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MathSofiya

  • 3 years ago

Statistics I'm trying to create a Frequency Table. Please let me know if I'm doing it right. One minute please...

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  1. MathSofiya
    • 3 years ago
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    \[\quad \begin{array}{|c|c|c|c|c|} \hline \frac{\textrm{class limits}}{\textrm{lower-upper}} & \frac{\textrm{class boundary}}{\textrm{lower-upper}} & \frac{\textrm{class}}{\textrm{midpoint}} &\text{tally}&\text{frequency}&\frac{\text{cumulative}}{\text{frequency}}\\ \hline 1 & & \\ \hline & & 4&6 \\ \hline & 5 & 4\\ \hline & & \\ \hline & 1 & \\ \hline & 1& 2\\ \hline \end{array}\] this is not the actual table. I'm new to making tables with Latex. Those numbers are just place holders so that my table doesn't collapse...now to my questions...

  2. MathSofiya
    • 3 years ago
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    I found the class width to be 6 \[\frac{43-0}{8}=5.375\] 34 21 37 31 10 24 39 10 17 18 32 17 3 10 6 5 6 6 13 22 25 3 5 2 9 3 0 4 29 26 5 5 24 15 3 8 16 9 10 3 12 10 10 10 11 12 13 1 9 43 13 14 32 24 15

  3. MathSofiya
    • 3 years ago
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    for the first column... so would I write 0-6 6-12 12-24 how do I know where to start the class limits? I know that I should add six to the upper and lower limit to get to the next limits, but where do i start?

  4. CliffSedge
    • 3 years ago
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    Well, one thing is you don't want the classes to overlap, so maybe 0-6, 7-13, 14-20, etc.

  5. CliffSedge
    • 3 years ago
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    It looks like discrete data, so you don't have to worry about continuity corrections.

  6. MathSofiya
    • 3 years ago
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    but I have to add 6 to the lower limit and to the upper limit. If I do what you did that would be adding 7 to each 0-6 7-13 Would that mean it doesn't matter where I start? could I do... 0-5 6-11 12-17 ?

  7. CliffSedge
    • 3 years ago
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    Yeah, I think since 0 is included in the data, 0-5, 6-11, .. would be better if you want to have 8 classes. It'll overshoot a little at the end, but will include all the data.

  8. MathSofiya
    • 3 years ago
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    Why would it overshoot at the end?

  9. CliffSedge
    • 3 years ago
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    8 classes with a width of 6 each would go to 47 (48-1 because starting count at 0)

  10. CliffSedge
    • 3 years ago
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    It's fine though, otherwise if you only had a width of 5, it would only go to 39 and that is less than your maximum value.

  11. MathSofiya
    • 3 years ago
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    will do. Thanks for your help!

  12. MathSofiya
    • 3 years ago
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    \[\quad \begin{array}{|c|c|c|c|} \hline \frac{\textrm{class limits}}{\textrm{lower-upper}} & \frac{\textrm{class boundary}}{\textrm{lower-upper}} & \frac{\textrm{class}}{\textrm{midpoint}} &\text{frequency}&\frac{\text{cumulative}}{\text{frequency}}&\frac{f}{n}\\ \hline 0-5 & 0.0-5.5 &13 &13& 2.75&0.218\\ \hline 6-11& 5.5-11.5& 15&28& 8.5 &0.27\\ \hline 12-17&11.5-17.5& 11&39& 14.5 &0.2\\ \hline 18-23&17.5-23.5& 3 &42& 20.5 &0.054 \\ \hline 24-29&23.5-29.5& 6 &48& 26.5 &0.109\\ \hline 30-35&29.5-35.5& 4 &52& 32.5 &0.072\\ \hline 36-41&35.5-41.5& 2 &54& 38.5 &0.036\\ \hline 42-47&41.5-47.5& 1 &55& 44.5 &0.018\\ \hline \end{array}\]

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