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 to be within 0.04 of the true proportion (i.e., between 0.16 and 0.24)?(Hint: Use the sampling distribution of the sample proportion in this case.) It is certain that the resulting sample proportion will be within 0.04 of the true proportion. There is roughly a 99.7% chance that the resulting sample proportion will be within 0.04 of the true proportion. There is roughly a 95% chance that the resulting sample proportion will be within 0.04 of the true proportion. There is roughly a 68% chance that the resulting sample proportion will be within 0.04 of the true proportion.

The question is asking for the probability that the sample proportion will be within 0.04 of the true proportion. This is a question about the sampling distribution of the sample proportion.

Step 1: Identify the parameters of the population distribution. The problem tells us that the true proportion (p) is 0.20.

Step 2: Identify the parameters of the sampling distribution. The central limit theorem tells us that the sampling distribution of the sample proportion is approximately normal with mean equal to the population proportion (p = 0.20) and standard deviation equal to sqrt[p(1-p)/n], where n is the sample size. In this case, n = 400, so the standard deviation is sqrt[0.20*0.80/400] = 0.02.

Step 3: Calculate the z-scores for the boundaries of the interval. The z-score is the number of standard deviations a value is from the mean. The lower boundary is 0.16 and the upper boundary is 0.24. The z-score for 0.16 is (0.16-0.20)/0.02 = -2. The z-score for 0.24 is (0.24-0.20)/0.02 = 2.

Step 4: Use the standard normal table to find the probabilities associated with these z-scores. The table tells us that the probability of a z-score being less than -2 is approximately 0.0228 and the probability of a z-score being less than 2 is approximately 0.9772.

Step 5: The probability that the z-score is between -2 and 2 is 0.9772 - 0.0228 = 0.9544. So, there is roughly a 95% chance that the resulting sample proportion will be within 0.04 of the true proportion.

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To answer this question, we need to use the concept of sampling distribution of the sample proportion.

First, we need to find the standard deviation of the sampling distribution. The formula for the standard deviation of the sample proportion is sqrt[(P(1 - P) / n], where P is the population proportion and n is the sample size.

In this case, P = 0.20 and n = 400.

So, the standard deviation is sqrt[(0.20 * 0.80) / 400] = 0.02.

Next, we need to find the z-scores for the boundaries of the interval (0.16 and 0.24). The z-score is calculated as (x - μ) / σ, where x is the sample proportion, μ is the population proportion, and σ is the standard deviation of the sampling distribution.

For x = 0.16, the z-score is (0.16 - 0.20) / 0.02 = -2. For x = 0.24, the z-score is (0.24 - 0.20) / 0.02 = 2.

Finally, we need to find the probability that the z-score is between -2 and 2. This is equivalent to finding the area under the standard normal curve between -2 and 2. According to the empirical rule (or the 68-95-99.7 rule), about 95% of the data falls within 2 standard deviations of the mean in a normal distribution.

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

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