Question

The article "Unmarried Couples More Likely to Be Interracial" (San 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 of couples in the sample who are racially or ethnically mixed will be greater than \(.10\) ?

Short answer

a. Mean = 0.07 and Standard deviation = 0.018;\nb. Yes, it is reasonable because both \(np = 7\) and \(nq = 93\) are greater than 5;\nc. The mean won't change, but the standard deviation will now be 0.012. \nd. Yes, again it's even more reasonable now as both \(np = 14\) and \(nq = 186\) are greater than 5;\ne. First, a z-score= 7.5 is computed then its corresponding probability equals to 0.0001.

Step-by-step solution

Calculate the mean and standard deviation for n = 100

The mean of the sampling distribution of a proportion (\(p\)) is simply the proportion in the population. The standard deviation, on the other hand, uses the formula \(\sqrt{pq/n}\), where \(p = 0.07\) is the proportion of mixed racially or ethnically couples, \(q = 1-p = 0.93\) is the remaining proportion and \(n = 100\) is the sample size. It is essential to apply these values into this formula correctly.

Evaluate the normal approximation for n = 100

To evaluate if the sampling distribution of \(p\) can be approximated to a normal distribution, the rule of thumb suggests that both \(np\) and \(nq\) should be greater than 5. Apply the given values into these calculations and compare with the suggested rule.

Calculate the mean and standard deviation for n = 200

Repeat step 1, but with \(n = 200\). Note how the standard deviation may change but the mean remains the same, this is because the sample size does not affect the proportion in the population, but it does impact how spread out the sample proportions are likely to be around that mean (standard deviation).

Evaluate the normal approximation for n = 200

Repeat step 2 for \(n = 200\). Pay attention to how the larger sample size may affect the approximation to the normal distribution.

Compute the probability for n = 200 and p > 0.10

First, calculate a z-score using the formula \((X-μ)/(σ/√n)\), where \(X = 0.10\) is the value for which we're looking to find the probability, and \(μ\) and \(σ\) are the mean and standard deviation of the distribution. Next, find the area to the right of the z-score on a standard normal table to get the required probability.