91Ó°ÊÓ

The article "Unmarried Couples More Likely to Be Interracial" ( \mathrm{\\{} S a n ~ Luis Obispo Tribune, March 13,2002 ) reported that \(7 \%\) of married couples in the United States are mixed racially or ethnically. Consider the population consisting of all married couples in the United States. a. A random sample of \(n=100\) couples will be selected from this population and \(p\), the proportion of couples that are mixed racially or ethnically, will be computed. What are the mean and standard deviation of the sampling distribution of \(p\) ? b. Is it reasonable to assume that the sampling distribution of \(p\) is approximately normal for random samples of size \(n=100\) ? Explain. c. Suppose that the sample size is \(n=200\) rather than \(n=100\), as in Part (b). Does the change in sample size change the mean and standard deviation of the sampling distribution of \(p ?\) If so, what are the new values for the mean and standard deviation? If not, explain why not. d. Is it reasonable to assume that the sampling distribution of \(p\) is approximately normal for random samples of size \(n=200\) ? Explain. e. When \(n=200\), what is the probability that the proportion

Step by step solution

01

Calculation Mean and Standard Deviation for n=100

The mean of the sampling distribution of \( p \) is equal to the population proportion \( P = 0.07 \). The standard deviation can be calculated by the formula: \( \sqrt{(P \cdot (1-P))/n} \). Substituting \( P = 0.07 \) and \( n = 100 \), the standard deviation comes out to be \( \sqrt{(0.07 \cdot (1-0.07))/100} = 0.025 \)

02

Determining Normality for n=100

The rule of thumb for determining if the distribution is approximately normal is if both \( n \cdot P \geq 10 \) and \( n \cdot (1-P) \geq 10 \) hold true. Substituting \( n = 100 \) and \( P = 0.07 \), we can see that \( n \cdot P = 7 \) and \( n \cdot (1-P) = 93 \). Since \( n \cdot P < 10 \), the sampling distribution of \( p \) for \( n = 100 \) is not approximately normal.

03

Calculation Mean and Standard Deviation for n=200

The mean stays the same \( P = 0.07 \) as it is population proportion. For the standard deviation formula, \( \sqrt{(P \cdot (1-P))/n} \), we substitute \( n = 200 \) getting \( \sqrt{(0.07 \cdot (1-0.07))/200} = 0.018 \).

04

Determining Normality for n=200

Same rules, substituting \( n = 200 \),we get \( n \cdot P = 14 \) and \( n \cdot (1 - P) = 186 \), this time it's \(\geq 10\) for both, thus the distribution can be considered approximately normal.

05

Calculate Probability for n=200

The question doesn't specify any value for the proportion \( p \) when \( n = 200 \), so it's impossible to calculate the exact probability without this information.

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Key Concepts

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

Statistical Inference

Statistical inference involves drawing conclusions about a population based on information gathered from a sample. It's a fundamental part of statistics that allows researchers to make claims or predictions about a larger group using a smaller subset. For instance, the proportion (\( P \text{ or } p \) ) of interracial married couples in the United States is used to understand the likelihood of sampling a mixed couple.
Statistical inference often relies on probability distributions of samples, such as the sampling distribution of a sample proportion. Sampling distribution becomes the bridge between the data collected from a sample and the inferences made about the entire population.
In our exercise, we calculate the mean and standard deviation of the sampling distribution for a given number of sampled couples (<200>) to infer the population's characteristics. This calculation aids in understanding the variability we might expect by chance when sampling from the population.

Normality Assumption

The normality assumption is the presumption that a given distribution approximates the normal (also known as Gaussian) distribution. This approximation unlocks many powerful statistical tools and simplifies calculations. However, not all distributions meet this assumption.
A key condition to determine if it's reasonable to assume normality for binomial distributions, which include proportions, is that both the successes (<20>nP20>) and failures (<20>n(1-P)20>) in the sample should be at least 10. In our exercise, when <20>n = 10020>, the product of <20>nP20> didn't meet this criterion. However, it did when <20>n = 20020>, leading us to believe the sampling distribution approximates normality.
Normality enablessimpler calculations for probabilities and is crucial for producing reliable confidence intervals and hypothesis tests, which are both cornerstones of statistical inference.

Standard Deviation of Proportion

The standard deviation of a proportion measures the variability of proportions that one would observe in many random samples from the same population. It quantifies the extent to which the proportion in any given sample might differ from the population proportion (<20>P20>).
In the exercise, when the sample size increased from <20>n = 10020> to <20>n = 20020>, the calculated standard deviation decreased from <20>0.02520> to <20>0.01820>. This decrease reflects a fundamental concept: larger samples lead to a reduction in variability and a