anonymous one year ago QUICK QUESTION need someone who knows statistics

1. anonymous

2. anonymous

3. anonymous

What's the problem here? @Michele_Laino answered your question just fine the first time around.

4. anonymous

I didn't understand the equation.

5. anonymous

6. anonymous

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

so from this you get "yes, the x^2 value was too high"?

8. anonymous

@SithsAndGiggles

9. anonymous

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

statistics is super not my thing. i hardly understand what you're talking about

11. anonymous

@SithsAndGiggles

12. anonymous

You'll have to be more precise. What do you not understand? I'll try my best to explain.

13. anonymous

i dont understand any of this. is like a foreign language.

14. anonymous

@sithsandgiggles

15. anonymous

i just need the answer right now @sithsandgiggles

16. Michele_Laino

Thanks!! :) @SithsAndGiggles