Question

Find the \(95 \%\) confidence interval of the difference in the distance that day students travel to school and the distance evening students travel to school. Two random samples of 40 students are taken, and the data are shown. Find the \(95 \%\) confidence interval of the difference in the means. $$ \begin{array}{lccc} & \bar{X} & \sigma & n \\ \hline \text { Day students } & 4.7 & 1.5 & 40 \\ \text { Evening Students } & 6.2 & 1.7 & 40 \end{array} $$

Short answer

The 95% confidence interval of the difference in means is [-2.233, -0.767].

Step-by-step solution

Define Parameters

Identify the sample means, standard deviations, and sample sizes for day and evening students. For day students, \( \bar{X}_1 = 4.7 \), \( \sigma_1 = 1.5 \), and \( n_1 = 40 \). For evening students, \( \bar{X}_2 = 6.2 \), \( \sigma_2 = 1.7 \), and \( n_2 = 40 \).

Calculate the Difference in Sample Means

Calculate the difference in the sample means: \( \bar{X}_1 - \bar{X}_2 = 4.7 - 6.2 = -1.5 \).

Find the Standard Error of the Difference of Means

Use the formula for standard error: \( SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{1.5^2}{40} + \frac{1.7^2}{40}} \). This gives us \( SE \approx 0.374 \).

Determine the Z-value for 95% Confidence Interval

For a 95% confidence interval, the critical z-value is 1.96.

Calculate the Margin of Error

Multiply the standard error by the z-value: \( ME = 1.96 \times 0.374 \approx 0.733 \).

Calculate the Confidence Interval

The 95% confidence interval is given by \((\bar{X}_1 - \bar{X}_2) \pm ME\). Thus, it is \(-1.5 \pm 0.733\), resulting in the interval \([-2.233, -0.767]\).

Key concepts

Difference of Means

The difference of means is a simple yet important concept in inferential statistics. It represents the difference between the averages (means) of two distinct groups. In our case, we have day students and evening students, each with their average travel distances to school. By calculating the difference of means, we seek to understand how one group's average behavior compares to another's.
For day students, the average distance (\( \bar{X}_1 \)) is 4.7 units, and for evening students (\( \bar{X}_2 \)), it is 6.2 units. Thus, the difference is found using the formula:

• Difference = \( \bar{X}_1 - \bar{X}_2 \)
• Calculating this gives us: \( 4.7 - 6.2 = -1.5 \)

This tells us that on average, evening students travel more by 1.5 units compared to day students. A negative value implies that the first group travels less than the second group, which is significant in deciding transportation policies.

Standard Error

Standard error is a measure of the variability of a sample statistic. In this case, it quantifies the precision of the difference between the two sample means we calculated. It tells us how much the sample mean difference would vary if we were to take many samples.
The formula to determine the standard error of the difference of means for two samples is:

• \( SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \)

Here, \( \sigma_1 \) and \( \sigma_2 \) are the standard deviations for day and evening students respectively, and \( n_1 \), \( n_2 \) are the sample sizes, both equal to 40.
Substituting the values, we get:

• \( SE = \sqrt{\frac{1.5^2}{40} + \frac{1.7^2}{40}} \)
• This results in \( SE \approx 0.374 \)

The smaller the standard error, the more reliable the difference of means. This value is crucial for constructing our confidence interval.

Z-value

A z-value, also known as a z-score or standard score, reflects how far from the mean a data point is in terms of standard deviations. In confidence intervals, the z-value helps us determine how much variation we can expect when estimating the population mean from a sample mean.
For a standard 95% confidence interval, the critical z-value is 1.96. This value indicates that approximately 95% of the data falls within 1.96 standard deviations from the mean in a normal distribution.
Alterations to the confidence level—say 90% or 99%—would affect this z-value:

• For a 90% confidence level, the z-value would be 1.645
• For a 99% confidence level, it would be 2.576

Here, the z-value of 1.96 is multiplied by the standard error to determine how wide or narrow our confidence interval will be.

Margin of Error

The margin of error provides a range that expresses the uncertainty surrounding a sample estimate. It's a crucial component of any confidence interval, as it takes into account potential errors from sampling over a population.
To calculate the margin of error for a difference of means, we multiply the z-value by the standard error:

• \( ME = z \times SE \)
• Given the z-value of 1.96 and SE of approximately 0.374, we find:
• \( ME = 1.96 \times 0.374 = 0.733 \)

This 0.733 indicates the range by which we estimate the true difference of means could vary in either direction from our sample estimate.
This way, when combined with the calculated difference of means, we obtain the confidence interval: [-2.233, -0.767]. This form tells us that while the average travel distance for evening students is greater, this difference falls within our calculated range with 95% certainty.