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Math Help - help calculating p-value (again)

  1. #1
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    help calculating p-value (again)

    I post too much don't I? haha anyway, any help with this problem is appreciated.

    "From a normal population, a sample with N=4 is taken with sample mean M= 81 and sample standard deviation s = 10.42. For a null hypothesis = 92, calculate a t-score for the sample mean. What is the p-value for a two-tailed test? Based on the p-value, is the population mean significantly different from 92?"
    Calculating a tscore for sample mean is: sample mean - null hypothesis/ standard error of the mean so 81-92/(10.42/(sqrt)5)= -2.36
    P(-2.36<t<0)=0.4585. Thus, P(t<-2.36)=0.5-0.4585= 0.0415. Since this is a two-tailed test (it asked how the mean is different not less than or greater than) I double this amount? So is my p-value 0.2075? The last part of the question I always get stuck on (Based on the p-value, is the mean significantly different from 92) but I want to go with yes. ha. There's another question, but I'm stuck

    "Based on the p-value you found above, can you conclude that the population mean is significantly different from 92 at 0.01 level? How about at 0.1 level?"
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  2. #2
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    Quote Originally Posted by peanutbutter View Post
    I post too much don't I? haha anyway, any help with this problem is appreciated.

    "From a normal population, a sample with N=4 is taken with sample mean M= 81 and sample standard deviation s = 10.42. For a null hypothesis = 92, calculate a t-score for the sample mean. What is the p-value for a two-tailed test? Based on the p-value, is the population mean significantly different from 92?"
    Calculating a tscore for sample mean is: sample mean - null hypothesis/ standard error of the mean so 81-92/(10.42/(sqrt)5)= -2.36

    Mr F says: n = 4 so won't it be (81-92)/(10.42/(sqrt)4)= -2.11 .....

    P(-2.36<t<0)=0.4585. Thus, P(t<-2.36)=0.5-0.4585= 0.0415.

    Mr F says: Where have you got these numbers from? They aren't correct (even allowing for the error in the t-value). Pr(t < -2.11) = 0.06268. I got this value using the t-distribution with 4 - 1 = 3 defrees of freedom.

    Since this is a two-tailed test (it asked how the mean is different not less than or greater than) I double this amount?

    Mr F says: Correct.

    So is my p-value 0.2075? The last part of the question I always get stuck on (Based on the p-value, is the mean significantly different from 92) but I want to go with yes. ha. There's another question, but I'm stuck
    "Based on the p-value you found above, can you conclude that the population mean is significantly different from 92 at 0.01 level? How about at 0.1 level?"
    The p-value is 2(0.06268) = 0.12536. p-value > 0.01 therefore the result is not significant at the 0.01 significance level.

    p-value > 0.1 therefore the result is not significant at the 0.1 significance level. (The result would be significant for a one-tailed test since then p-value = 0.06268 < 0.1 ......)
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  3. #3
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    I knew I would mess up somewhere. Actually N=5. I typed 4 because that's my degree of freedom since N=5 and also because I have no brain apparently
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