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Traveling Distances 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 is [-2.227, -0.773].

Step by step solution

01

Identify the Given Data

Identify the sample means, sample standard deviations, and sample sizes for both groups. For day students: \( \bar{X}_d = 4.7 \), \( \sigma_d = 1.5 \), and \( n_d = 40 \). For evening students: \( \bar{X}_e = 6.2 \), \( \sigma_e = 1.7 \), and \( n_e = 40 \).
02

Determine the Standard Error of the Difference in Means

Use the formula for the standard error of the difference between two independent means: \( \text{SE} = \sqrt{\frac{\sigma_d^2}{n_d} + \frac{\sigma_e^2}{n_e}} \). Substitute the values: \( \text{SE} = \sqrt{\frac{1.5^2}{40} + \frac{1.7^2}{40}} \). Calculate to get \( \text{SE} \approx 0.371 \).
03

Find the Critical Value for the Confidence Level

For a 95% confidence interval with large sample sizes (\( n > 30 \)), use the standard normal distribution (Z-distribution). The critical value (\(z_{0.025}\)) from the Z-table is approximately 1.96.
04

Calculate the Margin of Error

The margin of error (ME) is given by the formula: \( \text{ME} = z \times \text{SE} \). Substitute the values: \( \text{ME} = 1.96 \times 0.371 \). Calculate to get \( \text{ME} \approx 0.727 \).
05

Compute the Confidence Interval

The confidence interval is the difference in sample means plus or minus the margin of error: \((\bar{X}_d - \bar{X}_e) \pm \text{ME}\). Substitute the values: \( (4.7 - 6.2) \pm 0.727 \). Calculate the interval: \( -1.5 \pm 0.727 \), which gives \([-2.227, -0.773]\).

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Key Concepts

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

Understanding Standard Error
The standard error is a crucial concept in statistical analysis, especially when comparing means between two groups. It represents the variability of the sample mean estimate from the true population mean. In essence, it helps us understand how much our sample mean is expected to vary if we took multiple samples.

The formula to calculate the standard error for the difference between two independent sample means is: \[\text{SE} = \sqrt{\frac{\sigma_d^2}{n_d} + \frac{\sigma_e^2}{n_e}}\]where
  • \(\sigma_d\) and \(\sigma_e\) are the standard deviations of the samples, and
  • \(n_d\) and \(n_e\) are the sizes of the samples.

A smaller standard error means that our sample mean is a more precise estimate of the population mean. This precision helps increase our confidence in the estimations made from the data.
Introduction to Z-distribution
The Z-distribution, also known as the standard normal distribution, is fundamental in statistics for calculating probabilities and confidence intervals when dealing with large sample sizes. This bell-shaped curve has a mean of zero and a standard deviation of one.

When understanding the Z-distribution, it's important to consider its role in determining critical values for confidence intervals. When we say a 95% confidence interval, we mean that 95% of the possible sample means will fall within a certain range of the true population mean.
For a 95% confidence level, we find that the critical value, obtained from the Z-table, is approximately 1.96. This critical value tells us how many standard errors away from the sample mean we need to go to capture 95% of all possible sample outcomes.

So, using a Z-distribution to find the critical value is essential for calculating confidence intervals when the sample size is large (usually \(n > 30\)). It ensures that our statistical findings are robust and reliable.
Margin of Error: What it Means
The margin of error is a key component in constructing confidence intervals. It reflects the degree of uncertainty or potential error around our sample mean estimate.
To calculate the margin of error (ME), use the formula: \[\text{ME} = z \times \text{SE}\]where
  • \(z\) is the critical value from the Z-distribution, and
  • \(\text{SE}\) is the standard error of the difference between two means.

The margin of error tells us how far the true population parameter could reasonably deviate from the known sample statistic. A larger margin of error indicates more uncertainty in our estimate, while a smaller margin denotes more confidence and precision.

Therefore, the margin of error, combined with sample means and the Z-distribution, gives us a reliable way to express the range in which the true population parameter is likely to fall.
The Role of Statistical Analysis
Statistical analysis allows us to make sense of collected data, providing a structured approach to interpreting information. In the case of confidence intervals, statistical analysis helps us draw conclusions about a population parameter based on sample data.

When calculating confidence intervals, statistical analysis involves:
  • Determining the appropriate statistical methods and distributions to use, such as the Z-distribution in large sample scenarios.
  • Computing the necessary statistics, like the standard error and margin of error, to construct the confidence interval.
  • Interpreting the results to provide meaningful insights, like understanding the range where the true difference in means could fall.

Ultimately, statistical analysis transforms raw data into valuable knowledge, enabling data-informed decision making and helping us understand the variability and uncertainty present in the statistical processes we study.

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Most popular questions from this chapter

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. 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. Toy Assembly Test An educational researcher devised a wooden toy assembly project to test learning in 6-year-olds. The time in seconds to assemble the project was noted, and the toy was disassembled out of the child's sight. Then the child was given the task to repeat. The researcher would conclude that learning occurred if the mean of the second assembly times was less than the mean of the first assembly times. At \(\alpha=0.01,\) can it be concluded that learning took place? Use the \(P\) -value method, and find the \(99 \%\) confidence interval of the difference in means. $$ \begin{array}{l|ccccccc}{\text { Child }} & {1} & {2} & {3} & {4} & {5} & {6} & {7} \\ \hline \text { Trial } 1 & {100} & {150} & {150} & {110} & {130} & {120} & {118} \\ \hline \text { Trial 2} & {90} & {130} & {150} & {90} & {105} & {110} & {120}\end{array} $$

ACT Scores A random survey of 1000 students nationwide showed a mean ACT score of 21.4 . Ohio was not used. A survey of 500 randomly selected Ohio scores showed a mean of 20.8 . If the population standard deviation is \(3,\) can we conclude that Ohio is below the national average? Use \(\alpha=0.05 .\)

For Exercises 9 through \(24,\) 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. Winter Temperatures A random sample of daily high temperatures in January and February is listed. At \(\alpha=0.05,\) can it be concluded that there is a difference in variances in high temperature between the two months? $$ \begin{array}{l|ccccccccccc}{\text { Jan. }} & {31} & {31} & {38} & {24} & {24} & {42} & {22} & {43} & {35} & {42} \\ \hline \text { Feb. } & {31} & {29} & {24} & {30} & {28} & {24} & {27} & {34} & {27}\end{array} $$

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