Hmm...Random Sampling

**bbash30** · April 23rd 2008, 02:08 PM

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

**mr fantastic** · April 23rd 2008, 09:27 PM

Originally Posted by bbash30

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

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

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

Sample 2:

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

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

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

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

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

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

Decide on a significance level.

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

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

Is that value significant?

Key words: Difference sample proportions

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