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Chapter 8: Problem 10

Find \(z\) for each of the following confidence levels. a. \(90 \%\) b. \(95 \%\) c. \(96 \%\) d. \(97 \%\) e. \(98 \%\) f. \(99 \%\)

Short Answer

Expert verified
The z-scores for the confidence levels are: \n a. 90% = 1.645 \n b. 95% = 1.96 \n c. 96% = 2.05 \n d. 97% = 2.17 \n e. 98% = 2.33 \n f. 99% = 2.57

Step by step solution

01

Understand the concept of z-score and confidence level

The z-score represents the number of standard deviations away from the mean for a data point in a normal distribution. The confidence level is a percentage that quantifies the level of confidence that the statistical measure falls within a set range of values.
02

Use the Standard Normal Distribution Table

Since the exercise requires to find the z-score for different confidence levels, the z-table (or standard normal distribution table) will be used. This table shows the cumulative probabilities for a random variable Z, following the standard normal distribution. It's split into two parts - the positive z-score and the negative z-score. For calculating the z-score of a confidence level C, the z-score will be the value for which the cumulative probability is \((1+C)/2\$.
03

Find the z-scores

a. For 90% confidence level, find the z-score such that the cumulative probability equals \((1+0.9)/2 = 0.95\). From the z-table, the z-score corresponding to 0.95 is approximately 1.645. b. For 95% confidence level, find the z-score such that the cumulative probability equals \((1+0.95)/2 = 0.975\). From the z-table, the z-score corresponding to 0.975 is approximately 1.96. c. For 96% confidence level, find the z-score such that the cumulative probability equals \((1+0.96)/2 = 0.98\). From the z-table, the z-score corresponding to 0.98 is approximately 2.05. d. For 97% confidence level, find the z-score such that the cumulative probability equals \((1+0.97)/2 = 0.985\). From the z-table, the z-score corresponding to 0.985 is approximately 2.17. e. For 98% confidence level, find the z-score such that the cumulative probability equals \((1+0.98)/2 = 0.990\). From the z-table, the z-score corresponding to 0.990 is approximately 2.33. f. For 99% confidence level, find the z-score such that the cumulative probability equals \((1+0.99)/2 = 0.995\). From the z-table, the z-score corresponding to 0.995 is approximately 2.57.

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Key Concepts

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

Understanding the Z-Score
The z-score is a very useful concept in statistics, especially when you're dealing with normal distributions. A z-score is a measure that describes a data point's position in terms of standard deviations from the mean of a dataset. In simpler terms, it tells you how far and in which direction a data point is from the mean.
For example, a z-score of 2 means the data point is 2 standard deviations above the mean, while a z-score of -2 means 2 standard deviations below the mean.
  • This makes it possible to compare scores from different normal distributions.
  • This is crucial for determining how rare or common a particular data point is within a distribution.
  • Z-scores allow for meaningful comparisons between data points from different distributions, regardless of the original units.
The Standard Normal Distribution
The standard normal distribution is a type of continuous probability distribution that is symmetrical around its mean and forms the shape of a bell curve. This is why it is often referred to as the 'bell curve.' It has some important features:
  • The mean of the standard normal distribution is 0.
  • Its standard deviation is 1.
  • It is used as a reference to convert raw scores into z-scores, which allows for comparison across different datasets.

Using the standard normal distribution, you can predict how likely a z-score will occur by using the properties of the curve. The further the z-score is from 0, the less likely the corresponding data point is to occur by chance. This property is vital for hypothesis testing and constructing confidence intervals.
Exploring Cumulative Probability
Cumulative probability is the probability that a random variable is less than or equal to a given value. In the context of a standard normal distribution, cumulative probability allows us to find the likelihood that a random drawing from the population will have a z-score less than or equal to a certain value.
  • This concept is central to the functioning of z-tables, which list cumulative probabilities for different z-values.
  • To find a cumulative probability, you look for the z-score in the table, and the cumulative probability is given at the corresponding table location.
  • Cumulative probability helps in assessing the overall probability space covered by a certain range of z-scores.
This ability to measure and calculate cumulative probabilities is essential in deriving statistical insights and making informed decisions in statistics.
Decoding the Confidence Level
The confidence level represents how confident we can be that a particular interval contains the population parameter, often the mean. It is expressed as a percentage, which shows the proportion of times you expect this interval to contain the population parameter if you were to repeat your study many times.
  • Higher confidence levels mean there's a higher probability that the interval will contain the population parameter, but they also result in a wider interval.
  • Common confidence levels are 90%, 95%, and 99%.
  • To compute the related z-score, you can use the formula: \[(1 + \text{confidence level}) / 2\], and then find this value in the z-table.
The choice of confidence level is often a trade-off between precision and reliability, and selecting the right level is crucial depending on the context of the analysis.

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Most popular questions from this chapter

A researcher wants to determine a \(99 \%\) confidence interval for the mean number of hours that adults spend per week doing community service. How large a sample should the researcher select so that the estimate is within \(1.2\) hours of the population mean? Assume that the standard deviation for time spent per week doing community service by all adults is 3 hours.

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\)

A bank manager wants to know the mean amount owed on credit card accounts that become delinquent. A random sample of 100 delinquent credit card accounts taken by the manager produced a mean amount owed on these accounts equal to \(\$ 2640\). The population standard deviation was \(\$ 578\). a. What is the point estimate of the mean amount owed on all delinquent credit card accounts at this bank? b. Construct a \(97 \%\) confidence interval for the mean amount owed on all delinquent credit card accounts for this bank.

a. A sample of 100 observations taken from a population produced a sample mean equal to \(55.32\) and a standard deviation equal to \(8.4 .\) Make a \(90 \%\) confidence interval for \(\mu .\) b. Another sample of 100 observations taken from the same population produced a sample mean equal to \(57.40\) and a standard deviation equal to \(7.5 .\) Make a \(90 \%\) confidence interval for \(\mu .\) c. A third sample of 100 observations taken from the same population produced a sample mean equal to \(56.25\) and a standard deviation equal to \(7.9 .\) Make a \(90 \%\) confidence interval for \(\mu .\) d. The true population mean for this population is \(55.80 .\) Which of the confidence intervals constructed in parts a through c cover this population mean and which do not?

A department store manager wants to estimate at a \(98 \%\) confidence level the mean amount spent by all customers at this store. The manager knows that the standard deviation of amounts spent by all customers at this store is \(\$ 31\). What sample size should he choose so that the estimate is within \(\$ 3\) of the population mean?
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