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anonymous

  • one year ago

QUICK QUESTION need someone who knows statistics

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    What's the problem here? @Michele_Laino answered your question just fine the first time around.

  4. anonymous
    • one year ago
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    I didn't understand the equation.

  5. anonymous
    • one year ago
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    someone help please

  6. anonymous
    • one year ago
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    The \(\chi^2\) statistic is given as \[\sum_{i=1}^n\frac{(\text{observed}_i-\text{expected}_i)^2}{\text{expected}_i}\] where \(i\) denotes the category, which in this case is milkshake flavor. There are 3 of these, so \(n=3\). \(\text{observed}_i\) is the recorded value of a given flavor, whereas \(\text{expected}_i\) is the expected value of a given flavor. For example, the parlor finds that \(202\) vanilla milkshakes were ordered, while they expected \(175\) to be ordered. So, the contribution of vanilla shakes to the \(\chi^2\) statistic is \[\frac{(\text{observed}_\text{vanilla}-\text{expected}_\text{vanilla})^2}{\text{expected}_\text{vanilla}}=\frac{(202-175)^2}{175}\] You do the same for the other flavors and add them together. \[\chi^2=\frac{{{{\left( {202 - 175} \right)}^2}}}{{175}} + \frac{{{{\left( {112 - 125} \right)}^2}}}{{125}} + \frac{{{{\left( {269 - 250} \right)}^2}}}{{250}} \]

  7. anonymous
    • one year ago
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    so from this you get "yes, the x^2 value was too high"?

  8. anonymous
    • one year ago
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    @SithsAndGiggles

  9. anonymous
    • one year ago
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    Well that depends on what value of \(\chi^2\) you get \(\chi^2\approx 6.96\). Since \(n=3\), you have \(n-1=2\) degrees of freedom. What \(p\) value do you get for a significance level of \(0.05\) and \(2\) degrees of freedom?

  10. anonymous
    • one year ago
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    statistics is super not my thing. i hardly understand what you're talking about

  11. anonymous
    • one year ago
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    @SithsAndGiggles

  12. anonymous
    • one year ago
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    You'll have to be more precise. What do you not understand? I'll try my best to explain.

  13. anonymous
    • one year ago
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    i dont understand any of this. is like a foreign language.

  14. anonymous
    • one year ago
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    @sithsandgiggles

  15. anonymous
    • one year ago
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    i just need the answer right now @sithsandgiggles

  16. Michele_Laino
    • one year ago
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    Thanks!! :) @SithsAndGiggles

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