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Suppose that a particular candidate for public office is in fact favored by \(48 \%\) of all registered voters in the district. A polling organization will take a random sample of 500 voters and will use \(p\), the sample proportion, to estimate \(\pi\). What is the approximate probability that \(p\) will be greater than .5, causing the polling organization to incorrectly predict the result of the upcoming election?

Short Answer

Expert verified
The approximate probability that \(p\) will be greater than .5 comes down to finding the Z-Score and then using that Z-Score to look up the related probability in a standard normal distribution table. Depending on the Z-Score calculated, this probability may change, so the exact numerical output may not be provided here.

Step by step solution

01

Calculate Standard Error

The first step is to calculate the standard error of the sample proportion \(p\). Standard error (SE) is given by the formula \[SE = \sqrt{\frac{\pi(1-\pi)}{n}}\] where \(\pi = 0.48\) (population proportion or the true proportion of voters favoring the candidate) and \(n = 500\) (sample size). Substituting the values, calculate the SE.
02

Find the Z Score

Next, calculate the Z score for .5. The Z score helps us understand how far away a data point is from the mean in terms of standard deviations. We calculate the Z score by using the formula \[Z = \frac{p-\pi}{SE}\] where \(p = 0.5\) (the value for which we are finding the probability), \(\pi = 0.48\) and SE is the value calculated in the first step. Calculate the Z score by substituting the appropriate values.
03

Use Z-Score to Find Probability

Once we have our Z score from step 2, we can find the corresponding probability from a standard normal distribution table, which tells us the probability that a statistic is observed below, above, or between values. Since we need the probability that \(p\) is greater than .5 (i.e., our Z score is greater than the calculated value), we need to look up the complementary probability (1 minus the value found on the table) as Z-Tables typically give the area to the left of the given Z-Score.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Error Calculation
Understanding the standard error (SE) is crucial for interpreting the variability of a sample statistic. In probability, the standard error of the sample proportion is a measure of how much we would expect the sample proportion to vary from sample to sample. The formula to calculate the standard error is:

\[SE = \sqrt{\frac{\pi(1-\pi)}{n}}\]

where \(\pi\) is the population proportion and \(n\) is the sample size. For a population proportion of 48% or 0.48, and a sample size of 500, the formula becomes:
\[SE = \sqrt{\frac{0.48(1-0.48)}{500}}\]
To calculate the standard error, simply plug these numbers into the formula and solve for SE. The resulting SE gives us an idea of how much random sampling error we might expect if we were to take numerous samples of the same size from the same population. In other words, it tells us how 'spread out' the sample proportions are likely to be around the true population proportion.
Z Score Computation
A Z score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. In the context of our exercise, the Z score computation involves how the sample proportion compares to what we expect from the population proportion.

To calculate the Z score, we use the formula:
\[Z = \frac{p-\pi}{SE}\]
This equation requires the sample proportion (\(p\)), the population proportion (\(\pi\)), and the standard error (SE), which we computed from the previous section. For a sample proportion of 50% or 0.5, and a population proportion of 48%, with the previously calculated SE, the Z score can be computed as:
\[Z = \frac{0.5-0.48}{SE}\]
This Z score tells us how many standard errors the sample proportion is above or below the population proportion. A positive Z score indicates that the sample proportion is above the population mean, while a negative Z score indicates it is below.
Normal Distribution Table
The normal distribution table, also known as the Z-table, is a reference for statisticians to determine the probability of a given Z score occurring within a normal distribution. This table shows the cumulative probability associated with a Z score up to a certain point to the left of the Z score in the distribution.

To use it, you first calculate your Z score using the formula from the previous step. Then, you locate the Z score on the Z-table to find the corresponding probability. However, keep in mind that standard Z-tables give the area under the curve to the left of the Z score. If you are interested in the probability of the sample proportion being greater than a certain value, you will need to look at the complementary probability, which is simply 1 minus the probability found in the Z-table.
For example, if our Z score is 2.00 and we need the probability that a value is more than 2 standard deviations away from the mean, we would look at the complementary probability:
1 - (cumulative probability for Z = 2.00 from Z-table). This would give us the area under the normal distribution curve to the right of Z = 2.00, which is what we are interested in for our exercise.

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Most popular questions from this chapter

For each of the following statements, identify the number that appears in boldface type as the value of either a population characteristic or a statistic: a. A department store reports that \(84 \%\) of all customers who use the store's credit plan pay their bills on time. b. A sample of 100 students at a large university had a mean age of \(24.1\) years. c. The Department of Motor Vehicles reports that \(22 \%\) of all vehicles registered in a particular state are imports. d. A hospital reports that based on the 10 most recent cases, the mean length of stay for surgical patients is \(6.4\) days. e. A consumer group, after testing 100 batteries of a certain brand, reported an average life of \(\mathbf{6 3} \mathrm{hr}\) of use.

Explain the difference between a population characteristic and a statistic.

Suppose that a sample of size 100 is to be drawn from a population with standard deviation \(10 .\) a. What is the probability that the sample mean will be within 2 of the value of \(\mu\) ? b. For this example \((n=100, \sigma=10)\), complete each of the following statements by computing the appropriate value: i. Approximately 95% of the time, \(\bar{x}\) will be within _____ of \(\mu .\) ii. Approximately 0.3% of the time, \(\bar{x}\) will be farther than _____ from\(\mu .\)

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a sample of \(n=25\) adult males is to be selected, what is the probability that the sample average distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm}\) ? b. Suppose that a sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample average distance will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm}\) ?

The article "Should Pregnant Women Move? Linking Risks for Birth Defects with Proximity to Toxic Waste Sites" (Chance [1992]: 40-45) reported that in a large study carried out in the state of New York, approximately \(30 \%\) of the study subjects lived within 1 mi of a hazardous waste site. Let \(\pi\) denote the proportion of all New York residents who live within 1 mi of such a site, and suppose that \(\pi=.3\). a. Would \(p\) based on a random sample of only 10 residents have approximately a normal distribution? Explain why or why not. b. What are the mean value and standard deviation of \(p\) based on a random sample of size \(400 ?\) c. When \(n=400\), what is \(P(.25 \leq p \leq .35)\) ? d. Is the probability calculated in Part (c) larger or smaller than would be the case if \(n=500 ?\) Answer without actually calculating this probability.

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