Question

Construct a \(95 \%\) confidence interval for the difference in the population means. Then find a point estimate for the difference in the population means and calculate the margin of error: Compare your results. Can you conclude that there is a difference in the two population means? $$\begin{aligned}&\text { } n_{1}=n_{2}=50, \bar{x}_{1}=125.2, \bar{x}_{2}=123.7, s_{1}=5.6\\\ &s_{2}=6.8\end{aligned}$$

Short answer

And, can we conclude that there is a significant difference in the two population means? Answer: The point estimate for the difference in population means is 1.5, the margin of error is 1.55, and the 95% confidence interval for the difference in population means is (-0.05, 3.05). Since the confidence interval includes 0, we cannot conclude with 95% confidence that there is a significant difference in the two population means.

Step-by-step solution

The point estimate for the difference in population means can be found by taking the difference of the sample means, i.e., \(\bar{x}_{1} - \bar{x}_{2}\). Given that \(\bar{x}_{1} = 125.2\) and \(\bar{x}_{2} = 123.7\), we get: $$\bar{x}_{1} - \bar{x}_{2} = 125.2 - 123.7 = 1.5$$ #Step 2: Calculate the standard error of the difference in means#

The standard error of the difference in means can be found using the following formula: $$SE = \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}$$ Substitute the given values of \(s_{1}=5.6, s_{2}=6.8, n_{1}=50\) and \(n_{2}=50\) into the formula and calculate the standard error: $$SE = \sqrt{\frac{5.6^2}{50} + \frac{6.8^2}{50}} = \sqrt{0.6272} = 0.792$$ #Step 3: Calculate the critical value#

As we need a \(95\%\) confidence interval, the corresponding \(z\)-score is \(1.96\). This is because \(z = 1.96\) covers \(95\%\) of the area under the standard normal curve. #Step 4: Calculate the margin of error#

The margin of error can be calculated using the critical value and the standard error: $$\text{Margin of Error} = z \times SE = 1.96 \times 0.792 = 1.55$$ #Step 5: Construct the Confidence Interval #

Now, we will construct the confidence interval using the point estimate, and the margin of error: $$(\bar{x}_{1} - \bar{x}_{2}) \pm \text{Margin of Error} = 1.5 \pm 1.55 = (-0.05, 3.05)$$ #Step 6: Interpret the results#

The \(95\%\) confidence interval for the difference in population means is \((-0.05, 3.05)\). Given that this interval includes \(0\), we cannot conclude with \(95\%\) confidence that there is a significant difference in the two population means, as the true difference could be zero.

Key concepts

Population Means

In statistical terms, the population mean is the average of all individuals within a population. When conducting studies or surveys, we often want to infer the mean value of a characteristic for an entire population based on sample data.

To obtain estimates for two different groups or populations, we collect data samples and calculate the sample means. In the provided problem,

• Two groups are considered with sample means, denoted as \( \bar{x}_{1} = 125.2 \) and \( \bar{x}_{2} = 123.7 \).
• The primary goal is to compare these sample means to see how the population means might differ.

Understanding the population mean's concept is crucial in determining if any observed difference is due to random chance or a true underlying difference in means.

Standard Error

The standard error is a measure that indicates the amount of variability or dispersion of the sample mean estimates from the true population mean. It plays a critical role in constructing confidence intervals, as it provides insight into the precision of the sample mean as an estimate.

In our problem, the standard error for the difference in means is calculated using the formula:\[SE = \sqrt{\frac{s_{1}^2}{n_{1}} + \frac{s_{2}^2}{n_{2}}}\]Substituting the values provided:

• \( s_{1} = 5.6 \), \( s_{2} = 6.8 \)
• \( n_{1} = 50 \), \( n_{2} = 50 \)

The standard error is calculated to be \( 0.792 \).

This value represents how much we expect the difference in sample means to vary from the true difference in population means. Smaller standard errors indicate more precise estimates.

Margin of Error

The margin of error provides a range around the point estimate that is likely to contain the true population parameter. It reflects the uncertainty of the estimate due to sampling variability and is derived from the standard error and a critical value from a statistical distribution, typically the normal distribution for large samples.

In this exercise, the margin of error is calculated using the formula:\[\text{Margin of Error} = z \times SE\]With a critical value \( z = 1.96 \) for a \(95\%\) confidence interval and standard error \( SE = 0.792 \), we find:

• Margin of Error = \( 1.96 \times 0.792 = 1.55 \)

This margin tells us that our point estimate is likely to be within \( 1.55 \) units of the true population difference, with \(95\%\) confidence. Understanding and calculating the margin of error is essential for accurate hypothesis testing and estimation.

Point Estimate

A point estimate provides a single value that serves as a plausible guess or approximation for an unknown population parameter.

In the context of comparing population means, the point estimate for the difference is computed from the difference between the sample means.

• Here, we have \( \bar{x}_{1} - \bar{x}_{2} = 125.2 - 123.7 = 1.5 \).

This value indicates our best estimate of the difference in the means of the two populations based on our samples.

While a point estimate provides crucial information, it does not reflect uncertainty. Therefore, it is important to pair it with a confidence interval to understand the estimation precision. This combination of point estimate and confidence interval gives a fuller picture of what the true population difference might be.