| taw mass/g |
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|
|
| 18.46 |
|
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|
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| 19.24 |
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|
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|
|
|
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| 19.3 |
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|
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| 19.85 |
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| 20.12 |
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| 20.12 |
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| 20.33 |
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|
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| 20.35 |
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|
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| 20.52 |
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| 20.78 |
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| ratio |
1. What level of measurement is the data? |
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|
| 10 |
2. Find the sample size n for the data. |
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| 18.46 |
3. Find the minimum. |
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| 20.78 |
4. Find the maximum. |
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| 2.32 |
5. Find the range. |
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|
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|
|
| 20.12 |
6. Find the median. |
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| 20.12 |
7. Find the mode. |
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|
|
| 19.91 |
8. Find the sample mean x. |
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| 0.71 |
9. Find the sample standard deviation sx. |
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| 0.0356 |
10. Find the sample coefficient of variation CV. |
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|
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| 48.82 |
11. Determine the z-score for a taw that has a mass of 54.48 grams. |
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| unusual |
12. Is the z-score above an ordinary or unusual z-score? |
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| 0.22 |
13. Calculate the standard error of the mean for the taws. |
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|
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14.Calculate the 68% confidence interval for the population mean mass of a taw. |
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|
| p( |
19.68 |
< μ < |
20.13 |
) = 0.68 |
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| Are the taws actually normally distributed? |
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|
|
| Bins |
Freq |
Rel Freq |
|
| 18.92 |
1 |
0.1 |
|
| 19.39 |
2 |
0.2 |
|
| 19.85 |
1 |
0.1 |
|
| 20.32 |
2 |
0.2 |
|
| 20.78 |
4 |
0.4 |
|
|
10 |
1 |
|
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|
|
| No, the taws are distributed strongly |
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|
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| skewed left and bimodally |
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| Clue to the skew: |
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| When the mean ≠ median, a skew is present |
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| Skewness |
-0.97 |
|
|
| The skew function is a formula for calculating the skew in the data. The negative, non-zero value |
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| is a left skew as seen in the diagram. |
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| Less massive ("lighter") taws are unusual |
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