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Thread: Hmm...Random Sampling

  1. #1
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    Hmm...Random Sampling

    A random sample of 100 high school students were asked whether they would turn to their parents for help with a homework assignment in mathematics, and another random sample of 100 high school students were asked the same question with regard to a homework assignment in English. If 62 students in the first sample and 44 students in the second sample would turn to their parents for help, test whether or not there is a difference between the two sample proportions.

    Any ideas?...not sure where to start
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  2. #2
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    Quote Originally Posted by bbash30 View Post
    A random sample of 100 high school students were asked whether they would turn to their parents for help with a homework assignment in mathematics, and another random sample of 100 high school students were asked the same question with regard to a homework assignment in English. If 62 students in the first sample and 44 students in the second sample would turn to their parents for help, test whether or not there is a difference between the two sample proportions.

    Any ideas?...not sure where to start
    The sample sizes are large so a normal approximation in what follows is valid. Also the samples are independent.

    Sample 1:

    Let $\displaystyle X_1$ be the random variable number of students who turn to their parents for maths help.

    $\displaystyle X_1$ ~ $\displaystyle \text{Binomial} (n = n_1 = 100, \, p = p_1)$.

    Large sample: $\displaystyle \frac{X_1}{n_1} = \frac{X_1}{100}$ ~ $\displaystyle \text{Normal}\left( \mu_1 = p_1, \, \sigma_1^2 = \frac{p_1(1 - p_1)}{n_1} = \frac{p_1(1 - p_1)}{100} \approx \frac{0.62(1 - 0.62)}{100} = 0.002356\right)$.



    Sample 2:

    Let $\displaystyle X_2$ be the random variable number of students who turn to their parents for english help.

    $\displaystyle X_2$ ~ $\displaystyle \text{Binomial} (n = n_2 = 100, \, p = p_2)$.

    Large sample: $\displaystyle \frac{X_2}{n_2} = \frac{X_2}{100}$ ~ $\displaystyle \text{Normal}\left( \mu_2 = p_2, \, \sigma_2^2 = \frac{p_2(1 - p_2)}{n_2} = \frac{p_2(1 - p_2)}{100} \approx \frac{0.44(1 - 0.44)}{100} = 0.002464\right)$.




    Therefore $\displaystyle D = \frac{\left( \frac{X_1}{100} - \frac{X_2}{100}\right) - \left( p_1 - p_2\right)}{\sqrt{0.002356 + 0.002464}} = \frac{\left( \frac{X_1 - X_2}{100}\right) - \left( p_1 - p_2\right)}{0.0694} $ ~ Normal(0, 1).



    The null hypothesis is $\displaystyle p_1 - p_2 = 0$.

    The alternative hypothesis is $\displaystyle |p_1 - p_2| \neq 0$. This implies a two-sided test.

    Decide on a significance level.

    $\displaystyle X_1 = 62$ and $\displaystyle X_2 = 44$. Therefore under the null hypothesis $\displaystyle D = \frac{\left( \frac{62 - 44}{100}\right) - 0}{0.0694} = 2.5937$.


    Two sided: P = 2 Pr(D > 2.5) = 0.0124.

    Is that value significant?




    Key words: Difference sample proportions
    Last edited by mr fantastic; Apr 23rd 2008 at 08:46 PM.
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