People Good at Probability and Statistics

[Chi Town Sox]

Mean ACT for 43 males is 21 and standard deviation is 5. Mean for 56 women is 20.8 with a standard deviation of 4.7. At 1% level - I am trying to find the Null Hypothesis (claim) and its complement. P-Value and the decision (reject for fail to reject) as well as if there is significant or insignificant evidence.

 

Anyone know how this could be calculated? I haven't done this in years and have been trying to help her for hours!!!

 

Thanks!

Edited November 19, 2009 by Chi Town Sox

[CanOfCorn]

7.

[FlaSoxxJim]

42

[ChiSox_Sonix]

Mean ACT for 43 males is 21 and standard deviation is 5. Mean for 56 women is 20.8 with a standard deviation of 4.7. At 1% level - I am trying to find the Null Hypothesis (claim) and its complement. P-Value and the decision (reject for fail to reject) as well as if there is significant or insignificant evidence.

 

Anyone know how this could be calculated? I haven't done this in years and have been trying to help her for hours!!!

 

Thanks!

 

You can't make up the H0 and H1 hypotheses without a conclusion. You need one or the other to answer this.

 

You can find the P-value off of a chart if by using your N-values and the alpha value, which in this case is 'plus or minus' .001

 

But, there just isn't enough information here. I don't even see what the question is or what the claim is supposed to be...?

Edited November 19, 2009 by ChiSox_Sonix

[Chi Town Sox]

You can't make up the H0 and H1 hypotheses without a conclusion. You need one or the other to answer this.

 

You can find the P-value off of a chart if by using your N-values and the alpha value, which in this case is 'plus or minus' .001

 

But, there just isn't enough information here. I don't even see what the question is or what the claim is supposed to be...?

 

Thanks - this is the questions word by word

 

The mean ACT score for 43 male high school students is 21 and the standard deviation is 5. The mean ACT score for 56 female high school students is 20.8 and the standard deviation is 4.7. At the 1% level, can you reject the claim that male and female high school students have equal ACT scores?

 

They are asking for....

 

H(sub 0):

H(sub 0):

 

P: Value

 

Decision: (Reject or Fail to Reject). There is (significant or insignificant) evidence to reject the claim

 

Maybe, that is why I can't figure it out - is it even possible?

 

 

 

[FlaSoxxJim]

You can't make up the H0 and H1 hypotheses without a conclusion. You need one or the other to answer this.

 

You can find the P-value off of a chart if by using your N-values and the alpha value, which in this case is 'plus or minus' .001

 

But, there just isn't enough information here. I don't even see what the question is or what the claim is supposed to be...?

 

There is enough information to perform an F-Test comparing standard deviations. Don't ask me how to do the math by hand, I gave that up as a bad habit long ago. Then again, that is only going to test the validity of a key underlying assumption of parametric statistics — that the two samples are drawn from the same statistical population.

 

H₀ is no difference between boys and girls and H₁ is that there is a difference, right?

 

With the means differing by 0.2, large sample size, very large standard deviations, and a 1% confidence level to reject the null, I can pretty much guarantee no statistical difference here. But I refuse to do the math because SoxTalk promised there would be no math.

 

Edited November 19, 2009 by FlaSoxxJim

[Pants Rowland]

There is enough information to perform an F-Test comparing standard deviations. Don't ask me how to do the math by hand, I gave that up as a bad habit long ago. Then again, that is only going to test the validity of a key underlying assumption of parametric statistics — that the two samples are drawn from the same statistical population.

 

H₀ is no difference between boys and girls and H₁ is that there is a difference, right?

 

With the means differing by 0.2, large sample size, very large standard deviations, and a 1% confidence level to reject the null, I can pretty much guarantee no statistical difference here. But I refuse to do the math because SoxTalk promised there would be no math.

 

 

Who are you, Gerald Ford?

[FlaSoxxJim]

Who are you, Gerald Ford?

 

Sit, Liberty, Sit!

[kjshoe04]

Thanks - this is the questions word by word

 

The mean ACT score for 43 male high school students is 21 and the standard deviation is 5. The mean ACT score for 56 female high school students is 20.8 and the standard deviation is 4.7. At the 1% level, can you reject the claim that male and female high school students have equal ACT scores?

 

They are asking for....

 

H(sub 0):

H(sub 0):

 

P: Value

 

Decision: (Reject or Fail to Reject). There is (significant or insignificant) evidence to reject the claim

 

Maybe, that is why I can't figure it out - is it even possible?

I'm in class now until 9 pm but if it hasn't been answered I will give it a go. Just did this stuff in my stats class a couple of weeks ago.

[kev211]

Yep I'll give this a go. Just learned this in my stats class.

[kev211]

Thanks - this is the questions word by word

 

The mean ACT score for 43 male high school students is 21 and the standard deviation is 5. The mean ACT score for 56 female high school students is 20.8 and the standard deviation is 4.7. At the 1% level, can you reject the claim that male and female high school students have equal ACT scores?

 

They are asking for....

 

H(sub 0):

H(sub 0):

 

P: Value

 

Decision: (Reject or Fail to Reject). There is (significant or insignificant) evidence to reject the claim

 

Maybe, that is why I can't figure it out - is it even possible?

Literally I just learned how to do this like the last few days in my stats class.

 

I don't know what they mean H(sub 0) but I'm assuming its H(sub o) and H(sub a)

 

Ho: X1=X2

Ha: X1=/=X2

 

Sample 1: (Male)

N1= 43

X1= 21

S1=5

 

Sample 2: (Female)

N1= 56

X1=20.8

S2=1.8

 

Alpha = .01

 

Using P-Value

Finding the test Statistic

 

Z = (X1-X2)/SQRT(S1^2/N1+S2^2/N2)

Z=(21-20.8)/SQRT(5^2/43+1.8^2/56)

Z=(.2)/SQRT(.58139+.05785)

Z=(.2)/SQRT(.63925)

z=0.25

 

P(Z>0.25)

Looking at the table we fine the area to the right of Z = 0.25 to = .4013

Pvalue = .4013 Since its a two-tailed test double it

Pvalue = .8026

 

Since Alpha .01(signifigance level) is less then the P-Value of .8026 you fail to reject the claim that there is a difference of scores between males and females.

 

I just learned this so I may be wrong, but I'm pretty sure I'm right so maybe someone can check over my work.

 

 

[Texsox]

This is an amazing site.

[kjshoe04]

Literally I just learned how to do this like the last few days in my stats class.

 

I don't know what they mean H(sub 0) but I'm assuming its H(sub o) and H(sub a)

 

Ho: X1=X2

Ha: X1=/=X2

 

Sample 1: (Male)

N1= 43

X1= 21

S1=5

 

Sample 2: (Female)

N1= 56

X1=20.8

S2=1.8

 

Alpha = .01

 

Using P-Value

Finding the test Statistic

 

Z = (X1-X2)/SQRT(S1^2/N1+S2^2/N2)

Z=(21-20.8)/SQRT(5^2/43+1.8^2/56)

Z=(.2)/SQRT(.58139+.05785)

Z=(.2)/SQRT(.63925)

z=0.25

 

P(Z>0.25)

Looking at the table we fine the area to the right of Z = 0.25 to = .4013

Pvalue = .4013 Since its a two-tailed test double it

Pvalue = .8026

 

Since Alpha .01(signifigance level) is less then the P-Value of .8026 you fail to reject the claim that there is a difference of scores between males and females.

 

I just learned this so I may be wrong, but I'm pretty sure I'm right so maybe someone can check over my work.

I did it while I was in class since I was bored and I got the same stuff so hopefully were not both wrong.

Edited November 20, 2009 by kjshoe04

[FlaSoxxJim]

Literally I just learned how to do this like the last few days in my stats class.

 

I don't know what they mean H(sub 0) but I'm assuming its H(sub o) and H(sub a)

 

Ho: X1=X2

Ha: X1=/=X2

 

Sample 1: (Male)

N1= 43

X1= 21

S1=5

 

Sample 2: (Female)

N1= 56

X1=20.8

S2=1.8

 

Alpha = .01

 

Using P-Value

Finding the test Statistic

 

Z = (X1-X2)/SQRT(S1^2/N1+S2^2/N2)

Z=(21-20.8)/SQRT(5^2/43+1.8^2/56)

Z=(.2)/SQRT(.58139+.05785)

Z=(.2)/SQRT(.63925)

z=0.25

 

P(Z>0.25)

Looking at the table we fine the area to the right of Z = 0.25 to = .4013

Pvalue = .4013 Since its a two-tailed test double it

Pvalue = .8026

 

Since Alpha .01(signifigance level) is less then the P-Value of .8026 you fail to reject the claim that there is a difference of scores between males and females.

 

I just learned this so I may be wrong, but I'm pretty sure I'm right so maybe someone can check over my work.

 

Very nice, but I see two issues.

 

First, I think you got your final interpretation backwards. a non-significant calculated P-value will cause you to fail to reject the null (no difference), but you indicated you should "fail to reject the claim that there is a difference", so I think you just got it crossed in the end.

 

I'm also questioning your interpretation of one- versus two-tailed tests. Certainly, yes, this is a two-tailed comparison since there is no a priori reason to assume one gender is smarter than the other. But, that means you will have to split your rejection area between the two tails of the bell curve. Logically that suggests you should halve the look-up (table) value rather than doubling it because you can't place your entire rejection area on just one side of the distribution.

 

Am I misinterpreting something on that?

[southsider2k5]

There is enough information to perform an F-Test comparing standard deviations. Don't ask me how to do the math by hand, I gave that up as a bad habit long ago. Then again, that is only going to test the validity of a key underlying assumption of parametric statistics — that the two samples are drawn from the same statistical population.

 

H₀ is no difference between boys and girls and H₁ is that there is a difference, right?

 

With the means differing by 0.2, large sample size, very large standard deviations, and a 1% confidence level to reject the null, I can pretty much guarantee no statistical difference here. But I refuse to do the math because SoxTalk promised there would be no math.

 

 

LETS GO SHOPPING!

[kjshoe04]

Very nice, but I see two issues.

 

First, I think you got your final interpretation backwards. a non-significant calculated P-value will cause you to fail to reject the null (no difference), but you indicated you should "fail to reject the claim that there is a difference", so I think you just got it crossed in the end.

 

I'm also questioning your interpretation of one- versus two-tailed tests. Certainly, yes, this is a two-tailed comparison since there is no a priori reason to assume one gender is smarter than the other. But, that means you will have to split your rejection area between the two tails of the bell curve. Logically that suggests you should halve the look-up (table) value rather than doubling it because you can't place your entire rejection area on just one side of the distribution.

 

Am I misinterpreting something on that?

yes, you should halve the .01 to .005.

[kev211]

Very nice, but I see two issues.

 

First, I think you got your final interpretation backwards. a non-significant calculated P-value will cause you to fail to reject the null (no difference), but you indicated you should "fail to reject the claim that there is a difference", so I think you just got it crossed in the end.

 

I'm also questioning your interpretation of one- versus two-tailed tests. Certainly, yes, this is a two-tailed comparison since there is no a priori reason to assume one gender is smarter than the other. But, that means you will have to split your rejection area between the two tails of the bell curve. Logically that suggests you should halve the look-up (table) value rather than doubling it because you can't place your entire rejection area on just one side of the distribution.

 

Am I misinterpreting something on that?

Yeah I just typed it wrong I think. I got confused because you accept the null and reject the alternative but the wording of the last sentence of the paragraph threw me off.

 

I was taught that you double the P-value by my professor, so maybe doubling the P-value accomplishes the same thing as halving the significance level(Alpha)?

[knightni]

I had to take this class 4 times in college to get a "C."

 

[FlaSoxxJim]

Yeah I just typed it wrong I think. I got confused because you accept the null and reject the alternative but the wording of the last sentence of the paragraph threw me off.

 

I was taught that you double the P-value by my professor, so maybe doubling the P-value accomplishes the same thing as halving the significance level(Alpha)?

 

I don't think so (or at least I can't logically see how). If you are setting the significance threshold (alpha) at 0.01, then that means that only 1 percent of the population distribution balls in that rejection area — hence, your chance of getting a statistically significant result merely by chance (Type 1 error) is only 1 in 100. In a one-tailed test, the whole 0.01% of the bell curve is located at one end of the distribution or another. In a two-tailed test, that 1% rejection area has to be split to cover both ends of the distribution. Since there is now only a 1 in 200 chance now of obtaining a significant result by chance, the threshold table P-value has to be less than for a 1-tailed test and not more. The calculated F-value from the means comparison has to be correspondingly greater to achieve an effective alpha of 0.005.

 

None of that is to say I'm not impressed by how quickly you're picking this stuf up, because I am.

 

All in all, I'm with Barbie and SSMike on this. Math is hard — Let's go shopping!

[Texsox]

Every time I read the thread title I think of you know who . . .

[southsider2k5]

I don't think so (or at least I can't logically see how). If you are setting the significance threshold (alpha) at 0.01, then that means that only 1 percent of the population distribution balls in that rejection area — hence, your chance of getting a statistically significant result merely by chance (Type 1 error) is only 1 in 100. In a one-tailed test, the whole 0.01% of the bell curve is located at one end of the distribution or another. In a two-tailed test, that 1% rejection area has to be split to cover both ends of the distribution. Since there is now only a 1 in 200 chance now of obtaining a significant result by chance, the threshold table P-value has to be less than for a 1-tailed test and not more. The calculated F-value from the means comparison has to be correspondingly greater to achieve an effective alpha of 0.005.

 

None of that is to say I'm not impressed by how quickly you're picking this stuf up, because I am.

 

All in all, I'm with Barbie and SSMike on this. Math is hard — Let's go shopping!

 

[FlaSoxxJim]

A Doctor, a Rabbi, and a Statistician go duck hunting together. The three of them are sitting behind the duck blind when a beautiful prize duck flies out into the clearing in front of them. The Doctor stands up, aims and fires and misses two feet in front of the duck. Then the Rabbi jumps up, aims and fires and misses two feet behind the duck. Finally, the Statistician jumps up and

excitedly exclaims,

"Hooray, we got him!!"