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Math scores of 13 year olds, Part I. The National Assessment of Educational Progress tested a simple random sample of 1,000 thirteen year old students in both 2004 and 2008 (two separate simple random samples). The average and standard deviation in 2004 were 257 and 39 , respectively. In \(2008,\) the average and standard deviation were 260 and \(38,\) respectively. Calculate a \(90 \%\) confidence interval for the change in average scores from 2004 to 2008 , and interpret this interval in the context of the application. (Reminder: check conditions.) \(^{31}\)

Short Answer

Expert verified
The 90% confidence interval for the change in average scores is [0.168, 5.832]. The scores likely improved slightly.

Step by step solution

01

Understand the Problem

We are asked to find a 90% confidence interval for the change in average math scores from 2004 to 2008. This involves comparing the means from two independent samples: one from 2004 and one from 2008.
02

Check Conditions

Verify that the conditions necessary for constructing a confidence interval for the difference in means are met. Here, both samples are simple random samples, are independent, and the sample sizes (1000 each) are large enough for the Central Limit Theorem to apply.
03

Identify the Data and Formula

The 2004 data gives us a mean (\(\bar{x}_1 = 257\)) and standard deviation (\(s_1 = 39\)), and the 2008 data provides a mean (\(\bar{x}_2 = 260\)) and standard deviation (\(s_2 = 38\)). We'll use the formula for the confidence interval of the difference between two means:\[CI = \left(\bar{x}_2 - \bar{x}_1\right) \pm z^* \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]Where \(z^*\) is the z-value for a 90% confidence level.
04

Calculate the Standard Error of the Difference

First, compute the standard error (SE) of the difference between means:\[SE = \sqrt{\frac{39^2}{1000} + \frac{38^2}{1000}} = \sqrt{\frac{1521}{1000} + \frac{1444}{1000}} = \sqrt{1.521 + 1.444} = \sqrt{2.965} \approx 1.722\]
05

Find the z-value for a 90% Confidence Interval

Using a standard normal distribution, the z-value corresponding to a 90% confidence interval is 1.645.
06

Calculate the Margin of Error

The margin of error (ME) is calculated as:\[ME = z^* \cdot SE = 1.645 \times 1.722 \approx 2.832\]
07

Compute the Confidence Interval

The point estimate for the difference in means is \(\bar{x}_2 - \bar{x}_1 = 260 - 257 = 3\). The 90% confidence interval is:\[CI = 3 \pm 2.832 = [0.168, 5.832]\]
08

Interpret the Confidence Interval

We are 90% confident that the true difference in average scores, from 2004 to 2008, is between 0.168 and 5.832 points. This suggests a small but possibly significant improvement in scores over this period.

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Key Concepts

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

Standard Error
The standard error (SE) is a measure of how much the sample mean is expected to fluctuate from the true population mean. It's a crucial concept in statistics because it quantifies the variability in our estimate of a mean. In the context of our problem, we computed the standard error of the difference between two means to assess the uncertainty around our point estimate.

To calculate the standard error for the difference in means, you take the standard deviations of both samples and the sample sizes. The formula used is:
  • SE = \( \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \)
Where:
- \( s_1 \) and \( s_2 \) are the sample standard deviations for 2004 and 2008, respectively.
- \( n_1 \) and \( n_2 \) are the sample sizes, which in this exercise are both 1,000.

For our exercise, we derived an SE of about 1.722. This tells us that the average difference in scores would deviate by about this much, purely due to random sampling, if there were no true change in the mean scores from 2004 to 2008.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that, given a sufficiently large sample size, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution. This theorem is key for developing confidence intervals.

In our exercise, both samples had 1,000 students, which is large enough for the CLT to apply. This means we can assume the sampling distribution of the mean difference in scores is approximately normal. This approximation enables us to construct a confidence interval even without exact knowledge of the population distribution. The CLT justifies using the standard normal z-value to calculate the confidence interval.

Remember, the size of the sample is paramount when considering the applicability of the CLT. For smaller samples, the population distribution would need to be normal to use z-values for confidence intervals.
Difference in Means
The difference in means is simply the subtraction of one mean from another. In this case, we are looking at the difference between the average math scores of two distinct years: 2004 and 2008. The calculation helps us see if there is a change in the average scores over time.

The formula applied to find the confidence interval for the difference in means was:
  • CI = \(\left(\bar{x}_2 - \bar{x}_1\right) \pm z^* \cdot SE\)
Where:
- \(\bar{x}_2\) is the mean for 2008, and \(\bar{x}_1\) is the mean for 2004.
- \(z^*\) is the critical value for the chosen confidence level (90% in this exercise).
- SE is the standard error of the difference.

By calculating the difference in means, we assess whether any observed change, as indicated by the point estimate (3 points here), is statistically significant. The resulting confidence interval of [0.168, 5.832] provides a range for the true difference, suggesting a minor improvement in average scores between the years.

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Most popular questions from this chapter

Sleep habits of New Yorkers. New York is known as "the city that never sleeps". A random sample of 25 New Yorkers were asked how much sleep they get per night. Statistical summaries of these data are shown below. Do these data provide strong evidence that New Yorkers sleep more or less than 8 hours a night on average? \begin{tabular}{rrrrr} \hline \(\mathrm{n}\) & \(\bar{x}\) & \(\mathrm{~s}\) & \(\min\) & \(\max\) \\ \hline 25 & 7.73 & 0.77 & 6.17 & 9.78 \\ \hline \end{tabular} (a) Write the hypotheses in symbols and in words. (b) Check conditions, then calculate the test statistic, \(T,\) and the associated degrees of freedom. (c) Find and interpret the p-value in this context. Drawing a picture may be helpful. (d) What is the conclusion of the hypothesis test? (e) If you were to construct a \(95 \%\) confidence interval that corresponded to this hypothesis test, would you expect 8 hours to be in the interval?

Coffee, depression, and physical activity. Caffeine is the world's most widely used stimulant, with approximately \(80 \%\) consumed in the form of coffee. Participants in a study investigating the relationship between coffee consumption and exercise were asked to report the number of hours they spent per week on moderate (e.g., brisk walking) and vigorous (e.g., strenuous sports and jogging) exercise. Based on these data the researchers estimated the total hours of metabolic equivalent tasks (MET) per week, a value always greater than \(0 .\) The table below gives summary statistics of MET for women in this study based on the amount of coffee consumed. 3 (a) Write the hypotheses for evaluating if the average physical activity level varies among the different levels of coffee consumption. (b) Check conditions and describe any assumptions you must make to proceed with the test. (c) Below is part of the output associated with this test. Fill in the empty cells. (d) What is the conclusion of the test?

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Math scores of 13 year olds, Part II. Exercise 4.15 provides data on the average math scores from tests conducted by the National Assessment of Educational Progress in 2004 and 2008 . Two separate simple random samples were taken in each of these years. The average and standard deviation in 2004 were 257 and \(39,\) respectively. In \(2008,\) the average and standard deviation were 260 and \(38,\) respectively. (a) Do these data provide strong evidence that the average math score for 13 year old students has changed from 2004 to \(2008 ?\) Use a \(10 \%\) significance level. (b) It is possible that your conclusion in part (a) is incorrect. What type of error is possible for this conclusion? Explain. (c) Based on your hypothesis test, would you expect a \(90 \%\) confidence interval to contain the null value? Explain.

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