. You results are, for 4 mph, 67 mL, 69 mL, and 81 mL. For 6 mph, your results are 87, 82, and 91 mL. For 8 mph, your results are 101, 101, and 109 mL. List and calculate at least 6 useful statistics.

(disclaimer: this is a FICTIONAL experiment. The results are highly improbable, and the experiment is unlikely to be done. But the question is about statistics, not ideas.)

Do they have to be all different?

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> 72 mL (= (67 + 69 + 81)/3), 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2)}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is the second-highest value:[/b] [math]67, \underline{69}, 81[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = max - min = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]MAD = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = (|67 - 72.33| + |69 - 72.33| + |81 - 72.33|) / 3 = 5.8 mL[/math], 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y}}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

Jacobi wrote:Standard deviation measures what quality of a data distribution?

It measures the average variation that the actual values have from the mean value.

OrigamiPlanet wrote:

Jacobi wrote:Standard deviation measures what quality of a data distribution?

Answer (Click to Show Hidden Text)

It measures the average variation that the actual values have from the mean value.

Jacobi wrote:

OrigamiPlanet wrote:

Jacobi wrote:Standard deviation measures what quality of a data distribution?

Answer (Click to Show Hidden Text)

It measures the average variation that the actual values have from the mean value.

Not exactly...
Try again.

Jacobi wrote:

dxu46 wrote:You test the affect of running speed on perspiration. You results are, for 4 mph, 67 mL, 69 mL, and 81 mL. For 6 mph, your results are 87, 82, and 91 mL. For 8 mph, your results are 101, 101, and 109 mL. List and calculate at least 6 useful statistics.

(disclaimer: this is a FICTIONAL experiment. The results are highly improbable, and the experiment is unlikely to be done. But the question is about statistics, not ideas.)

Do they have to be all different?

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> 72 mL (= (67 + 69 + 81)/3), 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is, the second-highest value:[/b] [math]67, \underline{69}, 81[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = max - min = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]\textrm{MAD} = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = (|67 - 72.33| + |69 - 72.33| + |81 - 72.33|) / 3 = 5.8 mL[/math], 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y}}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

1. Mean Perspiration (mL): 4 mph -> [math]\frac{67+69+81}{3} =[/math] 72 mL, 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2)}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is the second-highest value:[/b] [math]67, \underline{69}, 81 \Rightarrow[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = \textrm{max} - \textrm{min} = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]MAD = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = \frac{|67 - 72.33| + |69 - 72.33| + |81 - 72.33|}{3} =[/math] 5.8 mL, 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y})}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

UTF-8 U+6211 U+662F wrote:

Jacobi wrote:

OrigamiPlanet wrote:

Answer (Click to Show Hidden Text)

It measures the average variation that the actual values have from the mean value.

Not exactly...
Try again.

Wait, what's wrong with that answer? That's pretty much the intent of a standard deviation. Did you just want the spread of the distribution?

Edit: Also, correcting this because the mismatched parentheses bother me...

Jacobi wrote:

dxu46 wrote:You test the affect of running speed on perspiration. You results are, for 4 mph, 67 mL, 69 mL, and 81 mL. For 6 mph, your results are 87, 82, and 91 mL. For 8 mph, your results are 101, 101, and 109 mL. List and calculate at least 6 useful statistics.

(disclaimer: this is a FICTIONAL experiment. The results are highly improbable, and the experiment is unlikely to be done. But the question is about statistics, not ideas.)

Do they have to be all different?

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> 72 mL (= (67 + 69 + 81)/3), 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is, the second-highest value:[/b] [math]67, \underline{69}, 81[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = max - min = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]\textrm{MAD} = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = (|67 - 72.33| + |69 - 72.33| + |81 - 72.33|) / 3 = 5.8 mL[/math], 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y}}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]

should be

6 Different Stats (Click to Show Hidden Text)

1. Mean Perspiration (mL): 4 mph -> [math]\frac{67+69+81}{3} =[/math] 72 mL, 6 mph -> 87 mL, 8 mph -> 104 mL
2. Standard Deviation Perspiration (mL): 4 mph -> [math]\sqrt{\frac{(67-72)^2 + (69 - 72)^2 + (81 - 72)^2)}{3}} =[/math] 7.6 mL, 6 mph -> 4.5 mL, 8 mph -> 4.6 mL
3. Median Perspiration (mL): 4 mph -> [b]Find middle of 3 values in order, that is the second-highest value:[/b] [math]67, \underline{69}, 81 \Rightarrow[/math] 69 mL, 6 mph -> 87 mL, 8 mph - > 101 mL
4. Range Perspiration (mL): 4 mph -> [math]R = \textrm{max} - \textrm{min} = 81-67 =[/math] 14 mL, 6 mph -> 9 mL, 8mph -> 8 mL.
5. Mean Absolute Deviation (mL): 4 mph -> [math]MAD = \frac{1}{n}\Sigma_i |x_i - \overline{x}| = \frac{|67 - 72.33| + |69 - 72.33| + |81 - 72.33|}{3} =[/math] 5.8 mL, 6 mph -> 3.1 mL, 8 mph -> 3.6 mL
6. Pearson Correlation between Hours and Milliliters: [math]r = \Sigma_i \frac{(x_i - \overline{x})(y_i - \overline{y})}{s_x s_y}[/math]
[math]\overline{x} = 6.00[/math]
[math]\overline{y} = 87.6[/math]
[math]s_y = 40.9[/math]
[math]s_x = 4.9[/math]
.
.
.
[math]r = 0.938[/math]