Asurvey 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?

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

Step-by-step explanation:

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

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

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

In this problem

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

has a pvalue of 0.0901

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

step-by-step explanation:

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