Question

Independent random samples of \(n_{1}=50\) and \(n_{2}=60\) observations were selected from populations 1 and \(2,\) respectively. The sample sizes and computed sample statistics are given in the table: $$\begin{array}{lcc} & \multicolumn{2}{c} {\text { Population }} \\\\\cline { 2 - 3 } & 1 & 2 \\\\\hline \text { Sample Size } & 5 & 60 \\\\\text { Sample Mean } & 100.4 & 96.2 \\\\\text { Sample Standard Deviation } & 0.8 & 1.3\end{array}$$ Find a \(90 \%\) confidence interval for the difference in population means and interpret the interval.

Short answer

Question: Given two independent random samples from different populations with the following information: \(n_1 = 50, \bar{x}_1 = 100.4, s_1 = 0.8\) and \(n_2 = 60, \bar{x}_2 = 96.2, s_2 = 1.3\), find the 90% confidence interval for the difference in population means and interpret the result. Answer: The 90% confidence interval for the difference in population means is approximately (3.8669, 4.5331), suggesting that we are 90% confident that the true difference between the population means lies within this range. In other words, the mean of population 1 is likely between 3.87 and 4.53 units higher than the mean of population 2.

Step-by-step solution

1. Identify the given information

We are given the following information: - \(n_1=50,\) \(\bar{x}_1=100.4,\) \(s_1=0.8\) (Population 1) - \(n_2=60,\) \(\bar{x}_2=96.2,\) \(s_2=1.3\) (Population 2) - Desired confidence level is \(90\%\)

2. Calculate the pooled standard error

Now we will calculate the pooled standard error, which is the standard deviation of the difference in sample means. It is calculated as follows: \( SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\) Substituting the values, we get: $SE = \sqrt{\frac{0.8^2}{50} + \frac{1.3^2}{60}} = \sqrt{\frac{0.64}{50}+\frac{1.69}{60}} = \sqrt{0.0128+0.0282} = \sqrt{0.041} \approx 0.2025$

3. Find the critical z-value

Next, we will find the critical z-value for a 90% confidence level. Since we have a two-tailed test, we must divide the remaining \(10\%\) (or 0.10) in half and distribute it equally on both sides of the distribution curve. Thus, we look for \(1-\frac{0.10}{2}=0.95\) in the standard normal z-table to find our critical value, which is approximately \(z_{0.95} = 1.645\).

4. Calculate the confidence interval

Now, we will calculate the confidence interval. To do this, we will utilize the formula: \(\text{Confidence Interval} = (\bar{x}_1 - \bar{x}_2) \pm z_{0.95} \cdot SE\) Substituting the values, we get: $\text{Confidence Interval} = (100.4 - 96.2) \pm 1.645 \cdot 0.2025 = 4.2 \pm 1.645 \cdot 0.2025 = 4.2 \pm 0.3331$ The 90% confidence interval for the difference in population means is approximately: \( (4.2 - 0.3331, 4.2 + 0.3331) = (3.8669, 4.5331)\)

5. Interpret the interval

The 90% confidence interval for the difference in population means (\((3.8669, 4.5331)\)) suggests that we are 90% confident that the true difference between the population means lies within this range. In other words, the mean of population 1 is likely between 3.87 and 4.53 units higher than the mean of population 2.

Key concepts

Independent Random Samples

Understanding the concept of independent random samples is crucial when dealing with statistical analysis. Imagine flipping two different coins; the outcome of one flip does not affect the outcome of the other. This is the essence of independence in statistical terms.

When two samples are independent, the observations in one sample do not influence or relate to the observations in the other. This is vital when trying to determine the difference in means between two populations. If we flip each coin 50 and 60 times, as in our textbook exercise, we're collecting two independent samples of sizes n₁=50 and n₂=60 respectively. By confirming the independence of the samples, we can proceed with analyzing the differences in their characteristics, such as means, without worrying about intertwining effects.

Sample Statistics

Sample statistics provide a snapshot of the larger population's characteristics by analyzing a smaller, more manageable subset. For instance, the sample mean (\(\bar{x}\)) and sample standard deviation (s) serve as estimates for the population's true mean (μ) and standard deviation (σ), respectively.

In our numerical example, we have two sample means, \(\bar{x}_1=100.4\) and \(\bar{x}_2=96.2\), indicating the average measurements of each sample. The sample standard deviations, \(s_1=0.8\) and \(s_2=1.3\), express the variability within each of those samples. Being sample statistics, they are essential placeholders until we gather more information to better estimate the population parameters.

Pooled Standard Error

The pooled standard error acts as a weighted average of the standard errors from two independent samples. It merges the variability observed in both samples to give a singular measurement of error when estimating the difference between the two means.

To calculate the pooled standard error, we use the formula \( SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \), effectively combining the individual sample variances and adjusting for their sample sizes. In our case, this culminates in a pooled standard error of approximately 0.2025, a value that will be key when determining the accuracy of our confidence interval for the difference in means.

Critical Z-Value

The critical z-value is a standardized score that specifies the number of standard deviations a data point is from the mean. When constructing confidence intervals, it's the z-value that correlates with the desired level of confidence.

For a 90% confidence level, we find the critical z-value by looking up the z-score that leaves 5% in each tail of the standard normal distribution, totaling the 10% of the probability that lies beyond our confidence interval. Thus, our critical z-value is 1.645. This number will be multiplied by the pooled standard error to determine the margin of error for our interval estimate.

Interpretation of Confidence Interval

Interpreting a confidence interval means understanding what range of values are reasonable estimates for the difference in population means based on our sample data. A 90% confidence interval implies we can say with 90% certainty that the true mean difference falls between our calculated interval limits.

In practical terms, for the given example, a 90% confidence interval ranging from 3.8669 to 4.5331 indicates that, if we were to take many samples and construct intervals in this same manner, 90% of them would contain the actual difference in population means. Therefore, we're relatively confident that the mean of population 1 is higher than that of population 2 by about 3.87 to 4.53 units.