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Chapter 9: Problem 11

The formula used to compute a large-sample confidence interval for \(p\) is $$ \hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

Short Answer

Expert verified
The appropriate z critical values are: For 95% confidence level it is approximately \(1.96\), for 80% confidence level it is approximately \(1.28\), for 90% confidence level it is approximately \(1.645\), for 85% confidence level it is approximately \(1.44\), and for 99% confidence level it is approximately \(2.575\).

Step by step solution

01

Finding the Critical Value for 95% Confidence Level

For a 95% confidence level, look for the z value in the standard normal distribution table such that the area between \(-z\) and \(z\) is 0.95. This translates to an area of 0.475 (since 0.95/2 = 0.475) above and below the mean in a standard normal distribution table. The corresponding z value is approximately \(1.96\).
02

Finding the Critical Value for 80% Confidence Level

For an 80% confidence level, look for the z value such that the area between \(-z\) and \(z\) is 0.80. This translates to an area of 0.40 (since 0.80/2 = 0.40) above and below the mean in a standard normal distribution table. The corresponding z value is approximately \(1.28\).
03

Finding the Critical Value for 90% Confidence Level

For a 90% confidence level, look for the z value such that the area between \(-z\) and \(z\) is 0.90. This translates to an area of 0.45 (since 0.90/2 = 0.45) above and below the mean in a standard normal distribution table. The corresponding z value is approximately \(1.645\).
04

Finding the Critical Value for 85% Confidence Level

For an 85% confidence level, look for the z value such that the area between \(-z\) and \(z\) is 0.85. This translates to an area of 0.425 (since 0.85/2 = 0.425) above and below the mean in a standard normal distribution table. The corresponding z value is approximately \(1.44\).
05

Finding the Critical Value for 99% Confidence Level

For a 99% confidence level, look for the z value such that the area between \(-z\) and \(z\) is 0.99. This translates to an area of 0.495 (since 0.99/2 = 0.495) above and below the mean in a standard normal distribution table. The corresponding z value is approximately \(2.575\).

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Key Concepts

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

Z Critical Value
The z critical value is crucial when constructing confidence intervals, especially in large sample sizes. It stems from the concept of the standard normal distribution that we use to determine how many standard deviations away from the mean our test statistic is. This value helps ensure the confidence interval accurately reflects the desired probability coverage. For example, if you need a 95% confidence level, you would use a z critical value of approximately 1.96.

Here's why:
  • Confidence intervals are derived from the probability that a sample statistic falls within a certain range of the population parameter.
  • The z critical value corresponds to the particular percent of this interval, based on the Normal Distribution.
  • Depending on the confidence level, different z critical values are utilized:
    • For 80% confidence level: z is approximately 1.28
    • For 85% confidence level: z is approximately 1.44
    • For 90% confidence level: z is approximately 1.645
    • For 95% confidence level: z is approximately 1.96
    • For 99% confidence level: z is approximately 2.575
Understanding these values greatly benefits when aiming for precision in statistical estimations.
Standard Normal Distribution
The Standard Normal Distribution is a cornerstone in statistics. It is a normal distribution with a mean of 0 and a standard deviation of 1. This distribution is fundamental because it provides a way to understand how sample data aligns with expected patterns. When statisticians assume a standard normal distribution, they can convert any normal random variable into a standard score (or z-score).

  • The importance of the standard normal distribution:
    • It allows for a uniform method to determine the probability of a statistic being within a certain range.
    • The z-score, formulated as \( z = \frac{(X - \mu)}{\sigma} \), measures how many standard deviations an element is from the mean.
  • Using the standard normal distribution aids in finding areas under the curve, which correspond to probabilities.
This concept is especially helpful in constructing confidence intervals, as it provides a clear picture of where data points lie in relation to the population mean.
Large-Sample Confidence Interval
A large-sample confidence interval is a method used to estimate a population parameter from a sample. The beauty of this method lies in its simplicity and the fact that it becomes more accurate as the sample size grows. It's particularly reliable when the sample size is large, often defined as 30 or more observations.

Some key points to remember about large-sample confidence intervals:
  • The formula used: \[ \hat{p} \pm z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] where:
    • \( \hat{p} \) is the sample proportion
    • \( z \) is the z critical value corresponding to the desired confidence level
    • \( n \) is the sample size
  • It assumes a normal distribution of the sample mean due to the Central Limit Theorem.
  • This confidence interval effectively provides a range of plausible values for the population proportion.
Understanding the concept of large-sample confidence intervals can enhance one's ability to interpret statistical data accurately in research and real-world applications.

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Most popular questions from this chapter

The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21,2006\()\) reported that \(37 \%\) of college freshmen and \(48 \%\) of college seniors carry a credit card balance from month to month. Suppose that the reported percentages were based on random samples of 1000 college freshmen and 1000 college seniors. a. Construct a \(90 \%\) confidence interval for the proportion of college freshmen who carry a credit card balance from month to month. b. Construct a \(90 \%\) confidence interval for the proportion of college seniors who carry a credit card balance from month to month. c. Explain why the two \(90 \%\) confidence intervals from Parts (a) and (b) are not the same width.

For each of the following choices, explain which would result in a wider large-sample confidence interval for \(p\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

Each person in a random sample of 20 students at a particular university was asked whether he or she is registered to vote. The responses \((\mathrm{R}=\) registered, \(\mathrm{N}=\) not registered) are given here: R \(R N R N N R R R N R R R R R N R R R N\) Use these data to estimate \(p\), the proportion of all students at the university who are registered to vote.

In a study of academic procrastination, the authors of the paper "Correlates and Consequences of Behavioral Procrastination" (Procrastination, Current Issues and New Directions [20001) reported that for a sample of 411 undergraduate students at a midsize public university preparing for a final exam in an introductory psychology course, the mean time spent studying for the exam was 7.74 hours and the standard deviation of study times was 3.40 hours. For purposes of this exercise, assume thar it is reasonable to regard this sample as represcntative of students taking introductory psychology at this university. a. Construct a \(95 \%\) confidence interval to estimate \(\mu\), the mean time spent studying for the final exam for students taking introductory psychology at this university. b. The paper also gave the following sample statistics for the percentage of study time that occurred in the 24 hours prior to the exam: \(n=411 \quad \bar{x}=43.18 \quad s=21.46\) Construct and interpret a \(90 \%\) confidence interval for the mean percentage of study time that occurs in the 24 hours prior to the exam.

The Gallup Organization conducts an annual survey on crime. It was reported that \(25 \%\) of all households experienced some sort of crime during the past year. This estimate was based on a sample of 1002 randomly selected households. The report states, "One can say with \(95 \%\) confidence that the margin of sampling error is 卤3 percentage points." Explain how this statement can be justified.
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