Suppose that 20% of the residents in a certain state support an increase in the

Question

Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly sample 400 state residents and will then compute the proportion in the sample that support a property tax increase. How likely is the resulting sample proportion be within .04 of the true proportion (i.e., bet 16 and .24)? It is certain that the resulting sample proportion will be within .04 of the proportion. There is roughly a 99.7% chance that the resulting sample proportion will be within .04 of the true proportion. There is roughly a 95% chance that the resulting sample proportion will be within .04 of the true proportion. There is roughly a 68% chance that the resulting sample proportion will be within .04 of the true proportion.

Explanation / Answer

Solution:

For the given scenario, we are given p = 20% = 0.20

The standard deviation for the sampling distribution of proportion is given as below:

SD = sqrt(pq/n)

Where, p = 0.20, q = 1 – p = 1 – 0.20 = 0.80 and n = 400

SD = sqrt(0.20*0.80/400) = sqrt(0.0004) = 0.02

Now, by using empirical rule (68-95-99.7 rule)

1SD = 1*0.02 = 0.02

About 68% chance that the resulting sample proportion will be within 0.02 of the true proportion.

2SD = 2*0.02 = 0.04

About 95% chance that the resulting sample proportion will be within 0.04 of the true proportion.

3SD = 3*0.02 = 0.06

About 99.7% chance that the resulting sample proportion will be within 0.06 of the true proportion.

So, correct answer:

There is roughly a 95% chance that the resulting sample proportion will be within 0.04 of the true proportion.

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