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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 Z critical values for the given confidence levels are approximately as follows: For a 95% confidence level, the Z critical value is 1.96. For an 80% confidence level, the Z critical value is 1.28. For a 90% confidence level, the Z critical value is 1.645. For an 85% confidence level, the Z critical value is 1.44. For a 99% confidence level, the Z critical value is 2.576.

Step by step solution

01

Identify the confidence levels

The exercise gives the following confidence levels: 95%, 80%, 90%, 85%, and 99%.
02

Look up the critical values

We will need to check a standard normal distribution table or use the percent point function (quantile function) to get the Z scores. In a normal distribution table, for a 95% confidence level, look at the value where an area of 0.975 lies from the mean. For an 80% confidence level, look at the value where an area of 0.90 lies from the mean. Similar for other confidence levels, we look at the value where an area of 0.95, 0.925 and 0.995 lies from the mean respectively. These will give us the Z scores corresponding to our confidence levels.
03

Record the z critical values

After you have found the relevant Z scores, record them as the Z critical values.

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Key Concepts

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

Z Critical Value
A Z critical value is a key component in the field of statistics that represents the number of standard deviations a data point must be from the mean to fall within a certain percentile in a standard normal distribution. When calculating confidence intervals, which provide a range of values that likely contain a population parameter, the Z critical value helps to define the width of the interval based on the confidence level chosen.

To determine a Z critical value, one must first decide on the desired confidence level—such as 95%, 90%, or 99%—which indicates how certain one is that the true population parameter falls within the confidence interval. The Z critical value corresponds to the confidence level and is found using the standard normal distribution. The higher the confidence level, the larger the Z critical value will be, resulting in a wider interval to increase the likelihood that it contains the true population parameter.

For example, the Z critical values for common confidence levels are approximately as follows:
  • 1.96 for 95%
  • 1.28 for 90%
  • 2.58 for 99%
These values are used in the formula provided in the exercise to calculate the margin of error, which is then used to construct the confidence interval around the sample proportion (\(\text{p}\)).
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution, which is one of the most important probability distributions in statistics due to its natural appearance in many biological, financial, and social phenomena. A standard normal distribution is a normal distribution with a mean of zero and a standard deviation of one.

In any standard normal distribution, Z-scores play a significant role. A Z-score represents how many standard deviations away a data point is from the mean. Because the standard normal distribution is symmetrical, half of the scores lie above the mean and half below. The area under the curve represents the probability of a score falling within a certain range, and tables or statistical software are often used to find the area, which corresponds to the percentile of the Z-score.

For example, a Z-score of 1.96 corresponds to the 97.5th percentile, meaning that 97.5% of the data falls at or below this Z-score in a standard normal distribution. This is crucial for calculating the confidence intervals as seen in the exercise, because it helps to determine the range of values within which we can be 'confident' that the population parameter lies.
Hypothesis Testing
In statistics, hypothesis testing is a method used to determine if there is enough statistical evidence in a sample of data to infer that a certain condition is true for the entire population. It involves making an initial assumption, known as the null hypothesis, and then using sample data to assess the likelihood that the null hypothesis is true.

To perform hypothesis testing, statistics like the Z-score are calculated from the sample data. One then determines the p-value, which is the probability of observing the sample statistic or something more extreme if the null hypothesis is true. The Z critical value assists in this decision-making process. If the calculated statistic from the sample data is more extreme than the Z critical value (lying in the rejection region), then the null hypothesis is rejected in favor of the alternative hypothesis.

For instance, if a null hypothesis states that the mean test score for a national test is 50, and hypothesis testing yields a Z-score far from zero, the confidence interval calculated with the relevant Z critical value may not contain 50, leading to the rejection of the null hypothesis in favor of the assumption that the actual mean score differs from 50.

Hypothesis testing is a cornerstone of statistical analysis and is deeply interconnected with the concepts of Z critical values and confidence intervals, playing a critical role in fields ranging from scientific research to quality control in manufacturing.

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Most popular questions from this chapter

The article "Nine Out of Ten Drivers Admit in Survey to Having Done Something Dangerous" (Knight Ridder Newspapers, July 8,2005 ) reported the results of a survey of 1100 drivers. Of those surveyed, 990 admitted to careless or aggressive driving during the previous 6 months. Assuming that it is reasonable to regard this sample of 1100 as representative of the population of drivers, use this information to construct a \(99 \%\) confidence interval to estimate \(p\), the proportion of all drivers who have engaged in careless or aggressive driving in the previous 6 months.

The formula used to compute a confidence interval for the mean of a normal population when \(n\) is small is $$\bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}}$$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(95 \%\) confidence, \(n=17\) b. \(90 \%\) confidence, \(n=12\) c. \(99 \%\) confidence, \(n=24\) d. \(90 \%\) confidence, \(n=25\) e. \(90 \%\) confidence, \(n=13\) f. \(95 \%\) confidence, \(n=10\)

Based on a representative sample of 511 U.S. teenagers age 12 to 17 , International Communications Research estimated that the proportion of teens who support keeping the legal drinking age at 21 is \(\hat{p}=0.64\) \((64 \%)\). The press release titled "Majority of Teens (Still) Favor the Legal Drinking Age" (www.icrsurvey.com. January 21, 2009) also reported a margin of error of \(0.04(4 \%)\) for this estimate. Show how the reported value for the margin of error was computed.

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, \(\begin{array}{lll}\text { Recording, } & \text { Videotaping-and } & \text { Firing-Employees" }\end{array}\) (American Management Association, 2005) summarized the results of a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet and 131 had fired workers for e-mail misuse. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of the Internet. b. What are two reasons why a \(90 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of e-mail would be narrower than the \(95 \%\) confidence interval computed in Part (a)?

Acrylic bone cement is sometimes used in hip and knee replacements to fix an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens under specified conditions, and the resulting mean and standard deviation were \(306.09\) Newtons and \(41.97\) Newtons, respectively. Assuming that it is reasonable to believe that breaking force under these conditions has a distribution that is approximately normal, estimate the mean breaking force for acrylic bone cement under the specified conditions using a \(95 \%\) confidence interval.

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