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? $$\begin{aligned}&\text { } n_{1}=35, n_{2}=45, \bar{x}_{1}=36.8, \bar{x}_{2}=33.6, s_{1}=4.9\\\ &s_{2}=3.4\end{aligned}$$

Short answer

Answer: We can conclude that there is a difference in the population means since the 95% confidence interval includes positive values. The point estimate of the difference is 3.2, and the margin of error is 2.453. We can be 95% confident that the true difference in population means lies between 0.747 and 5.653.

Step-by-step solution

Calculate the difference in the sample means and the degrees of freedom

First, calculate the difference in sample means: \(\bar{x}_1 - \bar{x}_2\) $$\bar{x}_1 - \bar{x}_2 = 36.8 - 33.6 = 3.2$$ Then, calculate the degrees of freedom (d.f.): $$d.f. = n_1 + n_2 - 2 = 35 + 45 - 2 = 78$$

Find the critical t-value for the 95% confidence level

We are given a confidence level of \(95\%\). To find the critical t-value, we can look it up in the t-table or use a t-distribution calculator. With a d.f.=78 and a two-tailed test, the critical t-value is: $$t_{\alpha/2} = 1.990$$

Calculate the standard error of the difference

We can find the standard error using the formula: $$\text{Standard Error} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$ We are given \(s_1=4.9\), \(s_2=3.4\), \(n_1=35\), and \(n_2=45\). Plugging in the values: $$\text{Standard Error} = \sqrt{\frac{4.9^2}{35} + \frac{3.4^2}{45}} \approx 1.233$$

Calculate the margin of error

To find the margin of error, multiply the critical t-value by the standard error: $$\text{Margin of Error} = t_{\alpha/2} \cdot \text{Standard Error} = 1.990 \times 1.233 \approx 2.453$$

Construct the 95% confidence interval

Now we can construct the confidence interval using our calculated values: $$\text{Confidence Interval} = (\bar{x}_1 - \bar{x}_2) \pm \text{Margin of Error} = 3.2 \pm 2.453$$ This gives us a 95% confidence interval of: $$[0.747, 5.653]$$

Conclude and compare the results

Since the 95% confidence interval includes positive values, we can conclude that there is a difference in the population means. The point estimate of the difference in population means is \(3.2\), and the margin of error is \(2.453\). This means we can be \(95\%\) confident that the true difference in population means lies between \(0.747\) and \(5.653\).

Key concepts

Population Means

In statistics, the population mean refers to the average of a set of characteristics in the entire population. When we talk about the difference in population means, we're interested in comparing two sets of data to see if there's a significant distinction in their average values.
The equation for the population mean is calculated as:

• For each group, sum up all the values.
• Divide this sum by the total number of observations in the group.

In our exercise, we have the sample means \(ar{x}_1 = 36.8\) and \(ar{x}_2 = 33.6\), representing the averages of two different samples from separate populations. By calculating \(ar{x}_1 - ar{x}_2\), we get the point estimate for the difference in population means: \(3.2\). This point estimate reflects the observed difference, providing a basis for further analysis with confidence intervals.

Margin of Error

The Margin of Error (MoE) helps us understand the range within which we expect the true difference in population means to lie, with a given level of confidence. It's crucial because it acknowledges that sample data may hold some uncertainty or variation.
To compute the margin of error:

• Identify the critical t-value, which correlates to your confidence level (e.g., 95%).
• Multiply this value by the standard error of the difference.

In our example, we found the margin of error to be approximately \(2.453\), calculated by multiplying the critical t-value \(1.990\) by the standard error of \(1.233\). This margin, added and subtracted from the point estimate, defines the limits of our confidence interval, thus offering insights into the reliability and accuracy of the estimated average difference.

t-Distribution

The t-distribution is a probability distribution used in statistics that is particularly useful when dealing with small sample sizes or when the population standard deviation is unknown. It is similar to the normal distribution but has thicker tails, allowing it to accommodate the variability that can occur in samples with small sizes or unknown parameters.
When constructing confidence intervals for the difference in population means, the t-distribution helps determine the critical t-value, which is pivotal in calculating the margin of error. This is particularly essential when sample sizes are not large enough to rely on a normal distribution assumption, like in our exercise with sample sizes of 35 and 45.
We used the t-distribution with \(78\) degrees of freedom to find the critical t-value, \(t_{\alpha/2} = 1.990\), which reflects the cut-off value necessary to ensure a 95% confidence level. This value directly impacts the confidence interval by providing a measure of the sample's reliability in estimating the population parameter.

Standard Error

The Standard Error (SE) measures the variability or spread of a sample statistic, such as the sample mean, from the true population statistic. When examining the difference between two means, the standard error provides an understanding of how much the observed difference could vary due to sampling randomness.
The formula for calculating the standard error of the difference between two means is:\[\text{Standard Error} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]where \(s_1\) and \(s_2\) are the standard deviations of the samples, and \(n_1\) and \(n_2\) are the sample sizes.
In our exercise, plugging in the sample standard deviations \(s_1 = 4.9\) and \(s_2 = 3.4\) with sample sizes \(n_1 = 35\) and \(n_2 = 45\) gives us a standard error of approximately \(1.233\). This value helps build the confidence interval and plays a crucial role in determining the margin of error, making it possible to estimate the reliability of our point estimate difference.