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Chapter 8: Problem 30

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

The amount of money spent by a customer at a discount store has a mean of $$\$ 100$$ and a standard deviation of $$\$ 30.$$ What is the probability that a randomly selected group of 50 shoppers will spend a total of more than $$\$ 5300 ?$$ (Hint: The total will be more than \(\$ 5300\) when the sample average exceeds what value?)

Water permeability of concrete can be measured by letting water flow across the surface and determining the amount lost (in inches per hour). Suppose that the permeability index \(x\) for a randomly selected concrete specimen of a particular type is normally distributed with mean value 1000 and standard deviation 150 . a. How likely is it that a single randomly selected specimen will have a permeability index between 850 and \(1300 ?\) b. If the permeability index is to be determined for each specimen in a random sample of size 10 , how likely is it that the sample average permeability index will be between 950 and \(1100 ?\) between 850 and 1300 ?

Newsweek (November 23, 1992) reported that 40\% of all U.S. employees participate in "self-insurance" health plans \((\pi=.40)\). a. In a random sample of 100 employees, what is the approximate probability that at least half of those in the sample participate in such a plan? b. Suppose you were told that at least 60 of the \(100 \mathrm{em}\) ployees in a sample from your state participated in such a plan. Would you think \(\pi=.40\) for your state? Explain.

Let \(x\) denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of \(x\) are \(\mu=2 \mathrm{~min}\) and \(\sigma=0.8\) min, respectively. a. If \(\bar{x}\) is the sample average time for a random sample of \(n=9\) students, where is the \(\bar{x}\) distribution centered, and how much does it spread out about the center (as described by its standard deviation)? b. Repeat Part (a) for a sample of size of \(n=20\) and again for a sample of size \(n=100\). How do the centers and spreads of the three \(\bar{x}\) distributions compare to one another? Which sample size would be most likely to result in an \(\bar{x}\) value close to \(\mu\), and why?

Consider the following population: \(\\{1,2,3,4\\}\). Note that the population mean is $$\mu=\frac{1+2+3+4}{4}=2.5$$ a. Suppose that a random sample of size 2 is to be selected without replacement from this population. There are 12 possible samples (provided that the order in which observations are selected is taken into account): \(\begin{array}{cccccc}1,2 & 1,3 & 1,4 & 2,1 & 2,3 & 2,4 \\ 3,1 & 3,2 & 3,4 & 4,1 & 4,2 & 4,3\end{array}\) Compute the sample mean for each of the 12 possible samples. Use this information to construct the sampling distribution of \(\bar{x}\). (Display the sampling distribution as a density histogram.) b. Suppose that a random sample of size 2 is to be selected, but this time sampling will be done with replacement. Using a method similar to that of Part (a), construct the sampling distribution of \(\bar{x}\). (Hint: There are 16 different possible samples in this case.) c. In what ways are the two sampling distributions of Parts (a) and (b) similar? In what ways are they different?
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