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Retaking the SAT. An SRS of 400 high school seniors gained an average of \(x^{-}\) \(\bar{x}=9\) points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation \(\sigma=40\). We want to estimate the mean change in score \(\mu\) in the population of all high school seniors. (a) Give a \(95 \%\) confidence interval for \(\mu\) based on this sample. (b) Based on your confidence interval in part (a), how certain are you that the mean change in score \(\mu\) in the population of all high school seniors is greater than 0? [Hint: Does the interval in part (a) include 0?]

Short Answer

Expert verified
(a) 95% CI: (5.08, 12.92). (b) Very certain, as the interval does not include 0.

Step by step solution

01

Identify Given Information

We have a sample size of 400 students, an average score increase \( \bar{x} = 9 \), and a population standard deviation \( \sigma = 40 \). We are asked to find a 95% confidence interval for the mean change in score \( \mu \).
02

Find the Standard Error (SE)

The standard error of the sample mean is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \] where \( n = 400 \) is the sample size. Thus, \[ SE = \frac{40}{\sqrt{400}} = \frac{40}{20} = 2 \].
03

Find the Z-Score for 95% Confidence

For a 95% confidence interval, the Z-score for a normal distribution is approximately 1.96 (since the confidence level encompasses the center percentage of the distribution).
04

Calculate the Confidence Interval

Using the formula for a confidence interval \( CI = \bar{x} \pm Z \cdot SE \), we calculate: \[ CI = 9 \pm 1.96 \times 2 \] This results in a confidence interval: \[ CI = 9 \pm 3.92 \], which simplifies to: \[ (5.08, 12.92) \].
05

Interpret the Confidence Interval in Relation to 0

The confidence interval \((5.08, 12.92)\) does not include 0. This suggests that we are quite confident that the true mean change in SAT scores is greater than 0.

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Key Concepts

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

SAT Mathematics
The SAT Mathematics exam is an important component of the SAT standardized test, which assesses mathematical reasoning and problem-solving abilities of high school students. This exam serves as a vital tool for college admissions in the United States. Students can take the SAT multiple times to improve their scores, with each attempt providing a new data point to analyze.
Typically, a student's SAT Math score reflects their understanding of key topics like algebra, geometry, and basic arithmetic. In the scenario we're analyzing, students had a second attempt, where on average, scores increased by 9 points. This kind of statistical data is useful for educational institutions to evaluate if certain strategies or study methods are effective.
Understanding the change in scores through statistical measures, such as a confidence interval, can help in determining how likely it is that changes in test scores are due to chance or actual improvement in understanding.
Standard Error
The Standard Error (SE) is a measure that indicates the amount of variability or dispersion of the sample mean from the true population mean. It helps us understand how accurately the sample mean represents the population mean.
The SE can be calculated using the formula:
  • \[ SE = \frac{\sigma}{\sqrt{n}} \]
where \( \sigma \) is the population standard deviation, and \( n \) is the sample size. In the exercise, with \( \sigma = 40 \) and \( n = 400 \), we computed the SE as 2. This means that the average score increase of 9 points is likely to be accurate within this range.
Having a small SE in this context indicates that the sample mean gives a reliable estimate of the population mean, allowing us to build a confidence interval.
Z-Score
A Z-Score is a numerical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. In the context of confidence intervals, the Z-Score is used to determine how far away the sample mean is from the population mean in terms of standard errors.
For a 95% confidence interval, which is common in statistical analysis, the corresponding Z-Score is approximately 1.96. This score indicates that the sample mean falls within 1.96 standard deviations from the true mean 95% of the time.
When calculating the confidence interval, we multiplied this Z-Score by the Standard Error to determine the range around the sample mean. This range gives us confidence about where the true population mean lies with respect to the sample mean.
Normal Distribution
Normal Distribution is a key concept in statistics that represents data that clusters around a mean or average. Often referred to as a "bell curve," it is symmetric about the mean, meaning most data points are close to the mean; fewer occurrences are at the extremes or tails.
In the given exercise, it is assumed that the change in SAT Mathematics scores follows a Normal Distribution. This assumption allows us to use Z-Scores and Standard Errors to calculate confidence intervals accurately.
Understanding that the data is normally distributed means we can apply these statistical methods to infer characteristics about the population from the sample data. It's crucial in many statistical analyses because the properties of the normal distribution underpin a lot of the inferential statistics methods used, such as forming confidence intervals or hypothesis testing.

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Most popular questions from this chapter

Number Skills of Eighth-Graders. The National Assessment of Educational Progress (NAEP) includes a mathematics test for eighth-grade students. \({ }^{3}\) Scores on the test range from 0 to 500 . Demonstrating the ability to use the mean to solve a problem is an example of the skills and knowledge associated with performance at the Basic level. An example of the knowledge and skills associated with the Proficient level is being able to read and interpret a stem-and-leaf plot- In \(2015,136,900\) eighth-graders were in the NAEP sample for the mathematics test. The mean mathematics score was \(x^{-} x=282\). We want to estimate the mean score \(\mu\) in the population of all eighth-graders. Consider the NAEP sample as an SRS from a Normal population with standard deviation \(\sigma=110 .\) (a) If we take many samples, the sample mean \(x^{-} \bar{x}\) varies from sample to sample according to a Normal distribution with mean equal to the unknown mean score \(\mu\) in the population. What is the standard deviation of this sampling distribution? (b) According to the 95 part of the \(68-95-99.7\) rule, \(95 \%\) of all values of \(x^{-} x\) fall within on either side of the unknown mean \(\mu\). What is the missing number? (c) What is the \(95 \%\) confidence interval for the population mean score \(\mu\) based on this one sample?

Why are larger samples better? Statisticians prefer large samples. Describe briefly the effect of increasing the size of a sample on the margin of error of a \(95 \%\) confidence interval.

Suppose that an SRS of 2500 eighth-graders has \(x^{-} x=285\). Based on this sample, a \(95 \%\) confidence interval for \(\mu\) is (a) \(4.31 \pm 0.086\) (b) \(285 \pm 4.31\). (c) \(282 \pm 4.31\).

To give a \(99.9 \%\) confidence interval for a population mean \(\mu\), you would use the critical value (a) \(z^{*}=1.960\). (b) \(z^{*}=2.576\). (c) \(x^{+}=3.291\). Use the following information for Exercises \(16.12\) through 16.14. A laboratory scale is known to have a standard deviation of \(\sigma=0.001\) gram in repeated weighings. Scale readings in repeated weighings are Normally distributed, with mean equal to the true weight of the specimen. Three weighings of a specimen on this scale give \(3.412,3.416\), and \(3.414\) grams.

Addicted to Coffee. A Gallup Poll in July 2015 found that \(26 \%\) of the 675 coffee drinkers in the sample said they were addicted to coffee. Gallup announced, "For results based on the sample of 675 coffee drinkers, one can say with \(95 \%\) confidence that the maximum margin of sampling error is \(\pm 5\) percentage points." (a) Confidence intervals for a percent follow the form $$ \text { estimate } \pm \text { margin of error } $$ Based on the information from Gallup, what is the \(95 \%\) confidence interval for the percent of all coffee drinkers who would say they are addicted to coffee? (b) What does it mean to have \(95 \%\) confidence in this interval?

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