Suppose that a pre-election poll of 500 people showed that 51% of the sample supported the incumbent senator. If the population proportion who supported the incumbent senator is really 48%, how likely is it that we would see poll results such as this or higher?

A) 0.006
B) 0.03
C) 0.0901
D) 0.9099
E) 0.9680

To answer this question, we can use a one-sample proportion hypothesis test with a significance level of 0.05. Our null hypothesis is that the population proportion who support the incumbent senator is 0.48, and our alternative hypothesis is that the population proportion is higher than 0.48.

We can calculate the standard error of the sample proportion as follows:

SE = sqrt[(0.48)(0.52)/500] = 0.034

We can then calculate the z-score for the sample proportion:

z = (0.51 - 0.48)/0.034 = 0.88

Using a standard normal distribution table, we can find that the probability of obtaining a z-score of 0.88 or higher is 0.1915.

However, because our alternative hypothesis is one-sided (we are testing for a higher proportion), we need to find the area under the curve to the right of the z-score. This probability is 1 - 0.1915 = 0.8085.

Therefore, the answer is D) 0.9099, which is the closest probability to 0.8085. This means that if the population proportion who supported the incumbent senator is really 48%, there is an 90.99% chance that we would see poll results of 51% or higher.

Learn more about alternative hypothesis here:

https://brainly.com/question/27335001

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