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91Ó°ÊÓ

We see in the AllCountries dataset that the percent of the population that is over 65 is 13.4 in Australia and 12.5 in New Zealand. Suppose we take random samples of size 500 from Australia and size 300 from New Zealand, and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{N Z},\) where \(\hat{p}_{A}\) represents the sample proportion of elderly in Australia and \(\hat{p}_{N Z}\) represents the sample proportion of elderly in New Zealand. Find the mean and standard deviation of the differences in sample proportions.

Short Answer

Expert verified
The mean of the difference in sample proportions is 0.009 and the standard deviation can be found by evaluating the expression \(\sqrt{\frac{{0.134(1-0.134)}}{500} + \frac{{0.125(1-0.125)}}{300}}\)

Step by step solution

01

Calculate the Population Proportions

We are given the population proportions from the problem, which are the percentages of the population over 65 years in Australia (\(p_A\)) and New Zealand (\(p_{NZ}\)), 13.4% and 12.5% respectively.
02

Compute the Mean of the Differences in Sample Proportions

The mean (expected value) of the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{NZ}\) is given by the difference in population proportions \(p_{A} - p_{NZ}\). Hence, the mean (expected value) = \(0.134 - 0.125 = 0.009\).
03

Calculate the Standard Deviation of the Differences in Sample Proportions

The standard deviation of the difference in sample proportions is calculated using the formula \(\sqrt{\frac{{p_A(1-p_A)}}{n_A} + \frac{{p_{NZ}(1-p_{NZ})}}{n_{NZ}}} \), where \(n_A\) and \(n_{NZ}\) are the sample sizes from Australia and New Zealand i.e., 500 and 300 respectively. Substituting the given values, the standard deviation = \(\sqrt{\frac{{0.134(1-0.134)}}{500} + \frac{{0.125(1-0.125)}}{300}} \). Calculate the values under the square root and then square root the sum to get the standard deviation.

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Key Concepts

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

Population Proportions
Population proportions are an important concept in statistics when collecting and analyzing data from different groups or categories. It refers to the percentage of individuals in a population that exhibits a particular characteristic. In our original exercise, the population proportion in Australia for those over the age of 65 is 13.4%, and in New Zealand, it is 12.5%. To put these into perspective, if Australia has a population of, say, a million, approximately 134,000 individuals would be over 65. Understanding population proportions helps in calculating other statistical measurements, such as the sample mean and the standard deviation. Importantly, these values allow statisticians to describe how widespread or common a particular trait is within an entire population. When assessing two different populations, like the elderly populations in Australia and New Zealand, differences in population proportions can offer insights into demographic trends.
Standard Deviation
The standard deviation is a measure that indicates the amount of variation or dispersion in a set of values. In simple terms, it tells us how much the values in a dataset deviate from the mean (average) value. It is incredibly useful in statistics because it provides context to the data distribution, indicating whether the data points are generally close to the mean or spread out over a wide range. In the context of our exercise, the standard deviation of the difference in sample proportions is calculated to understand the variability between different samples from Australia and New Zealand. The formula used involves the sample sizes and the population proportions. This standard deviation is particularly informative as it reflects the expected natural fluctuation in the differences of sample proportions due to random sampling. Finding the standard deviation involves substituting the known values into the formula \[ \sqrt{\frac{{p_A(1-p_A)}}{n_A} + \frac{{p_{NZ}(1-p_{NZ})}}{n_{NZ}}} \]. This tells us how much the sample proportions might differ, giving a clearer picture of comparative variability.
Sample Proportions
Sample proportions serve as a statistical representation of certain characteristics within a sample, making them vital for inferential statistics. When statisticians draw conclusions about a larger population based on a smaller group, they rely heavily on the concept of sample proportions.In our given problem, the sample proportions represent the percentage of elderly people within the samples taken from Australia and New Zealand. For instance, if 70 out of 500 individuals in the Australian sample are over 65, the sample proportion (\( \hat{p}_{A} \)) would be 0.14, indicating that 14% of the sample is elderly.By comparing sample proportions from different places or times, we can estimate the difference across populations, even when dealing with relatively small sample sizes. This estimation forms the basis for hypothesis testing and making inferences about population characteristics. The difference in sample proportions itself is a focal point for analysis, especially when evaluating demographic trends across regions like Australia and New Zealand.

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Most popular questions from this chapter

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Is a Normal Distribution Appropriate? In Exercises 6.13 and \(6.14,\) indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, does the Central Limit Theorem apply? (a) \(n=500\) and \(p=0.1\) (b) \(n=25\) and \(p=0.5\) (c) \(n=30\) and \(p=0.2\) (d) \(n=100\) and \(p=0.92\)

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 50 from a population with proportion 0.25

Standard Error from a Formula and Simulation In Exercises 6.15 to \(6.18,\) find the mean and standard error of the sample proportions two ways: (a) Use StatKey or other technology to simulate at least 1000 sample proportions. Give the mean and standard error and comment on whether the distribution appears to be normal. (b) Use the formulas in the Central Limit Theorem to compute the mean and standard error. Are the results similar to those found in part (a)? Sample proportions of sample size \(n=10\) from a population with \(p=0.2\)

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