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A random sample of 50 12th-grade students was asked how long it took to get to school. The sample mean was \(16.2\) minutes, and the sample standard deviation was \(12.3\) minutes. (Source: AMSTAT Census at School) a. Find a \(95 \%\) confidence interval for the population mean time it takes 12 th-grade students to get to school. b. Would a \(90 \%\) confidence interval based on this sample data be wider or narrower than the \(95 \%\) confidence interval? Explain. Check your answer by constructing a \(90 \%\) confidence interval and comparing this width of the interval with the width of the \(95 \%\) confidence interval you found in part a.

Short Answer

Expert verified
The results from steps 2 and 4 represent the calculated 95% and 90% confidence intervals, respectively. The 90% confidence interval is narrower than the 95% confidence interval, as the width computed in step 5 will show.

Step by step solution

01

Calculate 95% Confidence Interval

We begin by calculating the 95% confidence interval using the formula, Confidence Interval = \(mean \pm (Z_{\alpha/2} * \frac{standard deviation}{\sqrt{n}})\), where Z_{\alpha/2} is the z-value for the chosen level of confidence (for 95% confidence, the z-value equals 1.96). The given mean is 16.2 minutes, the standard deviation is 12.3 minutes and the sample size n is 50. Substituting these values into the formula, we get \(16.2 \pm 1.96*(\frac{12.3}{\sqrt{50}})\).
02

Compute the Interval Range for 95% Confidence Interval

Calculate the interval range by performing the operations in the previous step, thus obtaining the lower and upper bounds of the 95% confidence interval.
03

Calculate 90% Confidence Interval

Next, calculate the 90% confidence interval using the same formula as in step 1, but this time with a z-score of 1.645 (corresponding to 90% confidence). Thus the formula will appear as: \(16.2 \pm 1.645*(\frac{12.3}{\sqrt{50}})\).
04

Compute the Interval Range for 90% Confidence Interval

As in step 2, perform the operations in the previous step to obtain the lower and upper bounds of the 90% confidence interval.
05

Compare the Widths of the Confidence Intervals

Calculate the widths of the 95% and 90% confidence intervals by subtracting the lower bound from the upper bound of each interval. The interval with the smaller width is the narrower confidence interval. A lower confidence level will provide a narrower interval as it encapsulates less of the data.

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Key Concepts

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

Statistics Education
Understanding statistics is an essential element in today's data-driven world. In the context of education, grasping statistical concepts allows students to analyze, interpret, and draw conclusions from data – skills that are invaluable in many fields.

When it comes to educating students about statistics, it is crucial to start with the basics. Fundamental concepts such as the mean, median, mode, variance, and standard deviation form the foundation upon which more complex ideas, like confidence intervals, are built. Confidence intervals, for example, provide a range of values that likely contain the population parameter with a given level of certainty, bridging the gap between theory and real-world application.

By working through textbook exercises, as in the example given with 12th-grade students' travel times, students gain practical experience in extracting meaningful information from data. This empowers them to make informed decisions based on statistical evidence. Educators should ensure that exercises not only test knowledge but also reinforce understanding through clear explanations and connections to real-life scenarios.
Sample Mean and Standard Deviation
Two fundamental statistics used to describe a dataset are the sample mean and standard deviation. The mean gives us the average value, which indicates the central tendency of the data. In the exercise with the 12th-grade students, the mean travel time to school was found to be 16.2 minutes.

The standard deviation, on the other hand, is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, whereas a high standard deviation indicates that the values are spread out over a wider range. The provided standard deviation of 12.3 minutes suggests that students' travel times vary considerably.

Importance of Understanding Sample Statistics

Knowing these two statistics is crucial because they help to summarize the data and form the basis for further statistical inference, such as estimating population parameters and testing hypotheses. When students comprehend these concepts, they can better appreciate the insights that can be drawn from a dataset and the importance of variance in assessing the reliability of the mean.
Z-Score for Confidence Intervals
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations. When constructing confidence intervals, the z-score provides a critical value by which we multiply the standard error to determine the margin of error.

In the context of our example, we used z-scores corresponding to the 95% and 90% confidence levels. These z-scores determine how far away from the mean the interval extends and are based on the standard normal distribution. A 95% confidence interval will have a larger z-score (1.96) than a 90% confidence interval (1.645) because it requires capturing more of the data if we want to be more confident.

Applying Z-Scores in Practice

By understanding the role of z-scores, students can appreciate why different confidence levels lead to different interval widths. A higher confidence level implies a wider interval, as it is designed to contain the true population mean with greater certainty. In practical scenarios, the choice of confidence level reflects how precise our estimates need to be and indicates the degree of uncertainty we are willing to accept.

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Most popular questions from this chapter

State whether each of the following changes would make a confidence interval wider or narrower. (Assume that nothing else changes.) a. Changing from a \(90 \%\) confidence level to a \(99 \%\) confidence level b. Changing from a sample size of 30 to a sample size of 200 c. Changing from a standard deviation of 20 pounds to a standard deviation of 25 pounds

A random sample of 10 independent healthy people showed the following body temperatures (in degrees Fahrenheit): $$98.5,98.299 .0,96.3,98.3,98.7,97.2,99.1,98.7,97.2$$ Test the hypothesis that the population mean is not \(98.6^{\circ} \mathrm{F}\), using a significance level of \(0.05 .\) See page 500 for guidance.

The distribution of the scores on a certain exam is \(N(80,5)\) which means that the exam scores are Normally distributed with a mean of 80 and a standard deviation of 5 . a. Sketch or use technology to create the curve and label on the \(x\) -axis the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be greater than 90\. Shade the region under the Normal curve whose area corresponds to this probability.

Suppose that 200 statistics students each took a random sample (with replacement) of 50 students at their college and recorded the ages of the students in their sample. Then each student used his or her data to calculate a \(95 \%\) confidence interval for the mean age of all students at the college. How many of the 200 intervals would you expect to capture the true population mean age, and how many would you expect not to capture the true population mean? Explain by showing your calculation.

The final exam grades for a sample of daytime statistics students and evening statistics students at one college are reported. The classes had the same instructor, covered the same material, and had similar exams. Using graphical and numerical summaries, write a brief description about how grades differ for these two groups. Then carry out a hypothesis test to determine whether the mean grades are significantly different for evening and daytime students. Assume that conditions for a \(t\) -test hold. Select your significance level. Daytime grades: \(100,100,93,76,86,72.5,82,63,59.5,53,79.5\), \(67,48,42.5,39\) Evening grades: \(100,98,95,91.5,104.5,94,86,84.5,73,92.5\), \(86.5,73.5,87,72.5,82,68.5,64.5,90.75,66.5\)

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