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Chapter 15: Problem 21

Scores on the Evidence-Based Reading part of the SAT exam in a recent year were roughly Normal with mean 536 and standard deviation 102. You choose an SRS of 100 students and average their SAT Evidence-Based Reading scores. If you do this many times, the standard deviation of the average scores you get will be close to a. \(102 .\) b. \(102 / 100=1.02\). c. \(102 / \sqrt{100}=10.2\).

Short Answer

Expert verified
The correct answer is c: \(102 / \sqrt{100} = 10.2\).

Step by step solution

01

Understand the Question Context

We are dealing with the Evidence-Based Reading section of the SAT, which is normally distributed with a mean \( \mu = 536 \) and a standard deviation \( \sigma = 102 \). We are interested in the standard deviation of the average scores of a sample size of 100 students.
02

Recall the Central Limit Theorem

The Central Limit Theorem states that the distribution of the sample mean \( \bar{x} \) will be approximately normally distributed if the sample size is large, and its standard deviation (known as the standard error) is given by \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( n \) is the sample size.
03

Apply the Formula for Standard Error

We apply the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \) to determine the standard deviation of the average SAT scores for 100 students. Substituting our values, \( \sigma = 102 \) and \( n = 100 \), we get: \[ \sigma_{\bar{x}} = \frac{102}{\sqrt{100}} = \frac{102}{10} = 10.2 \]
04

Compare with Multiple Choice Options

Based on our calculation, the standard deviation of the average scores, \( \sigma_{\bar{x}} \), is 10.2. Among the options given, answer c, \( 102 / \sqrt{100} = 10.2 \), is the correct one.

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Key Concepts

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

Statistics
Statistics is a branch of mathematics dealing with data collection, analysis, interpretation, and presentation. It allows us to understand complex phenomena and make informed decisions based on data. Statistic professionals often deal with samples and populations to infer insights.
In **descriptive statistics**, we summarize data from a sample using measures like mean, median, and mode. These provide a snapshot of data trends.
In **inferential statistics**, we use data from a sample to make predictions or generalizations about a larger population. This involves concepts like hypothesis testing and confidence intervals.
Understanding statistics is essential when working with standard deviation, sample mean, and normal distribution, as seen in this SAT score problem.
Standard Deviation
Standard deviation measures how spread out the numbers in a data set are. It tells us the amount of variation or dispersion from the average (mean).
- A low standard deviation means the data points are close to the mean. - A high standard deviation indicates data points are spread over a wider range.
For the SAT scores discussed, the standard deviation is 102. This means scores typically vary by 102 points from the mean score of 536.
When examining a sample, like the SAT scores of 100 students, the formula for standard error helps us understand the distribution of the sample mean. Standard error reflects how much the sample mean deviates from the population mean, especially as sample size increases.
Sample Mean
The sample mean, denoted as \( \bar{x} \), is the average of observations in a sample. It is calculated by summing all the data points and dividing by the number of observations.
In our SAT score scenario, taking samples of 100 students and averaging their scores gives us the sample mean for that group. The Central Limit Theorem assures us that if we repeatedly draw large samples, the distribution of these sample means will approach a normal distribution.
In this context, the sample mean helps us understand typical performance in a subset of students. It allows educational researchers to draw conclusions about the performance of the larger group of test-takers.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a bell-shaped curve with specific properties. - It is symmetrical around the mean, with data points more concentrated in the center. - The mean, median, and mode of a normal distribution are equal.
In a practical sense, the SAT scores we are analyzing are approximately normally distributed, meaning most students' scores cluster around the mean score of 536 with fewer extreme scores.
The Central Limit Theorem plays a crucial role, indicating that with a sufficiently large sample size, the distribution of the sample mean will be normal even if the population distribution is not strictly normal.
Understanding the normal distribution is vital for interpreting test scores and making predictions about student performance across samples.

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Most popular questions from this chapter

A Sample of Young Men. A government sample survey plans to measure the mean total cholesterol level of an SRS of men aged 20-34. The researchers will report the mean \(x\) from their sample as an estimate of the mean total cholesterol level \(\mu\) in this population. a. Explain to someone who knows no statistics what it means to say that \(x\) is an "unbiased" estimator of \(\mu\). b. The sample result \(x\) is an unbiased estimator of the population truth \(\mu\) no matter what size SRS the study uses. Explain to someone who knows no statistics why a large sample gives more trustworthy results than a small sample.

Pollutants in Aut o Ex hausts (continued). The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life ( 150,000 miles of driving) of cars of a particular model varies Normally with mean 80 \(\mathrm{mg} / \mathrm{mi}\) and standard deviation \(4 \mathrm{mg} / \mathrm{mi}\). A company has 25 cars of this model in its fleet. What is the level \(L\) such that the probability that the average NOX + NMOG level \(x\) for the fleet is greater than \(L\) is only \(0.01\) ? (Hint: This requires a backward Normal calculation. See page 89 in Chapter 3 if you need a review.)

Guns in School. Researchers surveyed 14,765 American high school students (grades 9-12) and found that \(27.3 \%\) of those surveyed were in grade 9 . The percentage of all American high school students who are are in grade 9 is \(\mathbf{2 6 . 5} \%\). The percentage of those surveyed who were in grade 9 and had carried a gun to school was \(\mathbf{4 . 4 \%}\). Is each of the boldface numbers a parameter or a statistic?

An October 20, 2019, poll of Canadian adults who were registered voters found that \(31.6 \%\) said they would vote conservative in the October 2019 elections. Election records show that \(\mathbf{3 4 . 4} \%\) actually vot ed conservative. The boldface number is a a. sampling distribution. b. statistic. c. parameter.

Daily Activity. It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minutes per day that people spend standing or walking. I Among mildly obese people, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with mean 373 minutes and standard deviation 67 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with mean 526 minutes and standard deviation 107 minutes. A researcher records the minutes of activity for an SRS of five mildly obese people and an SRS of five lean people. a. What is the probability that the mean number of minutes of daily activity of the five mildly obese people exceeds 420 minutes? b. What is the probability that the mean number of minutes of daily activity of the five lean people exceeds 420 minutes?
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