Question

A survey asks a simple random sample of 500 adults in Ohio if they support an increasbin the state sales tax from 5% to 6%, with additional revenue going to education. Let p-hat denote the proportion in the sample who say they support the increase. Suppose that 53% of all adults in Ohio support the increase. What is the probability that less than half the sample will say they support the increase?

Answer 1

9.01% probability that less than half the sample will say they support the increase

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For a proportion p in a sample of size n, we have that
`\mu = p, \sigma = \sqrt{(p(1-p))/(n)}`

In this problem

`\mu = 0.53, \sigma = \sqrt{(0.53*0.47)/(500)} = 0.0223`

What is the probability that less than half the sample will say they support the increase?

This is the pvalue of Z when X = 0.5. So

`Z = (X - \mu)/(\sigma)`

`Z = (0.5 - 0.53)/(0.0223)`

`Z = -1.34`

`Z = -1.34` has a pvalue of 0.0901

9.01% probability that less than half the sample will say they support the increase

Answer 2

We need to get the z-score for the given data from z-score table. After which, use this formula
p-hat - z √p-hat (1 - p-hat)/n
To get the probability that less than of half of the sample say they support the increase