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Determine \(z_{\alpha}\) for the following values of \(\alpha\) : a. \(\alpha=.0055\) b. \(\alpha=.09\) c. \(\alpha=.663\)

Short Answer

Expert verified
a) \(z_{\alpha} = 2.55\); b) \(z_{\alpha} = 1.34\); c) \(z_{\alpha} = -0.42\).

Step by step solution

01

Understanding the Problem

To find the critical value \(z_{\alpha}\), we need to determine the point on the standard normal distribution curve for which the area to the right of \(z_{\alpha}\) is equal to the given \(\alpha\). The standard normal distribution is symmetric about zero and has a mean of zero and a standard deviation of one.
02

Finding \(z_{\alpha}\) for \(\alpha=0.0055\)

For \(\alpha=0.0055\), lookup the Z-table for the value closest to \(1 - 0.0055 = 0.9945\). The closest value is 0.9945 which corresponds to \(z = 2.55\). Therefore, \(z_{\alpha} = 2.55\).
03

Finding \(z_{\alpha}\) for \(\alpha=0.09\)

For \(\alpha=0.09\), lookup the Z-table for the value closest to \(1 - 0.09 = 0.91\). The closest value is 0.9100 which corresponds to \(z = 1.34\). Therefore, \(z_{\alpha} = 1.34\).
04

Finding \(z_{\alpha}\) for \(\alpha=0.663\)

For \(\alpha=0.663\), lookup the Z-table for the value closest to \(1 - 0.663 = 0.337\). The closest value is 0.3363 which corresponds to \(z = -0.42\). Therefore, \(z_{\alpha} = -0.42\).

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a key concept when it comes to statistics, especially in the context of statistical hypothesis testing and confidence intervals. It is essentially a normal distribution with a mean of zero and a standard deviation of one. This makes it a simple yet powerful model for representing data and probabilities.

What makes the standard normal distribution useful is its symmetry around zero, which means that the left side is a mirror image of the right side. This property allows us to easily calculate probabilities for ranges of data by looking up values in a Z-table.

Some important properties of the standard normal distribution include:
  • It is defined by the probability density function: \[ f(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \]
  • The total area under the curve is 1, indicating that it represents a complete set of probabilities.
  • Approximately 68% of values fall within one standard deviation of the mean, 95% fall within two, and 99.7% fall within three standard deviations.
When using the standard normal distribution, you can relate any other normal distribution back to it by converting other distributions to z-scores, which we'll cover later.
Critical Value
Critical values are essential in statistics for determining the threshold at which you decide to reject a null hypothesis. In the context of the standard normal distribution, a critical value is a point along the horizontal axis that marks the boundary for a specified tail area of the distribution.

To find a critical value, we generally follow these steps:
  • Specify the significance level (\( \alpha \)). This is the tail area beyond the critical value, or the probability of making a type I error.
  • Subtract \( \alpha \) from 1 to find the cumulative left area. For example, if \( \alpha = 0.05 \), the cumulative area is 0.95.
  • Use a Z-table or statistical software to find the z-score that corresponds to this cumulative area.
Critical values are especially useful for hypothesis tests and setting up confidence intervals, providing a statistically based "cutoff point" for decision making.
Z-score
A z-score is a statistical measurement that describes a data point's relationship to the mean of a group of points. Essentially, it tells you how many standard deviations away from the mean your data point is. The z-score is calculated using the formula:\[ z = \frac{X - \mu}{\sigma} \]Where:
  • \( X \) is the data point of interest.
  • \( \mu \) is the mean of the data set.
  • \( \sigma \) is the standard deviation of the data set.
Z-scores can be positive or negative, where a negative z-score indicates the data point is below the mean and a positive z-score means it is above the mean.

Using z-scores, we can normalize different data sets to make them comparable and use the standard normal distribution to find probabilities. For instance, in the exercise, we look up z-scores in the Z-table to find corresponding probabilities for different \( \alpha \) values. Understanding z-scores helps make sense of data variation and normal distribution curves.
Probability
Probability is a measure of the likelihood that an event will occur. It ranges from 0 to 1, where 0 indicates impossibility and 1 indicates certainty. In statistics, probability helps us make quantifiable predictions and inferences about data.

In the context of the standard normal distribution, probabilities help define critical z-scores and interpret statistical results. For example, with a confidence level of 95%, the probability that the z-score lies within the chosen z-scores (~1.96) is 0.95, leaving 0.05 divided equally in the tails.

When you look at z-scores in the Z-table, the values you see represent cumulative probabilities associated with z-scores on the standard normal distribution. These probabilities allow us to calculate the likelihood of a z-score occurring or to determine cut-off points for hypothesis testing.

In practical applications, understanding these probabilities assists in decision-making processes, such as determining whether observed results are statistically significant or could have occurred by random chance.

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Most popular questions from this chapter

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is \(.500\) in. A bearing is acceptable if its diameter is within \(.004\) in. of this target value. Suppose, however, that the setting has changed during the course of production, so that the bearings have normally distributed diameters with mean value 499 in. and standard deviation \(.002\) in. What percentage of the bearings produced will not be acceptable?

The article "Computer Assisted Net Weight Control" (Quality Progress, 1983: 22–25) suggests a normal distribution with mean \(137.2 \mathrm{oz}\) and standard deviation \(1.6 \mathrm{oz}\) for the actual contents of jars of a certain type. The stated contents was \(135 \mathrm{oz}\). a. What is the probability that a single jar contains more than the stated contents? b. Among ten randomly selected jars, what is the probability that at least eight contain more than the stated contents? c. Assuming that the mean remains at 137.2, to what value would the standard deviation have to be changed so that \(95 \%\) of all jars contain more than the stated contents?

Let \(\mathrm{Z}\) have a standard normal distribution and define a new rv \(Y\) by \(Y=\sigma Z+\mu\). Show that \(Y\) has a normal distribution with parameters \(\mu\) and \(\sigma\). [Hint: \(Y \leq y\) iff \(Z \leq\) ? Use this to find the cdf of \(Y\) and then differentiate it with respect to \(y\).]

The article 'Error Distribution in Navigation"' \((J\). of the Institute of Navigation, 1971: 429-442) suggests that the frequency distribution of positive errors (magnitudes of errors) is well approximated by an exponential distribution. Let \(X=\) the lateral position error (nautical miles), which can be either negative or positive. Suppose the pdf of \(X\) is $$ f(x)=(.1) e^{-.2|x|}-\infty

An individual's credit score is a number calculated based on that person's credit history that helps a lender determine how much he/she should be loaned or what credit limit should be established for a credit card. An article in the Los Angeles Times gave data which suggested that a beta distribution with parameters \(A=150, B=850\), \(\alpha=8, \beta=2\) would provide a reasonable approximation to the distribution of American credit scores. a. Let \(X\) represent a randomly selected American credit score. What are the mean value and standard deviation of this random variable? What is the probability that \(X\) is within 1 standard deviation of its mean value? b. What is the approximate probability that a randomly selected score will exceed 750 (which lenders consider a very good score)?

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