AP Statistics FRQ

A

Audentes

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I understand how to do Part A. For Part B, am I supposed to use a chi-square test for goodness-of-fit based on expected counts where 400-500 and 500-600 are each supposed to be 34% of 250 (85) and etc. to compare this to the normal distribution? Then the outer columns as 0.11*250 and 0.05*250, respectively? If I do this, I get a P value of 0 which means rejection of the null. Just would like to know if I'm doing this right.

And for Part C, I set up (249)(114.4)^2 / sd^2 = 207.2 and (249)(114.4)^2 / sd^2 = 294.6 to get an interval of 105.22 to 125.43.

Thanks
Audentes
 
Do you know how to compute the Chi-Square statistics?
[math]\chi^2= \sum\dfrac{(O_i-E_i)^2}{E_i}[/math]
where [imath]O_i[/imath] is the observed frequency and [imath]E_i[/imath] is the expected frequency. [imath][/imath]
 
BigBeachBanana said:
Do you know how to compute the Chi-Square statistics?
[math]\chi^2= \sum\dfrac{(O_i-E_i)^2}{E_i}[/math]
where [imath]O_i[/imath] is the observed frequency and [imath]E_i[/imath] is the expected frequency. [imath][/imath]
Click to expand...
Yes, it gives me a x2 value of approx. 81.05 which corresponds to a P-value of 0
 
Oh Noooo!! I entered 8 in my calculator instead of 82 for 500-600's observed value..
I just did it by hand and discovered this. Now I have a x2 value of 11.40538 and a p-value of 0.044.
BigBeachBanana said:
I got a different value for [imath]\chi^2[/imath]. How you did compute the expected frequency?
Click to expand...
 
Audentes said:
Oh Noooo!! I entered 8 in my calculator instead of 82 for 500-600's observed value..
I just did it by hand and discovered this. Now I have a x2 value of 11.40538 and a p-value of 0.044.
Click to expand...
I have slightly different answer by mostly likely due to rounding error. Can you draw the conclusion of the hypo test?
 
BigBeachBanana said:
I have slightly different answer by mostly likely due to rounding error. Can you draw the conclusion of the hypo test?
Click to expand...
Because the p-value 0.044 is less than the typical significance values of 0.05 and 0.10, we reject HsubO. There is convincing evidence that the population of SAT math scores is not normally distributed.

Is this correct?
 
Audentes said:
Because the p-value 0.044 is less than the typical significance values of 0.05 and 0.10, we reject HsubO. There is convincing evidence that the population of SAT math scores is not normally distributed.

Is this correct?
Click to expand...
You should note that this is a two-tailed test so you need to find the critical value of both lower and upper tail.
Secondly, alpha are chosen before hand not after the test. The second part didn't clearly state what is the alpha level, but mostly likely they're assuming alpha = 5%. This means you need to find the critical values of the 2.5% and 97.5%. If the value is between the 2 critical values, then you fail to reject the null hypo.
 
BigBeachBanana said:
You should note that this is a two-tailed test so you need to find the critical value of both lower and upper tail.
Secondly, alpha are chosen before hand not after the test. The second part didn't clearly state what is the alpha level, but mostly likely they're assuming alpha = 5%. This means you need to find the critical values of the 2.5% and 97.5%. If the value is between the 2 critical values, then you fail to reject the null hypo.
Click to expand...
Am I multiplying the p-value by 2, or am I finding the critical values on the x2 table and determining if the x2 value of 11.40 falls in between?
 
Audentes said:
Am I multiplying the p-value by 2, or am I finding the critical values on the x2 table and determining if the x2 value of 11.40 falls in between?
Click to expand...
Find the value of 2.5% and 97.5% with the corresponding df. If 11.4 falls between the 2 then fail to reject. Else, reject the null.
 
df=5 with tail probability of 0.025 gives a x2 value of 12.83. So 11.4 is between -12.83 and 12.83, meaning fail to reject?
BigBeachBanana said:
Find the value of 2.5% and 97.5% with the corresponding df. If 11.4 falls between the 2 then fail to reject. Else, reject the null.
Click to expand...
 
It doesn't look like they give you 97.5%. I'm not sue how you're supposed to conclude the result. The chi-square distribution is not symmetrical, unlike the normal distribution.
 
BigBeachBanana said:
It doesn't look like they give you 97.5%. I'm not sue how you're supposed to conclude the result. The chi-square distribution is not symmetrical, unlike the normal distribution.
Click to expand...
Alright, thank you for looking into it. Do you know if I could find it on the Casio fx-9750 Giii?

Also, if you could verify my approach to Part C, that would be great!
 
I suspect they're only want you to use right-tailed for part b), which is wrong in my opinion. But what you said in #7 would be correct under one-tailed test.

Here are the steps for part c).
1) Find [imath]s^2[/imath]
2) Find [imath]\sigma^2_1[/imath] and [imath]\sigma^2_2[/imath] from the given [imath]\chi^2_1[/imath] and [imath]\chi^2_2[/imath]
3) This gives you the upper and lower bounds of [imath]\sigma^2[/imath]
 
BigBeachBanana said:
I suspect they're only want you to use right-tailed for part b), which is wrong in my opinion. But what you said in #7 would be correct under one-tailed test.

Here are the steps for part c).
1) Find [imath]s^2[/imath]
2) Find [imath]\sigma^2_1[/imath] and [imath]\sigma^2_2[/imath] from the given [imath]\chi^2_1[/imath] and [imath]\chi^2_2[/imath]
3) This gives you the upper and lower bounds of [imath]\sigma^2[/imath]
Click to expand...
Okay, thank you so much.
 
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