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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 \(22 \%\) of the Canadian population is first generation, that is, they were bom outside Canada. \({ }^{6}\) The number \(X\) of first-generation Canadians in random samples of 1500 persons should therefore vary with the binomial ( \(n=1500, p=0.22\) ) distribution. (a) What are the mean and standard deviation of \(X\) ? (b) Use the Nomal approximation to find the probability that the sample will contain between 340 and 390 first-generation Canadians. Be sure to check that you can safely use the approximation.

Short Answer

Expert verified
Mean: 330, Standard Deviation: ~16.22, Probability: ~0.2775

Step by step solution

01

Calculate the Mean of X

To calculate the mean of a binomial distribution, we use the formula \( \mu = np \). Here, \( n = 1500 \) and \( p = 0.22 \). Thus, \( \mu = 1500 \times 0.22 = 330 \).
02

Calculate the Standard Deviation of X

The standard deviation of a binomial distribution is calculated using the formula \( \sigma = \sqrt{np(1-p)} \). So, \( \sigma = \sqrt{1500 \times 0.22 \times (1 - 0.22)} = \sqrt{1500 \times 0.22 \times 0.78} \approx 16.22 \).
03

Check Conditions for Normal Approximation

The normal approximation can be used if both \( np \geq 10 \) and \( n(1-p) \geq 10 \). For this problem, \( np = 330 \geq 10 \) and \( n(1-p) = 1170 \geq 10 \). Both conditions are satisfied, so we can use the normal approximation.
04

Use Continuity Correction for Normal Approximation

When using a normal approximation for a binomial distribution, apply continuity correction. The probability that the sample contains between 340 and 390 first-generation Canadians is represented as \( P(340 < X < 390) = P(339.5 < X < 390.5) \).
05

Convert to Standard Normal Variable

Convert \( X \) to the standard normal variable \( Z \) using \( Z = \frac{X - \mu}{\sigma} \). For \( X = 339.5 \), \( Z = \frac{339.5 - 330}{16.22} \approx 0.586 \). For \( X = 390.5 \), \( Z = \frac{390.5 - 330}{16.22} \approx 3.73 \).
06

Find Probability Using Z Table

Look up the probabilities for \( Z = 0.586 \) and \( Z = 3.73 \) using the Z-table or a calculator. \( P(Z < 0.586) \approx 0.7224 \) and \( P(Z < 3.73) \approx 0.9999 \). The probability of \( 339.5 < X < 390.5 \) is \( 0.9999 - 0.7224 = 0.2775 \).

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Key Concepts

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

Binomial Distribution
A binomial distribution is a common model used to predict the outcomes of experiments where there are only two possible outcomes - like success and failure. In this exercise, we are looking at the number of first-generation Canadians, which can be considered successes where the two outcomes are: First-generation Canadian or not.
The key parameters of a binomial distribution are:
  • \( n \): the number of trials or sample size
  • \( p \): the probability of success on a single trial
In our problem, each of the 1500 people is a trial (\( n = 1500 \)), and the success probability (being a first-generation Canadian) is \( 0.22 \). This makes our distribution \( B(n = 1500, p = 0.22) \).
This model helps us determine how likely any given number of successes (first-generation Canadians in this case) is based on the known success rate within the Canadian population.
Normal Approximation
Normal approximation is a technique used to approximate the binomial distribution with a normal distribution. It simplifies calculations by using a continuous distribution, suited for cases when the sample size is large.
Key points to use normal approximation:
  • The sample size \( n \) should be large enough.
  • Both \( np \geq 10 \) and \( n(1-p) \geq 10 \) must be satisfied.
In our exercise, \( np = 330 \) and \( n(1-p) = 1170 \) fulfill these conditions, so the normal approximation can safely be applied. This shift allows us to use techniques for continuous data to understand our binomial data, simplifying complex computations.
Survey Errors
Survey errors are variations that can cause significant discrepancies in survey-based results. They arise from different sources such as undercoverage – some members of the population being inadequately represented, and nonresponse – when not all individuals participate.
Checking for these errors often involves comparing the survey sample with known truths about the population. For instance, if a survey exaggerated the count of first-generation Canadians compared to the actual 22%, it indicates a survey error. Mitigating these errors involves random sampling and using stratification to ensure all segments of the population are well-represented.
Standard Deviation
Standard deviation represents the amount of variation or dispersion in a set of values. In the context of a binomial distribution, it tells us how much the number of successes (e.g., first-generation Canadians) might differ from the mean (expected number of successes).
The formula used is:\[\sigma = \sqrt{np(1-p)}\]where \( np \) is the mean, and \( 1-p \) is the probability of failure in a single trial.
For this exercise, the standard deviation \( \approx 16.22 \) indicates the expected variability around the mean number of 330 first-generation Canadians in the sample. Understanding standard deviation is crucial for gauging how much survey results might deviate due to chance.

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Most popular questions from this chapter

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