Question

Just before a referendum on a school budget, a local newspaper polls 400 voters in an attempt to predict whether the budget will pass. Suppose that the budget actually has the support of 52% of the voters. What’s the probability the newspaper’s sample will lead them to predict defeat? Be sure to verify that the assumptions and conditions necessary for your analysis are met.

Answer 1

Using the normal approximation to the binomial distribution, we calculate a z-score based on the sample proportion that would indicate defeat (less than 0.50) and find the corresponding probability. This probability represents the likelihood that the newspaper poll would incorrectly predict the defeat of a school budget that has 52% voter support.

Step-by-step explanation:

To determine the probability that the newspaper's sample will incorrectly predict the defeat of the school budget when it actually has the support of 52% of the voters, we can use the normal approximation to the binomial distribution. This is appropriate because the sample size is large (n=400) and the sample proportion is not too close to 0 or 1, which meets the conditions for the central limit theorem.

The sample proportion that indicates defeat is any proportion less than 0.50. We first calculate the standard error for the sampling distribution of the sample proportion:

SE = sqrt(p(1-p)/n) = sqrt(0.52*0.48/400) = 0.0249

Next, we find the z-score for the sample proportion of 0.50:

Z = (p_sample - p) / SE = (0.50 - 0.52) / 0.0249 = -0.8032

Using a standard normal distribution table or calculator, we find the probability corresponding to a z-score of -0.8032.

This probability represents the chance that the sample will lead to a prediction of defeat, although the actual support is 52%.