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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

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

Step by step solution

01

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 \( n_2 = 40 \).
02

Calculate the Difference in Sample Means

Calculate the difference in the sample means: \( \bar{X}_1 - \bar{X}_2 = 4.7 - 6.2 = -1.5 \).
03

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 \( SE \approx 0.374 \).
04

Determine the Z-value for 95% Confidence Interval

For a 95% confidence interval, the critical z-value is 1.96.
05

Calculate the Margin of Error

Multiply the standard error by the z-value: \( ME = 1.96 \times 0.374 \approx 0.733 \).
06

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 \([-2.233, -0.767]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The mean age of a random sample of 25 people who were playing the slot machines is 48.7 years, and the standard deviation is 6.8 years. The mean age of a random sample of 35 people who were playing roulette is 55.3 with a standard deviation of 3.2 years. Can it be concluded at \(\alpha=0.05\) that the mean age of those playing the slot machines is less than those playing roulette?

According to Nielsen Media Research, children (ages \(2-11\) ) spend an average of 21 hours 30 minutes watching television per week while teens (ages \(12-17\) ) spend an average of 20 hours 40 minutes. Based on the sample statistics shown, is there sufficient evidence to conclude a difference in average television watching times between the two groups? Use \(\alpha=0.01\) $$\begin{array}{lll} & \text { Children } & \text { Teens } \\\\\hline \text { Sample mean } & 22.45 & 18.50 \\\\\text { Sample variance } & 16.4 & 18.2 \\\\\text { Sample size } & 15 & 15\end{array}$$

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. The weights in ounces of a random sample of running shoes for men and women are shown. Calculate the variances for each sample, and test the claim that the variances are equal at \(\alpha=0.05\). Use the \(P\) -value method. $$ \begin{array}{rrr|rrr} && {\text { Men }} & {\text { Women }} \\ \hline 11.9 & 10.4 & 12.6 & 10.6 & 10.2 & 8.8 \\ 12.3 & 11.1 & 14.7 & 9.6 & 9.5 & 9.5 \\ 9.2 & 10.8 & 12.9 & 10.1 & 11.2 & 9.3 \\ 11.2 & 11.7 & 13.3 & 9.4 & 10.3 & 9.5 \\ 13.8 & 12.8 & 14.5 & 9.8 & 10.3 & 11.0 \end{array} $$

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In a random sample of 200 men, 130 said they used seat belts. In a random sample of 300 women, 63 said they used seat belts. Test the claim that men are more safety-conscious than women, at \(\alpha=0.01\). Use the \(P\) -value method.

Adults aged 16 or older were assessed in three types of literacy: prose, document, and quantitative. The scores in document literacy were the same for 19 - to 24 -year-olds and for 40 - to 49 -year-olds. A random sample of scores from a later year showed the following statistics. $$ \begin{array}{lccc} & & \text { Population } & \\ \text { Age group } & \begin{array}{l} \text { Mean } \\ \text { score } \end{array} & \begin{array}{c} \text { standard } \\ \text { deviation } \end{array} & \begin{array}{c} \text { Sample } \\ \text { size } \end{array} \\ \hline 19-24 & 280 & 56.2 & 40 \\ 40-49 & 315 & 52.1 & 35 \end{array} $$ Construct a \(95 \%\) confidence interval for the true difference in mean scores for these two groups. What does your interval say about the claim that there is no difference in mean scores?

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