Final answer:
The question deals with calculating a p-value for a sample proportion in a statistical test, specifically when you have a certain number of successes out of a total sample size. For a sufficiently large sample, normal distribution can be used to approximate the binomial distribution, allowing for the calculation and interpretation of the p-value against a given significance level.
Step-by-step explanation:
The subject of the question involves calculating the p-value for a sample proportion in a statistical test. When you have a sample size, such as n=100, with 28 successes, you can determine the sample proportion and subsequently use a z-test to calculate the p-value. In the context of hypothesis testing, the p-value is compared to a significance level (α) to make a decision about rejecting or failing to reject the null hypothesis.
For a larger sample size, like the one given with n=420,019 and x=172, you verify the conditions for using a normal distribution to approximate the binomial. Since np=142.8 and nq=419,876.2, the sample is sufficiently large, and the condition of having two independent outcomes with a fixed probability of success (p=0.00034) is met. Therefore, you can use the normal distribution to calculate the p-value and generalize the results to the population.
Increasing the sample size reduces the margin of error in calculating a confidence interval or conducting a hypothesis test, which in turn increases the accuracy of the conclusions drawn from the statistical analysis.