Independent random samples, each containing 800 observations, were selected from two binomial populations. The samples from populations 1 and 2 produced 320 and 400 successes, respectively.
Construct a 90% confidence interval for the true difference in population proportions of successes. In the interest of time you can assume that the sample sizes are large enough that it is appropriate to use a large-sample confidence interval.
To construct a confidence interval for the true difference in population proportions of successes, we can use the formula for a large-sample confidence interval for the difference in proportions.
The formula is:
CI = (p1 - p2) ± z * √((p1 * (1 - p1)/n1) + (p2 * (1 - p2)/n2))
Where:
CI is the confidence interval
p1 and p2 are the sample proportions of successes
n1 and n2 are the sample sizes
z is the z-score associated with the desired confidence level
In this case, we are given the sample sizes (n1 = 800, n2 = 800) and the number of successes in each sample (p1 = 320/800 = 0.4, p2 = 400/800 = 0.5).
We want to construct a 90% confidence interval, so the corresponding z-score can be found from the standard normal distribution table. The z-score for a 90% confidence level is approximately 1.645.
Now we can plug in the values into the formula to calculate the confidence interval:
CI = (0.4 - 0.5) ± 1.645 * √((0.4 * (1 - 0.4)/800) + (0.5 * (1 - 0.5)/800))
Simplifying the expression inside the square root:
CI = (0.4 - 0.5) ± 1.645 * √((0.4 * 0.6/800) + (0.5 * 0.5/800))
CI = -0.1 ± 1.645 * √(0.00024 + 0.0003125)
CI = -0.1 ± 1.645 * √0.0005525
CI = -0.1 ± 1.645 * 0.0235
Multiplying 1.645 by 0.0235:
CI = -0.1 ± 0.0387
Finally, subtracting and adding the result to the difference in sample proportions:
CI = (-0.1 - 0.0387, -0.1 + 0.0387)
CI = (-0.1387, -0.0613)
Therefore, the 90% confidence interval for the true difference in population proportions of successes is approximately (-0.1387, -0.0613).