Independent random samples of 280 and 350 observations were selected from
binomial populations 1 and 2, respectively. Sample 1 had 132 successes, and sample 2 had 178 successes.
Do the data present sufficient evidence to indicate that the proportion of
the successes in population 1 is smaller than the proportion of the successes in population 2?
Find the p-value of the test and make your conclusion.
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Denote the (binary - value 0 or 1) observations from the two populations `x_i` , `i=1,2,...,n_x` and `y_j`, `j=1,2,...,n_y`
where `n_x = 280` and `n_y =350`.
We are given that `sum_i x_i = 132` and that `sum_j y_j = 178`
so that we estimate the proportion of successes in the two populations, `p_x` and `p_y`, as
`hat(p)_x = sum_i x_i/n_x = 132/280 =0.471 ` and
`hat(p)_y = sum_j y_j/n_y = 178/350 = 0.509`
To test whether the proportion in population 1 is smaller than in population 2 we approximate the difference between the two proportions to be Normal and set up a Z statistic
`Z = (hat(p)_x-hat(p)_y)/hat(sigma)`
where we estimate `sigma` with the pooled variance estimator
`hat(sigma)^2 = ((n_x-1)hat(sigma)_x^2 + (n_y-1)hat(sigma)_y^2)/(n_x+n_y-2) = ((n_x-1)(hat(p)_x(1-hat(p)_x))/n_x + (n_y-1)(hat(p)_y(1-hat(p)_y))/n_y)/(n_x+n_y-2)` `= (0.248 + 0.249)/628 = 0.000792`
Therefore `hat(sigma) = sqrt(0.000792) = 0.0281`
`Z = (0.471-0.509)/0.0281 = -1.350`
This is not an especially low Z-score. In fact the p-value is `Phi^(-1)(-1.350) = 0.089 `
If the (one-sided) test is carried out at the 10% level then we would conclude that the data indicate that the proportion in population 1 is smaller than that in population 2. However, if we test at the 5% level (which is more standard) then there is insufficient evidence to say that the proportion in population 1 is smaller. (Pick one of these significance levels and give the result).
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