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Chapter 9: Problem 12

The formula used to compute a large-sample confidence interval for \(\pi\) is $$ p \pm(z \text { critical value }) \sqrt{\frac{p(1-p)}{n}} $$ 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: a. 1.96 for 95% confidence level, d. 1.28 for 80% confidence level, b. 1.64 for 90% confidence level, e. 1.44 for 85% confidence level, c. 2.57 for 99% confidence level.

Step by step solution

01

Identify Confidence Levels

For this exercise, the confidence levels provided are: 95%, 80%, 90%, 85%, and 99%.
02

Determine the Z Critical Value for each Confidence Level

The z-critical values represent standard deviations away from the mean (which is zero in a standard normal distribution). Z-value for a Upper Tail test can be found by subtracting the confidence level from 1, dividing by 2, and looking this probability up in Z-table:a. For a 95% confidence level, we need to find the z-value such that 95% of the distribution's area lies between -z and z. Subtract 0.95 from 1 yields 0.05, which divided by 2 equals 0.025. The z-value corresponding to 0.025 in the upper tail of the standard normal distribution table is 1.96.d. For an 80% confidence level, subtract 0.80 from 1 to get 0.20, which divided by 2 equals 0.10. The z-value corresponding to 0.10 in the upper tail is 1.28.b. For a 90% confidence level, subtract 0.90 from 1 to get 0.10, which divided by 2 equals 0.05. Find the z-value corresponding to 0.05 in the upper tail, which is 1.64.e. For an 85% confidence level, 1 - 0.85 equals 0.15, divided by 2 is 0.075. The z-value corresponding to 0.075 in the upper tail is 1.44.c. For a 99% confidence level, 1 - 0.99 equals 0.01, divided by 2 is 0.005. The z-value corresponding to 0.005 in the upper tail is 2.57.

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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 a crucial figure in statistics used to construct confidence intervals. It represents how many standard deviations away from the mean a data point is, within a standard normal distribution. The z critical value changes based on your chosen confidence level.

Here鈥檚 how you calculate it:
  • Subtract the confidence level from 1.
  • Divide the result by 2 to get the remaining probability in one tail of the distribution.
  • Find the corresponding z-value in a standard normal distribution table.
For example, a 95% confidence interval needs a z critical value of 1.96, which implies that 95% of the data is expected to fall within 1.96 standard deviations from the mean. Adjusting the confidence level changes the z critical value, as seen in the changes from 95% confidence level (1.96) to 99% (2.57), showing a wider range to increase certainty in estimation.
standard normal distribution
A standard normal distribution, often represented by a bell curve, is a critical concept in statistics. It has a mean of 0 and a standard deviation of 1. This distribution allows statisticians to determine the likelihood of a data point falling within a particular range.

Key aspects of the standard normal distribution include:
  • The bell-shaped curve, symmetric around the mean.
  • A predictable pattern where approximately 68% of values lie within 1 standard deviation, 95% within 2, and 99.7% within 3 standard deviations.
  • The use in determining z critical values for different confidence levels in hypothesis testing and creating confidence intervals.
By converting data into this standard form, you can easily understand how data points compare to the distribution mean, enabling efficient analysis using z-values or z-scores.
confidence levels
Confidence levels in statistics reflect how certain you are that a parameter lies within a specified range of values. It is an indication of the reliability of an estimate, often expressed as a percentage.

When you set a confidence level (like 95%, 90%, etc.), you're choosing the level of certainty for your estimate. This decision influences the z critical value, with higher confidence levels requiring larger z values, hence a broader confidence interval. Here is a brief overview of common confidence levels:
  • 95% Confidence Level: A z critical value of 1.96, implying that you can be 95% sure the parameter lies within this interval.
  • 90% Confidence Level: A z critical value of 1.64, providing a slightly narrower interval.
  • 99% Confidence Level: A z critical value of 2.57, offering broader range and more certainty.
Choosing an appropriate confidence level depends on the level of precision and certainty you desire. Higher confidence levels give more precise, but wider intervals, thus offering more assurance.
large-sample confidence interval
Constructing a large-sample confidence interval is a common statistical method used to estimate a population parameter. When you're working with large samples, typically over 30, you can assume that the sampling distribution of the sample proportion is approximately normal.

The general formula for a large-sample confidence interval is:
  • \[ p \pm (z \text{ critical value}) \sqrt{\frac{p(1-p)}{n}} \]
Where:
  • \( p \) is the sample proportion,
  • \( n \) is the sample size,
  • The z critical value corresponds to your chosen confidence level.
With this formula, you can determine the range in which the true population parameter will likely fall, within the set confidence level, such as 95% or 99%. This helps in making informed decisions by interpreting the statistical evidence drawn from the data.

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Most popular questions from this chapter

A random sample of \(n=12\) four-year-old red pine trees was selected, and the diameter (in inches) of each tree's main stem was measured. The resulting observations are as follows: $$ \begin{array}{llllllll} 11.3 & 10.7 & 12.4 & 15.2 & 10.1 & 12.1 & 16.2 & 10.5 \\ 11.4 & 11.0 & 10.7 & 12.0 & & & & \end{array} $$ a. Compute a point estimate of \(\sigma\), the population standard deviation of main stem diameter. What statistic did you use to obtain your estimate? b. Making no assumptions about the shape of the population distribution of diameters, give a point estimate for the population median diameter. What statistic did you use to obtain the estimate? c. Suppose that the population distribution of diameter is symmetric but with heavier tails than the normal distribution. Give a point estimate of the population mean diameter based on a statistic that gives some protection against the presence of outliers in the sample. What statistic did you use? d. Suppose that the diameter distribution is normal. Then the 90 th percentile of the diameter distribution is \(\mu+\) \(1.28 \sigma\) (so \(90 \%\) of all trees have diameters less than this value). Compute a point estimate for this percentile. (Hint: First compute an estimate of \(\mu\) in this case; then use it along with your estimate of \(\sigma\) from Part (a).)

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One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006 ). When asked 'Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes OK. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes OK to lie to avoid hurting someone's feelings. c. Based on the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the re-

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The use of the interval $$ p \pm(z \text { critical value }) \sqrt{\frac{p(1-p)}{n}} $$ requires a large sample. For each of the following combinations of \(n\) and \(p\), indicate whether the given interval would be appropriate. a. \(n=50\) and \(p=.30\) b. \(n=50\) and \(p=.05\) c. \(n=15\) and \(p=.45\) d. \(n=100\) and \(p=.01\) e. \(n=100\) and \(p=.70\) f. \(n=40\) and \(p=.25\) g. \(n=60\) and \(p=.25\) h. \(n=80\) and \(p=.10\)
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