Question

A study was conducted to compare the mean numbers of police emergency calls per 8 -hour shift in two districts of a large city. Samples of 100 8-hour shifts were randomly selected from the police records for each of the two regions, and the number of emergency calls was recorded for each shift. The sample statistics are listed here: $$\begin{array}{lcc} & \multicolumn{2}{c} {\text { Region }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 100 & 100 \\\\\text { Sample Mean } & 2.4 & 3.1 \\\\\text { Sample Variance } & 1.44 & 2.64\end{array}$$ Find a \(90 \%\) confidence interval for the difference in the mean numbers of police emergency calls per shift between the two districts of the city. Interpret the interval.

Short answer

Answer: The 90% confidence interval for the difference in the mean numbers of police emergency calls per shift between the two districts is (0.442, 0.958).

Step-by-step solution

Identify the given information

We are given the following information: - Sample size, \(n_1 = 100\) for Region 1 and \(n_2 = 100\) for Region 2 - Sample mean, \(\bar{x}_1 = 2.4\) for Region 1 and \(\bar{x}_2 = 3.1\) for Region 2 - Sample variance, \(s^2_1 = 1.44\) for Region 1 and \(s^2_2 = 2.64\) for Region 2 - Confidence level: 90%

Calculate the standard error

To calculate the standard error, we will use the following formula: $$SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=\sqrt{\frac{1.44}{100}+\frac{2.64}{100}}$$ SE = 0.157

Find the appropriate t-score for 90% confidence level

Since we have a 90% confidence interval, the significance level (\(\alpha\)) is \(1-0.90=0.10\). We will find the t-score using the following formula: $$t_{\frac{\alpha}{2}}=t_{0.05}$$ Since the sample sizes are large (both \(n_1\) and \(n_2\) are greater than 30), we can use the z-score instead of t-score. Using z-table, we find: $$z_{0.05}=1.645$$

Calculate the margin of error

Now, we will calculate the margin of error using the formula: $$E=z_{0.05}\times SE=1.645\times 0.157$$ E = 0.258

Calculate the lower and upper boundaries of the confidence interval

Now, we will calculate the confidence interval using the formula: $$CI=\left(\bar{x}_2-\bar{x}_1-E,\bar{x}_2-\bar{x}_1+E\right)=\left(3.1-2.4-0.258,3.1-2.4+0.258\right)$$ CI = (0.442 , 0.958)

Interpret the interval

The 90% confidence interval for the difference in the mean numbers of police emergency calls per shift between the two districts is (0.442, 0.958). This means that we are 90% confident that the difference in the mean number of emergency calls per shift is between 0.442 and 0.958 calls with Region 2 having more emergency calls than Region 1.

Key concepts

Sample Mean

The sample mean is a simple yet crucial concept in statistics. It represents the average of the sample data points collected from a population. In this exercise, the sample mean is used to compare the frequency of emergency calls in two districts over several shifts.

For instance, Region 1 has a sample mean of \(\bar{x}_1 = 2.4\) and Region 2 has a sample mean of \(\bar{x}_2 = 3.1\). These values indicate that, on average, each 8-hour shift in Region 1 receives 2.4 calls, while Region 2 receives 3.1 calls.

Sample means are calculated by summing all data points in a sample and dividing by the total number of points. It provides a quick summary of what is typical for the dataset.

Sample Variance

Sample variance measures the spread or variability of the sample data around the sample mean. It's a key statistic for determining how much the data points differ from the average.

In our example, the sample variance for Region 1 is \(s^2_1 = 1.44\) and for Region 2 it's \(s^2_2 = 2.64\). A higher variance means more spread out data. So, Region 2's emergency call rates vary more widely than those in Region 1.

The formula for calculating sample variance is: \[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} \]This indicates that each data point’s deviation from the mean is squared, summed up, and then divided by the number of data points minus one.

Standard Error

The standard error is an important statistic that indicates the accuracy with which a sample distribution represents a population. It informs us of the extent to which the sample mean estimates the population mean.

For the given dataset, the standard error (SE) was calculated using:\[ SE = \sqrt{ \frac{s^2_1}{n_1} + \frac{s^2_2}{n_2} } = \sqrt{ \frac{1.44}{100} + \frac{2.64}{100} } = 0.157 \]This calculation demonstrates how spread out the mean of our sample might be from the actual mean of the entire population.

The smaller the standard error, the closer the sample mean is likely to be to the population mean, increasing the reliability of our estimates.

Z-score

The Z-score is a statistical measure that describes a value's relation to the mean of a group of values, expressed in terms of standard deviations away from the mean.

Z-scores play a vital role when constructing confidence intervals for hypothesis testing. Given the large sample sizes in the exercise (over 30 for both regions), the Z-score is preferred over the t-score.

For a 90% confidence interval, the significance level is \(\alpha = 0.10\). The corresponding Z-score is found in the Z-table: \(z_{0.05} = 1.645\), because the level is divided by two for two tails of distribution.

This score helps us determine the margin of error and, subsequently, the confidence interval, illustrating where the population mean likely falls in relation to our sample mean.