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Checking for Survey Errors. One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About \(24 \%\) of the Canadian population is first generation-that is, they were born outside Canada. \(\underline{6}\) The number \(X\) of first-generation Canadians in random samples of 1500 persons should therefore vary with the binomial ( \(n=1500, p=0.24\) ) distribution. a. What are the mean and standard deviation of \(X\) ? b. Use the Normal approximation to find the probability that a sample will contain between 340 and 390 firstgeneration Canadians. Check that you can safely use the approximation.

Short Answer

Expert verified
The mean is 360, standard deviation is approximately 16.93, probability is about 0.851.

Step by step solution

01

Calculate the Mean

For a binomial distribution with parameters \(n\) and \(p\), the mean \(\mu\) is given by \(\mu = n \times p\). In this case, \(n = 1500\) and \(p = 0.24\), so the mean is \(\mu = 1500 \times 0.24 = 360\).
02

Calculate the Standard Deviation

The standard deviation \(\sigma\) for a binomial distribution is given by \(\sigma = \sqrt{n \times p \times (1-p)}\). Using \(n = 1500\) and \(p = 0.24\), find \(\sigma = \sqrt{1500 \times 0.24 \times 0.76} \approx 16.93\).
03

Check the Condition for Normal Approximation

For the normal approximation to the binomial distribution to be valid, both \(np\) and \(n(1-p)\) must be greater than 10. Here, \(np = 360\) and \(n(1-p) = 1140\), which are both greater than 10. Thus, the normal approximation is applicable.
04

Define the Range for Normal Approximation

We want to find the probability that the number of first-generation Canadians \(X\) is between 340 and 390. Applying continuity correction, we consider the range to be from 339.5 to 390.5.
05

Convert to the Standard Normal Distribution

To find this probability, convert the range to standard normal variables using \(Z = \frac{X - \mu}{\sigma}\). For \(X = 339.5\), \(Z = \frac{339.5 - 360}{16.93} \approx -1.21\), and for \(X = 390.5\), \(Z = \frac{390.5 - 360}{16.93} \approx 1.80\).
06

Calculate the Probability Using Z Scores

Using the standard normal distribution table, the probability \(P(-1.21 < Z < 1.80)\) can be found by calculating \(P(Z < 1.80) - P(Z < -1.21)\). This gives \(P(Z < 1.80) \approx 0.9641\) and \(P(Z < -1.21) \approx 0.1131\). Thus, the probability \(P(340 < X < 390)\) is \(0.9641 - 0.1131 = 0.851\).

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Key Concepts

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

Normal Approximation
Normal approximation is a method used to approximate the binomial distribution using the normal distribution, especially when dealing with large sample sizes. Binomial distributions describe the number of "successes" in a series of trials, which can be cumbersome to calculate by hand for large samples.

By applying the normal approximation, computations become more manageable.
  • Key requirement: both the expected number of successes ( p ext{np} ) and failures ( q ext{n(1-p)} ) must be greater than 10. This ensures that the normal distribution is a good fit for the data.
Using normal approximation makes it easier to calculate probabilities without undertaking complex computations typical of binomial probabilities.

Whenever we approximate a discrete distribution (like binomial) using a continuous one (like normal), remember to apply the continuity correction for accuracy, which leads us to the next topic.
Standard Deviation
Standard deviation (σσ) measures the spread or dispersion of a set of values. In a binomial distribution, standard deviation helps determine how much variation exists from the mean. It is crucial in predicting how often actual results will deviate from estimated ones.

For a binomial distribution with parameters (n,p), standard deviation is calculated as:
\[σ = \sqrt{n × p × (1 - p)}\]
This formula considers both the probability of success (p)and failure (1-p)at each trial, providing insight into the natural variability expected from random sample selection.

A larger standard deviation means the values are spread out over a wider range, whereas a smaller value indicates they are clustered closely around the mean. Knowing the standard deviation helps in deciding if specific outcomes are standard or indicate a potential survey error.
Survey Sampling
Survey sampling involves selecting a subset of individuals from a population to infer insights about the entire population. It's essential to minimize bias and error to ensure that survey results accurately represent the group being studied.

Key purposes of survey sampling include:
  • Understanding population characteristics economically and timely.
  • Evaluating survey methods against known statistics to check for biases.
  • Using random samples to reduce researcher bias and ensure sample diversity.
In survey sampling, undercoverage and nonresponse can introduce errors, so comparing survey data with known population facts is a good practice. This comparison helps indicate errors due to missing data or incorrect responses, allowing survey designers to refine their methodology.
Continuity Correction
Continuity correction is necessary when using a continuous distribution (like the normal distribution) to approximate a discrete one (such as the binomial distribution). It improves the accuracy of the probability estimates.

Here's why you need it:
  • Discrete variables assume specific values without intervals, while continuous variables assume any value within a range.
  • By adding or subtracting 0.5 (called continuity correction) to the discrete number, we adjust it for using the continuous normal distribution.
For example, when estimating the probability that a binomial variable is between two numbers, like in the exercise where first-generation Canadians should vary between 340 and 390 in sample size, you would actually look between 339.5 and 390.5 with the continuity correction. This correction fine-tunes the normal approximation and ensures more aligned probability outcomes.

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Most popular questions from this chapter

Preference for the Middle? When choosing an item from a group, researchers have shown that an important factor influencing choice is the item's location. This occurs in varied situations, such as shelf positions when shopping, when filling out a questionnaire, and even when choosing a preferred candidate during a presidential debate. Experimenters displayed five identical pairs of white socks by attaching them vertically to a blue background, which was then mounted on an easel for viewing. One hundred participants from the University of Chester were used as subjects and asked to choose their preferred pair of socks. 11 a. Suppose each subject selects a preferred pair of socks at random. What is the probability that a subject would choose the pair of socks in the center position? Assuming that the subjects make their choices independently, what is the distribution of \(X\), the number of subjects among the 100 who would choose the pair of socks in the center position? b. What is the mean of the number of subjects who would choose the pair of socks in the center position? What is the standard deviation? c. In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in the center. In this experiment, 34 subjects chose the pair of socks in the center. What is the probability that 34 or more subjects would choose the item in the center if each subject were selecting the preferred pair of socks at random? Use the Normal approximation. If your software allows, find the exact binomial probability and compare the two results. d. Do you feel that this experiment supports the "center stage effect"? Explain briefly.

Binomial Setting? A binomial distribution will be approximately correct as a model for one of these two sports settings and not for the other. Explain why by briefly discussing both settings. a. A National Football League kicker has made \(90 \%\) of his field goal attempts in the past. This season he attempts 20 field goals. The attempts differ widely in distance, angle, wind, and so on. b. A National Basketball Association player has made \(90 \%\) of his free-throw attempts in the past. This season he takes 150 free throws. Basketball free throws are always attempted from 15 feet away from the basket, with no interference from other players.

Roulette-Betting on Red. A roulette wheel has 38 slots, numbered 0,00 , and 1 to 36 . The slots 0 and 00 are colored green, 18 of the others are red, and 18 are black. The dealer spins the wheel and at the same time rolls a small ball along the wheel in the opposite direction. The wheel is carefully balanced so that the ball is equally likely to land in any slot when the wheel slows. Gamblers can bet on various combinations of numbers and colors. a. If you bet on "red," you win if the ball lands in a red slot. What is the probability of winning with a bet on red in a single play of roulette? b. You decide to play roulette four times, each time betting on red. What is the distribution of \(X\), the number of times you win? c. If you bet the same amount on each play and win on exactly two of the four plays, then you will "break even." What is the probability that you will break even? d. If you win on fewer than two of the four plays, then you will lose money. What is the probability that you will lose money?

Using Benford's Law. According to Benford's law (Example 12.7, page 281) the probability that the first digit of the amount of a randomly chosen invoice is a 1 or a 2 is \(0.477 .\) You examine 90 invoices from a vendor and find that 29 have first digits 1 or 2 . If Benford's law holds, the count of 1 s and \(2 \mathrm{~s}\) will have the binomial distribution with \(n=90\) and \(p=0.477\). Too few 1 s and 2 s suggests fraud. What is the approximate probability of 29 or fewer 1 s and \(2 s\) if the invoices follow Benford's law? Do you suspect that the invoice amounts are not genuine?

Multiple-Choice Tests. Here is a simple probability model for multiple-choice tests. Suppose each student has probability \(p\) of correctly answering a question chosen at random from a universe of possible questions. (A strong student has a higher \(p\) than a weak student.) Answers to different questions are independent. a. Stacey is a good student for whom \(p=0.75\). Use the Normal approximation to find the probability that Stacey scores between \(70 \%\) and \(80 \%\) on a 100 -question test. b. If the test contains 250 questions, what is the probability that Stacey will score between \(70 \%\) and \(80 \%\) ? You see that Stacey's score on the longer test is more likely to be close to her "true score."

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