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According to a 2018 Money magazine article, the average income in Kansas is $$\$ 53,906$$. Suppose the standard deviation is $$\$ 3000$$ and the distribution of income is rightskewed. Repeated random samples of 400 Kansas residents are taken, and the sample mean of incomes is calculated for each sample. a. The population distribution is right-skewed. Will the distribution of sample means be Normal? Why or why not? b. Find and interpret a \(z\) -score that corresponds with a sample mean of $$\$ 53,606 .$$ c. Would it be unusual to find a sample mean of $$\$ 54,500 ?$$ Why or why not?

Short Answer

Expert verified
a. Yes, the distribution of sample means will be approximately Normal. This is according to the Central Limit Theorem which states that for large samples, the distribution of sample means will become Normal, regardless of the population's shape. b. The z-score is -1, meaning the sample mean is one standard deviation below the population mean. c. It is not unusual to find a sample mean of \$54,500 because its z-score is 1.98, which is under the common threshold of 2 for qualifying as unusual.

Step by step solution

01

Understand Central Limit Theorem

Central Limit Theorem (CLT) states that if we have a population with mean \( \mu \) and standard deviation \( \sigma \) and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. The rule of thumb is that the sample size should be larger than 30 to invoke CLT.
02

Analyze Distribution of Sample Means

Even though the population distribution is right-skewed, the distribution of sample means will be approximately Normal because the sample size is large (400 > 30). This is according to the Central Limit Theorem.
03

Calculate z-score

Z-score is a measure of how many standard deviations an element is from the mean. In this context, we want to determine how many standard deviations a sample mean of $53,606 is from the population mean. The formula is \( z = \frac{x - \mu}{\sigma / \sqrt{n}} \) where \( x \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. We substitute our values into the formula: \( z = \frac{53606 - 53906}{3000 / \sqrt{400}} \), yielding \( z = -1 \). This means that the sample mean $53,606 is one standard deviation below the population mean.
04

Evaluate Unusualness of a Sample Mean

A z-score beyond -2 or 2 is often considered unusual. Following similar calculations as in Step 3, we find the z-score for a sample mean of $54,500. \( z = \frac{54500 - 53906}{3000 / \sqrt{400}} \) giving \( z = 1.98 \). This z-score is just under the threshold of being unusual. So, we would consider it not unusual to find a sample mean of $54,500.

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Key Concepts

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

Understanding Normal Distribution
When we talk about normal distribution, we're diving into one of the most fundamental concepts in statistics. Picture a symmetrical bell-shaped curve where most of the data cluster around the center, also known as the mean, and this distribution sees fewer occurrences of data as we move away from the center.

For a normal distribution, the mean, median, and mode are all the same, lying at the center of the curve. The significance of the normal distribution lies in its predictability; it allows statisticians and researchers to make inferences about populations using sample data. What's particularly interesting is that real-world phenomena such as heights, blood pressure readings, and yes, even incomes, tend to form a normal distribution if the sample is large enough.

The Central Limit Theorem plays a crucial role as it tells us that averages of samples (sample means) will form a normal distribution, no matter the shape of the population distribution, given that the samples are of sufficient size (usual rule of thumb is more than 30). This is why, even with a skewed income distribution in Kansas, the distribution of sample means can be expected to be normal if we take large enough samples.
Calculating and Interpreting the Sample Mean
The idea of the sample mean is intuitive: it's simply the average value of a sample. But when interpreted through the lens of Central Limit Theorem, the sample mean becomes a powerful tool to understand large datasets. If you collect incomes of 400 people in Kansas and average them, that's your sample mean. Do this many times, and the means will distribute normally around the population mean, due to the Central Limit Theorem.

Practically, when given a sample mean, such as the average income of \(53,606 in one of the samples, we use it to gauge how that sample compares to the overall population. It's about context and comparison. If our sample mean significantly deviates from the population mean, which is \)53,906 according to the data, that tells us something about the specific sample. For instance, is it poorer or wealthier than average? Understanding the sample mean in combination with other measures gives us a more complete picture of the dataset.
Decoding the Z-Score
Dive into the concept of the z-score, and you're unlocking a way to understand where a particular value stands in relation to the average. A z-score tells us how many standard deviations a value is from the mean. The formula is given by: \[\[\begin{align*}z = \frac{x - \mu}{\sigma / \sqrt{n}}\end{align*}\]\] In the context of the Kansas income example, if we use the z-score formula, we see that a sample mean income of \(53,606 is one standard deviation below the population mean (yielding a z-score of -1). This z-score helps us quantify how typical or atypical this sample mean is. A z-score of 0 would mean the sample mean is exactly the same as the population mean. As we move further from 0, the sample mean becomes more atypical.

Therefore, when the textbook asks whether it would be unusual to find a sample mean of \)54,500, calculating the z-score (which turns out to be 1.98) gives us a way to answer. Since a z-score of 1.98 is close to but less than the common 'unusual' threshold of 2, we could say it's at the higher end of typical. It's not common, but also not so rare as to raise eyebrows.

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Most popular questions from this chapter

A random sample of 25 baseball play ers from the 2017 Major League Baseball season was taken and the sample data was used to construct two confidence intervals for the population mean. One interval was \((22.0,42.8)\). The other interval was \((19.9,44.0)\). (Source: mlb.com) a. One interval is a \(95 \%\) interval, and one is a \(90 \%\) interval. Which is which, and how do you know? b. If a larger sample size was used, for example, 40 instead of 25 , how would this affect the width of the intervals? Explain.

State whether each situation has independent or paired (dependent) samples. a. A researcher wants to know whether pulse rates of people go down after brief meditation. She collects the pulse rates of a random sample of people before meditation and then collects their pulse rates after meditation. b. A researcher wants to know whether professors with tenure have fewer posted office hours than professors without tenure do. She observes the number of office hours posted on the doors of tenured and untenured professors.

State whether each of the following changes would make a confidence interval wider or narrower. (Assume that nothing else changes.) a. Changing from a \(90 \%\) confidence level to a \(99 \%\) confidence level b. Changing from a sample size of 30 to a sample size of 200 c. Changing from a standard deviation of 20 pounds to a standard deviation of 25 pounds

Choose a test for each situation: one-sample \(t\) -test, two-sample \(t\) -test, paired \(t\) -test, and no \(t\) -test. a. A random sample of students who transfered to a 4 -year university from community colleges are asked their GPAs. Our goal is to determine whether the mean GPA for transfer students is significantly different from the population mean GPA for all students at the university. b. Students observe the number of office hours posted for a random sample of tenured and a random sample of untenured professors. c. A researcher goes to the parking lot at a large grocery chain and observes whether each person is male or female and whether they return the cart to the correct spot before leaving (yes or no).

The weights of four randomly and independently selected bags of tomatoes labeled 5 pounds were found to be \(5.1\), \(5.0,5.3\), and \(5.1\) pounds. Assume Normality. a. Find a \(95 \%\) confidence interval for the mean weight of all bags of tomatoes. b. Does the interval capture \(5.0\) pounds? Is there enough evidence to reject a mean weight of \(5.0\) pounds?

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