91Ó°ÊÓ

Police are tested for their ability to correctly recognize and identify a suspect based on a witness or victim's verbal description of the suspect. Scores on the identification test range from 0 to 100 (perfect score). Three cities in Massachusetts are under study. The mean score for all police in the three cities is requested. However, funding will only permit \(m=150\) police to be tested. There is no preliminary study to estimate sample standard deviation of scores in each city. City A has \(N_{1}=183\) police, City B has \(N_{2}=371\) police, and City \(\mathrm{C}\) has \(N_{3}=255\) police. In most cases \(n_{i}\) is not a whole number so we round to the nearest whole number. Remember our total sample size is \(m=n_{1}+n_{2}+n_{3}\) (a) Use the method of proportional sampling to compute \(n_{1}, n_{2},\) and \(n_{3},\) the size of the random sample to be taken from each of the three cities. Round each \(n_{i}\) to the nearest whole number and make sure \(m=n_{1}+n_{2}+n_{3}\) (b) Suppose you actually conducted the specified number of tests in each city and obtained the following result: \(\overline{x_{1}}=96\) is the mean test score from city \(\mathrm{A}, \overline{x_{2}}=85\) is the mean test score from City \(\mathrm{B},\) and \(\overline{x_{3}}=88\) is the mean test score from City C. Use the weighted average \(\mu \approx \frac{n_{1}}{m} \overline{x_{1}}+\frac{n_{2}}{m} \overline{x_{2}}+\frac{n_{3}}{m} \overline{x_{3}}\) to get your best estimate for the population mean test score of all police in all three cities. Estimates for \(\mu\) are usually better if we have a preliminary study with reasonably accurate estimates for the sample standard deviation \(s_{i}\) of each stratum (see Problems 28 and 29 ). However, in the case where there is no preliminary study we can use proportional sampling and still obtain good results.

Step by step solution

01

Calculate Proportional Sampling for City A

First, we calculate the proportion of total police that are in City A. This is given by \( P_1 = \frac{N_1}{N_1 + N_2 + N_3} = \frac{183}{183 + 371 + 255} \). Calculating this gives \( P_1 = \frac{183}{809} \approx 0.226 \). Therefore, the sample size for City A is \( n_1 = m \times P_1 = 150 \times 0.226 \approx 33.9 \). Rounding 33.9 gives us \( n_1 = 34 \).

02

Calculate Proportional Sampling for City B

Next, we calculate the proportion of total police that are in City B. This is given by \( P_2 = \frac{N_2}{N_1 + N_2 + N_3} = \frac{371}{809} \approx 0.458 \). Therefore, the sample size for City B is \( n_2 = m \times P_2 = 150 \times 0.458 \approx 68.7 \). Rounding 68.7 gives us \( n_2 = 69 \).

03

Calculate Proportional Sampling for City C

For City C, calculate the proportion as \( P_3 = \frac{N_3}{N_1 + N_2 + N_3} = \frac{255}{809} \approx 0.316 \). The sample size for City C is \( n_3 = m \times P_3 = 150 \times 0.316 \approx 47.4 \). Rounding 47.4 gives us \( n_3 = 47 \).

04

Verify Total Sample Size

Add the rounded sample sizes: \( n_1 + n_2 + n_3 = 34 + 69 + 47 = 150 \). This matches the specified total sample size \( m = 150 \), so the calculation is correct.

05

Compute Weighted Average for Population Mean

We calculate the weighted average using \[ \mu \approx \frac{n_1}{m} \overline{x_1} + \frac{n_2}{m} \overline{x_2} + \frac{n_3}{m} \overline{x_3} \]. Substituting the values, \( \mu \approx \frac{34}{150} \times 96 + \frac{69}{150} \times 85 + \frac{47}{150} \times 88 \). Simplifying the calculation gives \( \mu \approx 21.76 + 39.14 + 27.57 = 88.47 \). Therefore, our best estimate for the population mean test score is approximately 88.47.

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Key Concepts

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

Proportional Sampling

Proportional sampling is a method used when you want to select samples from different groups that represent the same proportion as their presence in the overall population. This technique is particularly useful when the population is divided into distinct subgroups. By applying proportional sampling, you aim to ensure that each subgroup is fairly represented in your sample.To calculate the sample size from each subgroup, you first determine the proportion of each group within the entire population. For instance, in our police testing example, the total number of police from all cities combined is 809. City A has 183 police officers, so its proportion is calculated by \[ P_1 = \frac{183}{183 + 371 + 255} = \frac{183}{809} \], which approximates to 0.226. Multiplying this proportion by the total sample size, 150, gives \[ n_1 = 150 \times 0.226 \approx 34 \].You follow a similar process for Cities B and C. This ensures that the sample from each group is proportional to its size in the entire population, making the sample a miniature version of the population.

Weighted Average

The concept of a weighted average is key when the data comes from various subgroups of differing sizes or importance. In essence, it means that some elements of the sample have more influence over the average than others.When estimating a population mean from several subgroups, such as different cities, weighted averages take into account each subgroup's size and mean. The formula for this is:\[ \mu \approx \frac{n_1}{m} \overline{x_1} + \frac{n_2}{m} \overline{x_2} + \frac{n_3}{m} \overline{x_3} \] Here, \( n_1, n_2, n_3 \) are the sample sizes for each city, \( \overline{x_1}, \overline{x_2}, \overline{x_3} \) are the mean scores from each city, and \( m \) is the total sample size.For the example given, the weighted average lets you take into account that City B, with a larger sample size, contributes more to the final estimate than City A. After doing the math, the best estimate of the overall population mean score comes out to approximately 88.47.

Population Mean Estimation

Estimating the population mean is a crucial part of statistical analysis, especially when assessing characteristics across different groups. It gives insight into the overall population without having to survey every individual. In our scenario, despite varying group sizes and no preliminary standard deviation data, proportional sampling and weighted averages provide a robust method for estimating the mean. The steps ensure that the estimate accounts for:

• The relative size of each group, addressing representation bias,
• The performance of each subgroup, making the estimate more reflective of actual conditions.

An accurate mean helps in making informed decisions, as it reflects the aggregate status of the tested characteristic—in this case, the mean test score of police across the three cities. While having standard deviation data enhances estimates, these methods demonstrate that you can still achieve reliable insights in its absence.