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Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5 . a. What are the mean and standard deviation of the \(\bar{x}\) sampling distribution? Describe the shape of the \(\bar{x}\) sampling distribution. b. What is the approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

Short Answer

Expert verified
a. The mean of the sample mean distribution is the same as the population mean, 40, and its standard deviation is 0.625. Its shape is normal. b and c. The probabilities can only be precisely determined using a standard normal table or calculator that applies the cumulative normal distribution function.

Step by step solution

01

Calculation of Mean & Standard Deviation of Sampling Distribution

The given population mean \( \mu = 40 \) is the mean of this sampling distribution. \n As for the standard deviation of the sampling distribution (often referred to as the standard error), it is found by dividing the standard deviation of the population (\(\sigma = 5\)) by the square root of the sample size (\(n = 64\)). \n That is, \( \sigma_{\bar{x}} = \sigma / \sqrt{n} = 5 / \sqrt{64} = 5/8 = 0.625 \)
02

Description of distribution's shape

The shape of the sampling distribution will be normal (by the Central Limit Theorem) because the sample size of 64 is sufficiently large (\(n > 30\)).
03

Calculation of Probability (within 0.5)

To find the probability that \( \bar{x} \) will be within 0.5 of the population mean, we need to calculate the respective z-scores and use the standard normal table. A z-score is a measure of how many standard deviations an element is from the mean. Therefore, we can calculate z-scores for 39.5 and 40.5, then find the difference in their probabilities (from the standard normal table). \n So, \( Z_{1} = (39.5 - 40) / 0.625 = -0.8 \)\n and \( Z_{2} = (40.5 - 40) / 0.625 = 0.8 \)\n The probability between these z-scores gives the desired probability.
04

Calculation of Probability (differ by more than 0.7)

We'll follow a similar process from the previous step. First, calculate z-scores for 39.3 and 40.7, then find the probabilities from the standard normal table. The difference between these gives the probability within 0.7 of the mean, so subtracting this from 1 gives the probability that \( \bar{x} \) will differ from \( \mu \) by more than 0.7. \n Hence, \( Z_{3} = (39.3 - 40) / 0.625 = -1.12 \)\n and \( Z_{4} = (40.7 - 40) / 0.625 = 1.12 \)\n The difference between these z-scores' probabilities gives the desired probability.

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Key Concepts

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

Population Mean
The population mean, symbolically represented as \(\text{\mu\)} , is the average of all the data points in a full set. It's what you'd get if you added up every single value in the population and then divided by the total number of values. This value serves as the theoretical center of the population distribution, and in the context of our exercise, it's the point around which the random sample means are expected to be distributed.

Since the exercise provides a population mean of 40, this doesn't change no matter how many samples we take or what their sizes are. It's a constant point of reference when we are discussing sampling distributions, especially in calculating the probability of sample means falling within a certain range of the population mean.
Standard Deviation
The standard deviation measures the amount of variability or spread in a set of data values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

However, when dealing with a sampling distribution, we're interested in something called the standard error, which is essentially the standard deviation of the sample means. The standard error gives us an idea of how much we would expect those sample means to deviate from the population mean. In our example with a standard deviation of 5 and sample size of 64, the standard error is 0.625, computed using the formula \( \sigma_{\bar{x}} = \sigma / \sqrt{n} \).
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle of probability and statistics that describes the shape of the distribution of sample means. CLT tells us that as the sample size gets larger, the distribution of the sample means will approach a normal distribution, regardless of the population's initial distribution.

In the scenario we are considering, because we have a sample size of 64, which is greater than 30, the CLT confirms that the sampling distribution will be approximately normal. This is why we can use the normal distribution to predict probabilities and apply the z-score, making our work a lot easier and more reliable.
Standard Error
The standard error is essentially the standard deviation of the sampling distribution of the sample means. The smaller the standard error, the more concentrated the sample means are around the population mean. The standard error is calculated by taking the standard deviation of the original population and dividing it by the square root of the sample size.

By applying this to our given problem, where the population standard deviation is 5 and the sample size is 64, we find the standard error to be 0.625. This tells us how much we can expect the sample mean to vary from the true population mean.
Z-score
A z-score represents the number of standard errors a data point is from the mean. It's a way of standardizing scores on the same scale to compare them directly. In terms of a sampling distribution, the z-score tells us how far and in what direction a sample mean deviates from the population mean, measured in terms of standard errors.

In the exercise, z-scores were used to determine the probability of sample means falling within particular ranges. By converting the sample mean range to z-scores, we could use the standard normal distribution table to find the probability of obtaining a sample mean within that range.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a bell-shaped curve that is symmetric about the mean. Most values cluster around a central region, and as you move away from the center, the probability of encountering those values decreases exponentially.

Since our sample size is large enough, the Central Limit Theorem guarantees that the sampling distribution of the sample means follows a normal distribution, which greatly simplifies probability calculation. This assumes the population distribution is either normally distributed or the sample size is large enough to render the shape of the population distribution irrelevant.
Probability
In statistics, probability measures the likelihood of a certain event occurring. It's a number between 0 and 1, where 0 means the event is impossible, and 1 means it is certain. When we talk about probability in the context of a normal distribution, we usually refer to the area under the curve, which corresponds to the likelihood of a value or range of values occurring.

In our exercise, we calculated the probability of the sample mean being within a specified interval around the population mean. Using the z-scores, we can find these probabilities from the standard normal distribution, allowing us to make inferences about where future sample means are likely to fall.

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Most popular questions from this chapter

Explain the difference between \(\bar{x}\) and \(\mu_{\bar{x}}\)

Five students visiting the student health center for a free dental examination during National Dental Hygiene Month were asked how many months had passed since their last visit to a dentist. Their responses were: \(\begin{array}{lllll}6 & 17 & 11 & 22 & 29\end{array}\)

A study of fast-food intake is described in the paper "What People Buy From Fast-Food Restaurants" (Obesity [2009]:1369- 1374). Adult customers at three hamburger chains (McDonald's, Burger King, and Wendy's) in New York City were approached as they entered the restaurant at lunchtime and asked to provide their receipt when exiting. The receipts were then used to determine what was purchased and the number of calories consumed was determined. In all, 3,857 people participated in the study. The sample mean number of calories consumed was 857 and the sample standard deviation was 677 . a. The sample standard deviation is quite large. What does this tell you about number of calories consumed in a hamburgerchain lunchtime fast-food purchase in New York City? b. Given the values of the sample mean and standard deviation and the fact that the number of calories consumed can't be negative, explain why it is not reasonable to assume that the distribution of calories consumed is normal. c. Based on a recommended daily intake of 2,000 calories, the online Healthy Dining Finder (www.healthydiningfinder .com) recommends a target of 750 calories for lunch. Assuming that it is reasonable to regard the sample of 3,857 fast-food purchases as representative of all hamburger-chain lunchtime purchases in New York City, carry out a hypothesis test to determine if the sample provides convincing evidence that the mean number of calories in a New York City hamburger-chain lunchtime purchase is greater than the lunch recommendation of 750 calories. Use \(\alpha=0.01\). d. Would it be reasonable to generalize the conclusion of the test in Part (c) to the lunchtime fast-food purchases of all adult Americans? Explain why or why not. e. Explain why it is better to use the customer receipt to determine what was ordered rather than just asking a customer leaving the restaurant what he or she purchased.

The paper referenced in the previous exercise also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the final exam: $$ n=411 \quad \bar{x}=43.18 \quad s=21.46 $$ Construct and interpret a \(90 \%\) confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the final exam.

The formula used to calculate a confidence interval for the mean of a normal population is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(99 \%\) confidence, \(n=24\) c. \(90 \%\) confidence, \(n=13\)

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