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Chapter 5: Problem 34

Find the \(z^{*}\) values based on a standard normal distribution for each of the following. (a) An \(80 \%\) confidence interval for a proportion. (b) An \(84 \%\) confidence interval for a slope. (c) A \(92 \%\) confidence interval for a standard deviation.

Short Answer

Expert verified
The three \(z^{*}\) values for the specified confidence intervals are: 1.28 for 80%, 1.41 for 84%, and 1.75 for 92%.

Step by step solution

01

- Find \(z^{*}\) value for 80% confidence interval

In the standard normal distribution table, find the z-score corresponding to the area above 10% and below 90% (since we set the middle 80% as our confidence interval). After consulting the table, this \(z^{*}\) value for an 80% confidence interval is found to be approximately 1.28.
02

- Find \(z^{*}\) value for 84% confidence interval

Similarly, for an 84% confidence interval, find the z-score corresponding to the area above 8% and below 92%. The calculated \(z^{*}\) value in this case is approximately 1.41.
03

- Find \(z^{*}\) value for 92% confidence interval

Lastly, for a 92% confidence interval, find the z-score in the table which corresponds to the area above 4% and below 96%. The calculated \(z^{*}\) value in this case is approximately 1.75.

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Key Concepts

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

Z-Score
Understanding the z-score is fundamental in statistics. A z-score is a numerical measurement that describes a value's relationship to the mean of a group of values.

Z-score is measured in terms of standard deviations from the mean. If a z-score is 0, it indicates that the data point's score is identical to the mean score. A z-score of 1.0 signifies that the data point is one standard deviation above the mean. Likewise, a z-score of -1.0 signifies that the data point is one standard deviation below the mean. This metric is crucial when calculating confidence intervals as it helps determine how many standard deviations an element is away from the mean.
Standard Normal Distribution
The standard normal distribution, also known as the z-distribution, is a special kind of normal distribution where the mean is 0 and the standard deviation is 1. It's used to determine the likelihood that a statistic falls within a given range.

A key feature of the standard normal distribution is its symmetrical shape, where approximately 68% of scores fall within one standard deviation of the mean, about 95% fall within two standard deviations, and almost all (99.7%) lie within three standard deviations. This distribution is integral in calculating z-scores and their corresponding confidence intervals.
Proportion Confidence Interval
The proportion confidence interval is a type of interval estimate used to infer about a population proportion. It's based on the principle that a sample proportion, such as the percentage of people who favor a specific policy, can serve as a point estimate for what the proportion is in the larger population.

To compute this interval, statisticians use the standard error of the proportion and the z-score associated with the desired level of confidence. The importance of this interval lies in its ability to provide a range of plausible values for the population proportion, with a specified level of confidence that this range contains the true proportion.
Slope Confidence Interval
Within the context of regression analysis, the slope confidence interval helps us understand the potential range for the slope coefficient in the population from which our sample was drawn.

The interval estimate lets us assert, with a particular confidence level, that the true slope lies within this interval. Calculating this interval involves using the standard error of the slope estimate, and a t-distribution or standard normal distribution鈥檚 z-score, depending on sample size. This is critical for understanding the relationship between variables in a regression model.
Standard Deviation Confidence Interval
Finally, a standard deviation confidence interval provides a range of values that is likely to include the population standard deviation. Unlike the means or proportions, the standard deviation confidence intervals are calculated slightly differently because the data must conform to a Chi-square distribution when the population standard deviation is unknown.

This interval is important for quality control, risk assessment, and any statistical context where variability is of interest. In terms of our textbook problem, the calculated z-score of 1.75 for a 92% confidence interval corresponds to the level of certainty that we have in this interval estimate for the population's standard deviation.

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Most popular questions from this chapter

Where Is the Best Seat on the Plane? A survey of 1000 air travelers \(^{21}\) found that \(60 \%\) prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is \(S E=0.015 .\) Use a normal distribution to find and interpret a \(99 \%\) confidence interval for the proportion of air travelers who prefer a window seat.

Incentives for Quitting Smoking: Do They Work? Exercise 5.31 describes a study examining incentives to quit smoking. With no incentives, the proportion of smokers trying to quit who are still abstaining six months later is about 0.06 . Participants in the study were randomly assigned to one of four different incentives, and the proportion successful was measured six months later. Of the 498 participants in the group with the least success, 47 were still abstaining from smoking six months later. We wish to test to see if this provides evidence that even the smallest incentive works better than the proportion of 0.06 with no incentive at all. (a) State the null and alternative hypotheses, and give the notation and value of the sample statistic. (b) Use a randomization distribution and the observed sample statistic to find the p-value. (c) Give the mean and standard error of the normal distribution that most closely matches the randomization distribution, and then use this normal distribution with the observed sample statistic to find the p-value. (d) Use the standard error found from the randomization distribution in part (b) to find the standardized test statistic, and then use that test statistic to find the p-value using a standard normal distribution. (e) Compare the p-values from parts (b), (c), and (d). Use any of these p-values to give the conclusion of the test.

How Often Do You Use Cash? In a survey \(^{13}\) of 1000 American adults conducted in April 2012 . \(43 \%\) reported having gone through an entire week without paying for anything in cash. Test to see if this sample provides evidence that the proportion of all American adults going a week without paying cash is greater than \(40 \%\). Use the fact that a randomization distribution is approximately normally distributed with a standard error of \(S E=0.016 .\) Show all details of the test and use a \(5 \%\) significance level.

Penalty Shots in World Cup Soccer A study \(^{11}\) of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only \(41 \%\) of the time. The article notes that this is "slightly worse than random chance." We use these data as a sample of all World Cup penalty shots ever. Test at a \(5 \%\) significance level to see whether there is evidence that the percent guessed correctly is less than \(50 \%\). The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is \(S E=0.043 .\)

To Study Effectively, Test Yourself! Cognitive science consistently shows that one of the most effective studying tools is to self-test. A recent study \(^{10}\) reinforced this finding. In the study, 118 college students studied 48 pairs of Swahili and English words. All students had an initial study time and then three blocks of practice time. During the practice time, half the students studied the words by reading them side by side, while the other half gave themselves quizees in which they were shown one word and had to recall its partner. Students were randomly assigned to the two groups, and total practice time was the same for both groups, On the final test one week later, the proportion of items correctly recalled was \(15 \%\) for the reading-study group and \(42 \%\) for the self-quiz group. The standard error for the difference in proportions is about 0.07 . Test whether giving self-quizzes is more effective and show all details of the test. The sample size is large enough to use the normal distribution.
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