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Chapter 9: Problem 9

Determine the critical value \(z_{\alpha / 2}\) that corresponds to the given level of confidence. \(98 \%\)

Short Answer

Expert verified
The critical value \(z_{\alpha / 2}\) for a 98% confidence level is approximately 2.33.

Step by step solution

01

Identify the given confidence level

The given level of confidence is 98%. This means we are 98% confident that our interval estimate will contain the true population parameter.
02

Convert confidence level to significance level

The significance level \(\alpha\) is calculated as \(\alpha = 1 - \text{confidence level}\). So, \alpha = 1 - 0.98 = 0.02\.
03

Divide the significance level by 2

Since the critical value \(z_{\alpha / 2}\) corresponds to the area in each tail of the standard normal distribution curve, divide the significance level by 2: \alpha / 2 = 0.02 / 2 = 0.01\.
04

Find the critical value

Using the standard normal distribution table or a calculator, find the critical value that corresponds to an area of 0.01 in the tail. The critical value \(z_{\alpha / 2}\) for \alpha / 2 = 0.01\ is approximately 2.33.

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Key Concepts

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

confidence level
The confidence level is a crucial concept in statistics that shows how confident we are in a statistical estimate. In other words, it tells us how often we can expect our interval estimate to contain the true population parameter if we were to take many samples.
The confidence level is typically represented as a percentage. Common confidence levels include 90%, 95%, and 98%.
  • A 98% confidence level means we can be 98% sure our interval will include the true population parameter. There is a 2% chance it will not.
  • The higher the confidence level, the more confident we are in our estimate. However, higher confidence levels usually result in wider intervals.
Understanding confidence levels helps us decide how precise we want our estimates to be.
significance level
In statistics, the significance level \alpha\ (alpha) represents the probability of rejecting the null hypothesis when it is actually true. It is defined as \( \alpha = 1 - \text{confidence level} \) .

In the given exercise, the confidence level is 98%, so:

\( \alpha = 1 - 0.98 = 0.02 \)

This means there is a 2% chance of making a mistake in rejecting a true null hypothesis.

  • The significance level indicates how stringent we are with our hypothesis testing. A lower alpha means more stringent criteria.
  • For instance, a 0.05 significance level is commonly used, leading to a 5% risk of concluding that an effect exists when it does not.
Knowing the significance level helps us understand the risk involved in our statistical hypothesis tests.
standard normal distribution
The standard normal distribution is a special type of normal distribution with a mean of 0 and a standard deviation of 1. It is crucial for statistical calculations because it provides a common framework.

The standard normal distribution is symmetrical and has a bell shape. The total area under the curve is 1, representing the total probability.
  • The z-values in this distribution tell us how many standard deviations a point is from the mean.
  • Critical values in hypothesis testing are often determined using the standard normal distribution.
In the exercise, we use the standard normal distribution table to find the critical value that corresponds to a specific significance level.
critical value calculation
Calculating the critical value is an essential step in hypothesis testing that defines the boundary for deciding whether to reject the null hypothesis.
Here's how we determine the critical value in the given exercise:
  • The confidence level is 98%, meaning \( \alpha = 0.02 \).
  • We then divide the significance level by 2 to get \alpha / 2 = 0.01\.
  • Next, we use a standard normal distribution table or a calculator to find the z-value for \alpha / 2 = 0.01 \.
  • The critical value, \( z_{\alpha / 2} = 2.33 \), tells us where our data would fall to be considered unusual compared to a normal population.
This critical value helps us make decisions in statistical tests, indicating the point beyond which we reject the null hypothesis.

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Most popular questions from this chapter

True or False: To construct a confidence interval about the mean, the population from which the sample is drawn must be approximately normal.

A USA Today/Gallup poll asked 1006 adult Americans how much it would bother them to stay in a room on the 13 th floor of a hotel. Interestingly, \(13 \%\) said it would bother them. The margin of error was 3 percentage points with \(95 \%\) confidence. Which of the following represents a reasonable interpretation of the survey results? For those not reasonable, explain the flaw. (a) We are \(95 \%\) confident that the proportion of adult Americans who would be bothered to stay in a room on the 13th floor is between 0.10 and 0.16 . (b) We are between \(92 \%\) and \(98 \%\) confident that \(13 \%\) of adult Americans would be bothered to stay in a room on the 13th floor. (c) In \(95 \%\) of samples of adult Americans, the proportion who would be bothered to stay in a room on the 13 th floor is between 0.10 and 0.16 . (d) We are \(95 \%\) confident that \(13 \%\) of adult Americans would be bothered to stay in a room on the 13 th floor.

Travelers pay taxes for flying, car rentals, and hotels. The following data represent the total travel tax for a 3-day business trip in eight randomly selected cities. Note: Chicago has the highest travel taxes in the country at 101.27 dollar. In Problem 32 from Section \(9.2,\) it was verified that the data are normally distributed and that \(s=12.324\) dollars. Construct and interpret a \(90 \%\) confidence interval for the standard deviation travel tax for a 3 -day business trip. $$ \begin{array}{llll} \hline 67.81 & 78.69 & 68.99 & 84.36 \\ \hline 80.24 & 86.14 & 101.27 & 99.29 \\ \hline \end{array} $$

A simple random sample of size \(n<30\) for \(a\) quantitative variable has been obtained. Using the normal probability plot, the correlation between the variable and expected z-score, and the boxplot, judge whether a t-interval should be constructed. $$ n=9 ; \text { Correlation }=0.997 $$

True or False: The shape of the chi-square distribution depends on its degrees of freedom.
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