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Professor Smith conducted a class exercise in which students ran a computer program to generate random samples from a population that had a mean of 50 and a standard deviation of \(9 \mathrm{~mm}\). Each of Smith's students took a random sample of size \(n\) and calculated the sample mean. Smith found that about \(68 \%\) of the students had sample means between 48.5 and \(51.5 \mathrm{~mm} .\) What was \(n ?\) (Assume that \(n\) is large enough that the Central Limit Theorem is applicable.)

Short Answer

Expert verified
The sample size \(n\) is 36.

Step by step solution

01

Understanding the Concept of Standard Error

Recognize that the Central Limit Theorem describes how the sample mean distribution would behave. With large enough samples, it approximates a normal distribution where the standard error of the mean is equal to the population standard deviation divided by the square root of the sample size, given by the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \).
02

Identifying the Boundaries of the Interval

Understand that about 68% of the student's sample means lie within one standard error from the mean. This range corresponds to one standard deviation for a normal distribution. Therefore, the value \(48.5 \text{mm}\) is one standard error below the mean, and the value \(51.5 \text{mm}\) is one standard error above the mean, the mean being \(50 \text{mm}\).
03

Calculating the Standard Error

Based on the range provided, calculate the standard error using either of the boundary values. Since the boundaries are equidistant from the mean, the standard error can be computed as \( \sigma_{\bar{x}} = \frac{51.5 - 50}{1} = \frac{50 - 48.5}{1} = 1.5 \text{mm}\).
04

Solving for the Sample Size \(n\)

Use the equation for standard error to solve for the sample size \(n\). Substituting the given population standard deviation and calculated standard error into \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), we get \( 1.5 = \frac{9}{\sqrt{n}} \). Solving this equation for \(n\) involves squaring both sides and then dividing by the square of the standard error. \( n = \left( \frac{9}{1.5} \right)^2 = 6^2 = 36 \).

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Key Concepts

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

Standard Error
When diving into the field of statistics, understanding the concept of standard error is crucial. It is a measure that tells us how much the sample mean, or any other statistic, deviates from the true population mean.

The formula to calculate the standard error of the mean is given by \( \frac{\sigma}{\sqrt{n}} \) where \( \sigma \) is the population standard deviation, and \( n \) is the sample size. The standard error decreases when the sample size increases, meaning that larger samples provide a more accurate approximation of the population mean.

In an educational setting, like Professor Smith's exercise, the standard error is used to understand that the sample means will distribute around the population mean of 50 mm in a certain pattern. By observing that 68% of the sample means were within 1.5 mm of the population mean, students can infer the standard error. This is important because it means that the larger your sample, the smaller the standard error and the closer your sample mean is to the population mean.
Sample Size Calculation
Sample size calculation plays a pivotal role in designing studies and experiments. It's a statistical technique used to determine how many observations or subjects are needed to ensure that the findings are reliable.

To calculate the sample size, we need to define a few parameters including the desired margin of error, the standard deviation of the population, and the level of confidence. For instance, Professor Smith's class exercise was illustrating sample size calculation by utilizing the Central Limit Theorem. The exercise involved determining the sample size \( n \) required so that the sample means would fall within a specified range of the population mean. From the calculation steps given, it's evident that the sample size impacts the precision of study outcomes significantly.

Understanding this relationship helps students and researchers alike to design studies that are adequately powered to detect true effects or differences, consequently preventing wasted resources on studies that are too small or unnecessarily large.
Normal Distribution
The normal distribution is a fundamental concept in statistics often represented by a symmetrical, bell-shaped curve. It is defined by its mean and standard deviation, with about 68% of the data lying within one standard deviation from the mean.

In Professor Smith's exercise, the normal distribution is assumed due to the application of the Central Limit Theorem, which states that the distribution of sample means will tend to be normal, regardless of the population's distribution, given a sufficiently large sample size. With the given sample means of 48.5 mm and 51.5 mm, these values represent one standard deviation from the population mean of 50 mm, which is characteristic of a normal distribution.

This concept is not just academic; it is used in various real-world scenarios, such as quality control, stock market analysis, and many areas of scientific research, highlighting its relevance across diverse fields and its importance in statistical education.

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Most popular questions from this chapter

Consider taking a random sample of size 36 from a population in which \(52 \%\) of the people have type \(A\) blood. What is the probability that the sample proportion with type A blood will be greater than \(0.54 ?\) Use the normal approximation to the binomial with continuity correction.

The skull breadths of a certain population of rodents follow a normal distribution with a standard deviation of \(15 \mathrm{~mm}\). Let \(\bar{Y}\) be the mean skull breadth of a random sample of 81 individuals from this population, and let \(\mu\) be the population mean skull breadth. (a) Suppose \(\mu=50 \mathrm{~mm} .\) Find \(\operatorname{Pr}\\{\bar{Y}\) is within \(\pm 2 \mathrm{~mm}\) of \(\mu\\}\). (b) Suppose \(\mu=100 \mathrm{~mm} .\) Find \(\operatorname{Pr}\\{\bar{Y}\) is within \(\pm 2 \mathrm{~mm}\) of \(\mu\\}\). (c) Suppose \(\mu\) is unknown. Can you find \(\operatorname{Pr}\\{\bar{Y}\) is within \(\pm 2 \mathrm{~mm}\) of \(\mu\\} ?\) If so, do it. If not, explain why not.

Suppose that every day for 3 months Bill takes a random sample of 20 college students, records the number of calories they consume on that day, finds the average of the 20 observations, and adds the average to his histogram of the sampling distribution of the mean. Suppose also that every day for 2 months Susan takes a random sample of 30 college students and records the number of calories they consume on that day (which is fairly symmetric), finds the average of the 30 observations, and adds the average to her histogram of the sampling distribution of the mean. (a) Can we expect Bill's distribution and Susan's distribution to have the same shape? Why or why not? If not, how will the shapes differ? (b) Can we expect Bill's distribution and Susan's distribution to have the same center? Why or why not? If not. how will the centers differ? (c) Can we expect Bill's distribution and Susan's distribution to have the same spread? Why or why not? If not, how will the spreads differ?

A certain cross between sweet-pea plants will produce progeny that are either purple flowered or white flowered; is \(p=\frac{9}{16}\). Suppose \(n\) progeny are to be examined, and let \(\hat{P}\) be the sample proportion of purple- flowered plants. It might happen, by chance, that \(\hat{P}\) would be closer to \(\frac{1}{2}\) than to \(\frac{9}{16}\). Find the probability that this misleading event would occur if (a) \(n=1\). (b) \(n=64\). (c) \(n=320\). (Use the normal approximation without the continuity correction.)

The basal diameter of a sea anemone is an indicator of its age. The density curve shown here represents the distribution of diameters in a certain large population of anemones; the population mean diameter is \(4.2 \mathrm{~cm},\) and the standard deviation is \(1.4 \mathrm{~cm} .{ }^{4}\) Let \(\bar{Y}\) represent the mean diameter of 25 anemones randomly chosen from the population. (a) Find the approximate value of \(\operatorname{Pr}\\{4 \leq \bar{Y} \leq 5\\}\). (b) Why is your answer to part (a) approximately correct even though the population distribution of diameters is clearly not normal? Would the same approach be equally valid for a sample of size 2 rather than \(25 ?\) Why or why not?

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