Suppose population proportion1 = 0.4, sample size1 = 100, population proportion2 = 0.5, sample size2 = 120.
whats the formula? i think this is just a plug and chug
z = (p1 -p2)/sqrt(p1/n1 + p2/n))
Post a null hypothesis that would use a t test statistical analysis
To compare two population proportions, you can use the two-sample proportion test, sometimes called the z-test. The formula for the z-test statistic is:
z = (p̂1 - p̂2) / √((p̂(1-p̂)(1/n1) + p̂(1-p̂)(1/n2)))
Where:
- p̂1 and p̂2 are the sample proportions of the two groups.
- p̂ is the pooled sample proportion, calculated as (x1 + x2) / (n1 + n2), where x1 and x2 are the number of successes in each sample, and n1 and n2 are the sample sizes.
- n1 and n2 are the sample sizes of the two groups.
In your case, with population proportion1 of 0.4 and sample size1 of 100, and population proportion2 of 0.5 and sample size2 of 120, you can calculate the z-test statistic as follows:
p̂1 = (0.4 * 100) / 100 = 0.4
p̂2 = (0.5 * 120) / 120 = 0.5
p̂ = (x1 + x2) / (n1 + n2) = (0.4 * 100 + 0.5 * 120) / (100 + 120) = 0.45
z = (0.4 - 0.5) / √((0.45(1-0.45)(1/100) + 0.45(1-0.45)(1/120)))
Note that in practice, you would need to calculate the standard error of the difference and perform hypothesis testing to determine if the difference in proportions is statistically significant. Additionally, assumptions like independence of samples and large sample sizes should be considered.