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Hypothesis Testing in Statistics

  1. Jun 19, 2009 #1
    1. The problem statement, all variables and given/known data

    A random sample of 30 rats has 18 females and 12 males. If last year's proportion of females treated was 0.65, do the above data confirm that this year the proportion of females is different than last year?

    2. Relevant equations

    p = treated / total
    If using standard normal: z = (value - mean) / (standard deviation / squareroot(sample size) )

    3. The attempt at a solution

    I *think* the null hypothesis is p = 0.65 and the alternative hypothesis is p ≠ 0.65
    What I'm stuck on now is I don't know what test statistic to use to find the p value to decide if I should reject the null hypothesis. Not sure if normal or binomial is supposed to be used and how to do the calculations for this.
     
  2. jcsd
  3. Jun 19, 2009 #2

    EnumaElish

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    Homework Helper

    As you can see from this, in most cases the normal is a good approximation to the binomial. Your null hypothesis is p = 0.65, so normal mean = 0.65 * 30 = 19.5 and normal variance = p*(1-p)*30 under the null hypothesis.
     
    Last edited: Jun 19, 2009
  4. Jun 19, 2009 #3
    say that if I can't use the normal approximation, how would I do this?
     
  5. Jun 19, 2009 #4

    statdad

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    Why wouldn't you use the the usual test for one proportion?

    [tex]
    H_0 \colon p = 0.65, \quad H_a \colon p \ne 0.65
    [/tex]

    with test statistic

    [tex]
    Z = \frac{\hat p - 0.65}{\sqrt{\dfrac{0.65 \cdot 0.35}{30}}}
    [/tex]

    Your sample size, together with the number of "successes", will allow you to use the normal distribution for your test.
     
  6. Jun 19, 2009 #5
    I was thinking about using binomial in case the sample size was really really small (small proportions test) but I'll worry about that later.

    Thanks for showing me how I was supposed to approach this problem. :)
     
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