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Chapter 7: Problem 43

For a population, \(N=205,000, \mu=66\), and \(\sigma=7\). Find the \(z\) value for each of the following for \(n=49 .\) a. \(\bar{x}=68.44\) h. \(\bar{x}=58.75\) c. \(\bar{x}=62.35\) d. \(\bar{x}=71.82\)

Short Answer

Expert verified
The z-values for \(\bar{x}\) are computed by substituting each sample mean into the formula. Exact values will depend on the computation result.

Step by step solution

01

Substitute values into the formula

First, substitute the given values \(\mu=66\), \(\sigma=7\), and \(n=49\) into the formula. This will result in \(z = \frac{{\bar{x} - 66}}{{7 / \sqrt{49}}}\). Now, we will substitute each \(\bar{x}\) into this formula to find corresponding z-values.
02

Calculate z for \(\bar{x}=68.44\)

Substitute \(\bar{x}=68.44\) into the formula. It becomes \(z = \frac{{68.44 - 66}}{{7 / \sqrt{49}}}\). Solving this will give the z-score for \(\bar{x}=68.44\).
03

Calculate z for \(\bar{x}=58.75\)

Next, substitute \(\bar{x}=58.75\) into the formula. We get \(z = \frac{{58.75 - 66}}{{7 / \sqrt{49}}}\). Solving this will give the z-score for \(\bar{x}=58.75\).
04

Calculate z for \(\bar{x}=62.35\)

Next, replace \(\bar{x}=62.35\) into the formula. It will become \(z = \frac{{62.35 - 66}}{{7 / \sqrt{49}}}\). Solving this will provide the z-score for \(\bar{x}=62.35\).
05

Calculate z for \(\bar{x}=71.82\)

Finally, substitute \(\bar{x}=71.82\) into the formula. So, it becomes \(z = \frac{{71.82 - 66}}{{7 / \sqrt{49}}}\). Solving this will give the z-score for \(\bar{x}=71.82\).

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Key Concepts

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

Understanding Population Mean
The population mean, denoted by \(\mu\), is the average of all the data points in a population. It represents the central value around which data points are distributed. For example, in our exercise, the population mean is 66, which implies that, on average, the numbers in this whole population circle around this value.
To find the population mean, you add up all the data points and divide by the total number of points. It provides a snapshot of the population as a whole and is different from the sample mean, which represents just a part of the population.

In a practical scenario:
  • The population mean helps in understanding the general tendency of a large group.
  • It serves as a useful tool for comparing different populations or subgroups.
Always remember, the population mean gives you the big picture, not the minute details.
Significance of Standard Deviation
Standard deviation, symbolized as \(\sigma\), measures how much individual data points deviate, on average, from the population mean. In our exercise, the standard deviation is 7, meaning each data point varies by 7 units from the mean value of 66, on average.
Understanding standard deviation is crucial because:
  • It helps in assessing the spread of the data around the mean.
  • Larger standard deviations indicate more spread out data points, while smaller deviations suggest data points are clumped together.
Standard deviation can tell us about the reliability of the mean. A small standard deviation indicates that the mean is a good representation of the data. In contrast, a large standard deviation might suggest more variability and less reliability.
The Role of Sample Size
The sample size, denoted by \(n\), refers to the number of observations or data points used in the sample from the population. In our case, the sample size is 49. It's a critical factor in statistical analysis because it influences the accuracy and reliability of the conclusions drawn from the data.

Let's explore why sample size matters:
  • Larger sample sizes generally lead to more precise estimates of the population parameters, like the mean and standard deviation.
  • It reduces the margin of error and increases the power of a statistical test.
A well-chosen sample size can represent the entire population accurately, providing useful insights without examining every single data point in the whole population.
Understanding Normal Distribution
Normal distribution is a statistical concept where data is symmetrically distributed, forming the familiar 'bell curve'. In our scenario, if we imagine plotting the data points, they would cluster more at the mean (66) and taper off as you move away.

Its characteristics include:
  • The mean, median, and mode are all equal and located at the center.
  • Data symmetrically spreads about the mean, with predictable percentages within each standard deviation.
Understanding normal distribution is crucial because it helps in making various inferences about the population.
Due to its commonality, many statistical methods and testing procedures use normal distribution as a basis for inference.

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Most popular questions from this chapter

The credit card debts of all college students have a distribution that is skewed to the right with a mean of \(\$ 2840\) and a standard deviation of \(\$ 672\). Find the probability that the mean credit card debt for ? random sample of 36 college students would be a. between \(\$ 2600\) and \(\$ 2950\) b. less than \(\$ 3060\)

According to a 2008 survey by the Royal Society of Chemistry. \(30 \%\) of adults in Great Britain said that they typically run the water for a period of 6 to 10 minutes while they take a shower (http://www.rsc.org/AboutUs/News/PressReleases/2008/EuropeanShowerHabits.asp). Assume that this percentage is true for the current population of adults in Great Britain. Let \(\hat{p}\) be the proportion in a random sample of 180 adults from Great Britain who typically run the water for a period of 6 to 10 minutes while they take a shower. Find the mean and standard deviation of the sampling distribution of \(\hat{p}\) and describe its shape.

A population of \(N=4000\) has a population proportion equal to \(.12 .\) In each of the following cases, which formula will you use to calculate \(\sigma_{\tilde{p}}\) and why? Using the appropriate formula, calculate \(\sigma_{j}\) for each of these cases. a. \(n=800\) b. \(n=30\)

An article in the Daily Herald of Everett, Washington, noted that the average cost of going to a minor league baseball game for a family of four was \(\$ 55\) in 2009 (http://www.heraldnet.com/article/ 20090412/BIZ/704129929/1006/SPORTS03). Suppose that the standard deviation of such costs is \$13.25. Let \(\bar{x}\) be the average cost of going to a minor league baseball game for 33 randomly selected families of four in 2009 . Find the mean and the standard deviation of the sampling distribution of \(\bar{x}\).

A certain elevator has a maximum legal carrying capacity of 6000 pounds. Suppose that the population of all people who ride this elevator have a mean weight of 160 pounds with a standard deviation of 25 pounds. If 35 of these people board the elevator, what is the probability that their combined wcight will exceed 6000 pounds? Assume that the 35 people constitute a random sample from the population.
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