Question

random samples were selected from two binomial populations, with sample sizes and the number of successes given in Exercises \(11-12 .\) Construct a \(98 \%\) lower confidence bound for the difference in the population proportions. $$\begin{array}{lcc}\hline & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 500 & 500 \\\\\text { Number of Successes } & 120 & 147 \\\\\hline\end{array}$$

Short answer

Answer: -0.12856

Step-by-step solution

Identify the given information

We are given the following information: - Sample size for Population 1 (n1) = 500 - Number of successes for Population 1 (x1) = 120 - Sample size for Population 2 (n2) = 500 - Number of successes for Population 2 (x2) = 147 - Confidence level = 98%

Calculate the sample proportions and their difference

First, we will calculate the sample proportions (p-hat) for each population and their difference: For Population 1: p-hat1 = x1 / n1 = 120 / 500 = 0.24 For Population 2: p-hat2 = x2 / n2 = 147 / 500= 0.294 Difference in sample proportions: D = p-hat1 - p-hat2 = 0.24 - 0.294 = -0.054

Determine the critical value for the given confidence level

Using a standard normal Z table or calculator, we find the critical value (Z) that corresponds to a 98% lower confidence level: Since we want the lower bound, we will be looking for Z value at 1% level (100%-98%=1%): Z = -2.33

Calculate the standard error

Next, we need to calculate the standard error (SE) for the difference in the sample proportions: SE = sqrt( (p-hat1 * (1-p-hat1) / n1) + (p-hat2 * (1-p-hat2) / n2) ) SE = sqrt( (0.24 * 0.76 / 500) + (0.294 * 0.706 / 500) ) = 0.032

Calculate the margin of error and lower confidence bound

Now we can calculate the margin of error (ME) and the lower confidence bound (LCB) for the difference in the population proportions: Margin of error (ME) = Z * SE = -2.33 * 0.032 = -0.07456 Lower confidence bound (LCB) = D + ME = -0.054 + (-0.07456) = -0.12856

Interpret the result

Based on the given information and calculations, the 98% lower confidence bound for the difference in the population proportions is -0.12856. This means we can be 98% confident that the true difference in population proportions between Population 1 and Population 2 is at least -0.12856.

Key concepts

Binomial Population

In the context of statistics, a **binomial population** is a population where each trial results in a success or a failure. This type of population distribution is characterized by two parameters: the number of trials (n) and the probability of success (p) on each trial.
Binomial distributions are key when we're dealing with experiments or processes that result in binary outcomes, such as yes/no, success/failure, or true/false situations.
In the provided exercise, each population represents a binomial population. We have two sets of trials with specific outcomes, measured as the number of successes.

• Population 1: 500 trials and 120 successes
• Population 2: 500 trials and 147 successes

Understanding binomial populations helps in determining probabilities and testing hypotheses about population proportions, which is crucial in statistical analysis.

Sample Proportions

**Sample proportions** are estimates of the true population proportions, obtained from sample data. In this exercise, sample proportions help to summarize the data from each binomial population.
To calculate the sample proportion for Population 1, divide the number of successes by the sample size: \[ \hat{p}_1 = \frac{120}{500} = 0.24 \] Similarly, for Population 2: \[ \hat{p}_2 = \frac{147}{500} = 0.294 \] These proportions provide a simplified representation of the data, serving as a base for further statistical analysis, like confidence intervals. They are vital in comparing the traits between two populations.
The difference in these sample proportions (\( D = \hat{p}_1 - \hat{p}_2 = -0.054 \)) offers insight into variations in success rates between the two populations.

Standard Error

The **standard error (SE)** measures the variability of a sample statistic—here, the difference in sample proportions. It tells us how much these sample-based estimates might differ from the true population parameters.
In our exercise, the SE of the difference in sample proportions is calculated by:\[ SE = \sqrt{ \left( \frac{\hat{p}_1(1-\hat{p}_1)}{n_1} \right) + \left( \frac{\hat{p}_2(1-\hat{p}_2)}{n_2} \right) } \]Plugging in the sample proportions:\[ SE = \sqrt{ \left( \frac{0.24 \times 0.76}{500} \right) + \left( \frac{0.294 \times 0.706}{500} \right) } = 0.032 \]This value indicates how much the observed difference in sample proportions (\(-0.054\)) can be expected to vary from the true difference in population proportions due to random sampling variability.

Critical Value

The **critical value** in a statistical test defines the boundary for deciding if a test statistic indicates a result that is both significant and associated with a low probability under the null hypothesis. It is derived from a specified confidence level, which provides the test's assurance of correctness.
For our exercise, we want a 98% confidence interval. This leaves us with 1% on each tail of the distribution since we're examining a lower confidence bound.
Consulting a Z-table (standard normal distribution), the critical value corresponding to 0.01 (1% level) is:\[ Z = -2.33 \]This negative critical value is used because we're focused on the lower tail of the distribution for our confidence bound. It multiplies with the standard error to calculate the margin of error, impacting the range of the confidence interval.
Combining all these elements allows us to set the confidence interval to assess the true difference in population proportions.