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Math Help - confidence level

  1. #1
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    confidence level

    A program was developed to increase memory and retention. 9 participants were scored before and after the program. Their paired results:

    (80,85),(78,73),(80,86),(83,90),(93,95),(90,87),(7 9,75),(91,92),(95,98)

    where the first number is the before average and the second number is the average following the program. At a confidence level of 95%, can we conclude that the program increases memory?

    I'm stuck. Can I use the formula \bar{x}=+/-t_{\alpha/2}*\frac{s}{\sqrt{n}}? The two sets of data are confusing me. Thank you for helping.
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  2. #2
    MHF Contributor matheagle's Avatar
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    This is what is called a paired difference experiment.
    Subtract one value from the other and do the one sample t-test
    as you are used to doing.
    So your data is now {5,-5,6,7,..} and test to see if the population mean (change in memory) is positive.

    Here's a test procedure for it, the confidence interval is the same as before.
    http://www.vias.org/tmdatanaleng/cc_test_diffttest.html
    BUT I would test H_a: \mu>0, since you want to prove that memory is increased after this.

    A very good example of a paired difference experiment is weight reduction.
    You can test to see if people do lose a certain amount of weight over a period of time.
    Since you use the same subject before and after the samples are dependent.
    You wouldn't want to use two different sets of data, that doesn't make sense.
    Weigh 20 people before an 20 different people a month later is illogical.
    Last edited by matheagle; June 29th 2009 at 08:36 PM.
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  3. #3
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    oooook. thanks very much. so I have the differences which now are basically my observation values of (5,-5,6,7,2,-3,-3,1,3) for a new mean of 1.4444 and standard deviation of 4.303.

    I'm not sure what to subtract from the mean in the equation for the t test. I'm assuming it is H0 which would be 0 in this case? My H1 would then be p>0. So now I have

    t=\frac{1.4444-0}{\frac{4.303}{\sqrt{9}}}=1.007

    t_{0.025,8}=2.306 I don't know what critical region formula to use but it makes sense to me to be t \ge t_{0.025,8} which would not be true and thus failing to reject H0 (probability of no improvement)

    confidence interval is 1.444 +/- 1.96\frac{4.303}{\sqrt{9}}

    (1.444-2.81129, 1.444+2.81129)=(-1.3672, 4.25)
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  4. #4
    MHF Contributor matheagle's Avatar
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    You're testing a mean \mu, not a proportion/probabilty, p.

    I think this should be a one sided test.
    You want to 'prove' that the average memory INCREASES.

    H_0: \mu =0 vs. H_a: \mu>0.

    The rejection region would be (t_{n-1, \alpha},\infty) NOT \alpha/2.
    All the probability is in the one side.

    BUT some (lame) teachers will be happy with a two-sided CI and then they'll ask you to see if 0 is in that interval.
    However a one-sided CI is not lame, it's the same as a one-sided hypothesis
    test. But I would just go with the one-sided hypothesis test.
    Last edited by matheagle; June 30th 2009 at 04:23 PM.
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  5. #5
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    what t test would I then use?

    for the critical region if I read the chart right I got 1.860 as the value, so 1.860-\infty is the critical region? If the t test value is greater than or equal to 1.860, H0 is rejected? I'm confused because the only formulas in my book are for H0: p=p0

    When I finally straighten the rest out can I use that same confidence interval formula \bar{x} \pm 1.96\frac{s}{\sqrt{n}}?

    Thank you for helping me. It really means a lot to have someone willing to explain math to me.
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  6. #6
    MHF Contributor matheagle's Avatar
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    I'm sure you have test procedures for means and not just p's.
    The sample size is small so you cannot use normality in testing a proportion.
    You'd have to use a binomial table in that case.
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  7. #7
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    I do.
    For means I have (for known variance) H_{0}: \mu=\mu_{0} H_{1}: \mu>\mu_{0}

    z \ge z_{\alpha} OR \bar{x} \ge \mu_{0}+z_{\alpha}\sigma/\sqrt{n}

    For unknown variance

    t \ge t_{\alpha}(n-1) OR \bar{x} \ge \mu_{0}+t_{\alpha}(n-1)s/\sqrt{n}
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  8. #8
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by chrissy72 View Post
    I do.
    For means I have (for known variance) H_{0}: \mu=\mu_{0} H_{1}: \mu>\mu_{0}

    z \ge z_{\alpha} OR \bar{x} \ge \mu_{0}+z_{\alpha}\sigma/\sqrt{n}

    For unknown variance

    t \ge t_{\alpha}(n-1) OR \bar{x} \ge \mu_{0}+t_{\alpha}(n-1)s/\sqrt{n}
    The second one is exactly this...

    \bar{x} \ge \mu_{0}+t_{\alpha}(n-1)s/\sqrt{n}

    is the same as {\bar{x}-\mu_{0}\over s/\sqrt n}>t_{\alpha}(n-1)

    WHERE the test stat is {\bar{x}-\mu_{0}\over s/\sqrt n}

    and the rejection region is (t_{\alpha}(n-1), \infty)

    and \mu_{0}=0 since under the null hypothesis you ARE assumng that \mu is 0.
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  9. #9
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    so for this question i use the mean of the values before the program ( \mu_{0}) and the mean of the values after ( \bar{x}) then s would be the SD of \bar{x}? I though I had to use the info calculated from the difference (of which I got one number wrong that changed all my answers)
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  10. #10
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    I did the t test but how do I incorporate the confidence interval of 95% and how does that tell me whether memory increased or not? From the t test the null hypothesis was not rejected (H0: \mu_{0}=0).
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