Question

A CBS News/New York Times opinion poll asked 1,190 adults whether they would prefer balancing the federal budget over cutting taxes; 59% of those asked said "yes." Suppose that in fact 62% of all adults favor balancing the budget over cutting taxes. If this poll was not biased, what is the probability you would get p^=0.59 or less?

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a 0.0166 = 1.66% probability of a sample proportion of 0.59 or less.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sampling proportions of a proportion p in a sample of size n has mean $\mu = p$ and standard error $s = \sqrt{\frac{p(1 - p)}{n}}$

In this problem:

• 1,190 adults were asked, hence $n = 1190$

• In fact 62% of all adults favor balancing the budget over cutting taxes, hence $p = 0.62$.

The mean and the standard error are given by:

$\mu = p = 0.62$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.62(0.38)}{1190}} = 0.0141$

The probability of a sample proportion of 0.59 or less is the p-value of Z when X = 0.59, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.59 - 0.62}{0.0141}$

$Z = -2.13$

$Z = -2.13$ has a p-value of 0.0166.

0.0166 = 1.66% probability of a sample proportion of 0.59 or less.

You can learn more about the normal distribution and the central limit theorem at brainly.com/question/24663213