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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

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. $$ n=13 ; \text { Correlation }=0.966 $$

The following small data set represents a simple random sample from a population whose mean is \(50 .\) $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline \end{array} $$ (a) A normal probability plot indicates that the data could come from a population that is normally distributed with no outliers. Compute a \(95 \%\) confidence interval for this data set. (b) Suppose that the observation, 41 , is inadvertently entered into the computer as 14 . Verify that this observation is an outlier (c) Construct a \(95 \%\) confidence interval on the data set with the outlier. What effect does the outlier have on the confidence interval? (d) Consider the following data set, which represents a simple random sample of size 36 from a population whose mean is 50\. Verify that the sample mean for the large data set is the same as the sample mean for the small data set from part (a). $$ \begin{array}{llllll} \hline 43 & 63 & 53 & 50 & 58 & 44 \\ \hline 53 & 53 & 52 & 41 & 50 & 43 \\ \hline 47 & 65 & 56 & 58 & 41 & 52 \\ \hline 49 & 56 & 57 & 50 & 38 & 42 \\ \hline 59 & 54 & 57 & 41 & 63 & 37 \\ \hline 46 & 54 & 42 & 48 & 53 & 41 \\ \hline \end{array} $$ (e) Compute a \(95 \%\) confidence interval for the large data set. Compare the results to part (a). What effect does increasing the sample size have on the confidence interval? (f) Suppose that the last observation, 41 , is inadvertently entered as 14 . Verify that this observation is an outlier. (g) Compute a \(95 \%\) confidence interval for the large data set with the outlier. Compare the results to part (e). What effect does an outlier have on a confidence interval when the data set is large?

A school administrator is concerned about the amount of credit-card debt that college students have. She wishes to conduct a poll to estimate the percentage of full-time college students who have credit-card debt of \(\$ 2000\) or more. What size sample should be obtained if she wishes the estimate to be within 2.5 percentage points with \(94 \%\) confidence if (a) a pilot study indicates that the percentage is \(34 \% ?\) (b) no prior estimates are used?

Construct the appropriate confidence interval. A simple random sample of size \(n=25\) is drawn from a population that is normally distributed. The sample variance is found to be \(s^{2}=3.97\). Construct a \(95 \%\) confidence interval for the population standard deviation.

In March 2014, Harris Interactive conducted a poll of a random sample of 2234 adult Americans 18 years of age or older and asked, "Which is more annoying to you, tailgaters or slow drivers who stay in the passing lane?" Among those surveyed, 1184 were more annoyed by tailgaters. (a) Explain why the variable of interest is qualitative with two possible outcomes. What are the two outcomes? (b) Verify the requirements for constructing a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane. (c) Construct a \(90 \%\) confidence interval for the population proportion of all adult Americans who are more annoyed by tailgaters than slow drivers in the passing lane.
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