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Chapter 9: Problem 70

Find \(\alpha,\) the area of one tail, and the confidence coefficients of \(z\) that are used with each of the following levels of confidence. a. \(1-\alpha=0.80\) b. \(1-\alpha=0.98\) c. \(1-\alpha=0.75\)

Short Answer

Expert verified
The calculation results are: a. for \(1-\alpha=0.80\), \(\alpha = 0.20\) and \(z = 1.28\); b. for \(1-\alpha=0.98\), \(\alpha = 0.02\) and \(z = 2.33\); and c. for \(1-\alpha=0.75\), \(\alpha = 0.25\) and \(z = 1.15\).

Step by step solution

01

Finding \(\alpha\)

Given \(1-\alpha\), we can calculate \(\alpha\) by re-arranging the equation to \(\alpha = 1 - (1 - \alpha)\), giving: \n\na. \(\alpha = 0.20\) for \(1-\alpha=0.80\); \nb. \(\alpha = 0.02\) for \(1-\alpha=0.98\); and \nc. \(\alpha = 0.25\) for \(1-\alpha=0.75\).
02

Finding the Z-scores

We use a standard normal distribution (z-distribution) table to find the z-scores for the corresponding areas that represent \(\alpha./2\), because one-tailed and two-tailed tests both assume that the test statistic follows a normal distribution, symmetric about the center. \n\nWe will therefore look for areas of: \na. 0.80 + 0.20/2 = 0.9 \nb. 0.98 + 0.02/2 = 0.99 \nc. 0.75 + 0.25/2 = 0.875 \n\nUsing the z-table or a calculator that returns the z-scores, we get: \na. \(z = 1.28\) for an area of 0.90; \nb. \(z = 2.33\) for an area of 0.99; and \nc. \(z = 1.15\) for an area of 0.875.

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Key Concepts

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

Z-Scores
When we talk about z-scores, we are discussing a way to measure how far away a data point is from the mean in a standard normal distribution. The key idea is to express this distance in units of standard deviation.
  • A z-score tells you how many standard deviations a point is above or below the mean.
  • A positive z-score means the data point is above the mean, while a negative z-score means it's below the mean.
  • This makes it a powerful tool for standardizing data, allowing comparisons across different data sets.
In the context of confidence intervals, z-scores help us understand the strength of evidence against a null hypothesis. By looking at how extreme your test statistic is, you can determine if the observed results are within the expected range of variability.
Standard Normal Distribution
The standard normal distribution is a crucial concept in statistics, playing a key role in conducting hypothesis tests and constructing confidence intervals. It is a special case of the normal distribution.
  • The mean of this distribution is 0, and its standard deviation is 1.
  • It's a bell-shaped curve, symmetric about the mean.
  • This standard shape allows statisticians to use z-tables, which provide the probability for data points lying at particular z-scores.
Because of its symmetry and standard properties, you can apply it to any normally distributed data by converting raw scores to z-scores. This conversion helps in determining probabilities and critical z-values for constructing confidence intervals.
Tail Area
The tail area in the context of a normal distribution refers to the area under the curve that lies outside the desired range of confidence. It represents the probability of observing a statistic as extreme or more extreme than the observed one, assuming the null hypothesis is true.
  • The tail is what we leave out when choosing a confidence level.
  • A smaller tail area indicates a higher confidence level and vice versa.
  • For a given confidence level of \(1-\alpha\), the tail area is \alpha\.
Understanding this is necessary when making statistical inferences since it shows how common or rare the observed data is when viewed under the null hypothesis framework. Analyzing tail areas helps determine the limits within which a certain percentage of the population lies.
Confidence Level
The confidence level in statistical terms represents the proportion of times that an estimation procedure will capture the true parameter value across repeated sampling from the population. It's often denoted as \(1-\alpha\).
  • A commonly used confidence level is 95%, implying that if you were to take 100 different samples, 95 of them would yield intervals containing the true mean.
  • This level helps express how reliable your estimate is: higher confidence levels mean wider intervals and imply more confidence in containing the true parameter.
  • Calculating the confidence level also involves z-scores and understanding the standard normal distribution.
In practice, setting your confidence level depends on the context of the study, determining how much risk of error you're willing to accept when making predictions or decisions based on sample data.

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Most popular questions from this chapter

Three nationwide poll results are described below. USA Today Snapshot/Rent.com, August 18,2009 \(N=1000\) adults 18 and over; \(\mathrm{MoE} \pm 3 .\) (MoE is margin of error. "What renters look for the most when seeking an apartment:" Washer/dryer\(-39\%,\) Air Conditioning \(-30 \%,\) Fitness Center- \(10 \%,\) Pool \(-10 \%\) USA Today/Harris Interactive Poll, February \(10-15,2009 ; N=1010\) adults; MoE 卤3. "Americans who say people on Wall Street are "as honest and moral as other people." Disagree \(-70 \%\) Agree \(-26 \%,\) Not sure/refuse to answer \(-4 \%\) American Association of Retired Persons Bulletin/AARP survey, July 22-August 2, 2009; \(N=1006\) adults age 50 and older; \(\mathrm{MoE} \pm 3\). The American Association of Retired Persons Bulletin Survey reported that \(16 \%\) of adults, 50 and older, said they are likely to return to school. Each of the polls is based on approximately 1005 randomly selected adults. a. Calculate the \(95 \%\) confidence maximum error of estimate for the true binomial proportion based on binomial experiments with the same sample size and observed proportion as listed first in each article. b. Explain what caused the values of the maximum errors to vary. c. The margin of error being reported is typically the value of the maximum error rounded to the next larger whole percentage. Do your results in part a verify this? d. Explain why the round-up practice is considered "conservative." e. What value of \(p\) should be used to calculate the standard error if the most conservative margin of error is desired?

The May \(30,2008,\) online article "Live with Your Parents After Graduation?" quoted a 2007 survey conducted by Monster-TRAK.com. The survey found that \(48 \%\) of college students planned to live at home after graduation. How large of a sample size would you need to estimate the true proportion of students that plan to live at home after graduation to within \(2 \%\) with \(98 \%\) confidence?

Determine the \(p\) -value for each of the following hypothesis-testing situations. a. \(H_{o}: p=0.5, H_{a}: p \neq 0.5, z \star=1.48\) b. \(H_{o}: p=0.7, H_{a}: p \neq 0.7, z \star=-2.26\) c. \(H_{o}: p=0.4, H_{a}: p>0.4, z \star=0.98\) d. \(H_{o}: p=0.2, H_{a}: p<0.2, z \star=-1.59\)

Determine the critical region and critical value(s) that would be used to test the following using the classical approach: a. \(H_{o}: \sigma=0.5\) and \(H_{a}: \sigma>0.5,\) with \(n=18\) and \(\alpha=0.05\) b. \(H_{o}: \sigma^{2}=8.5\) and \(H_{a}: \sigma^{2}<8.5,\) with \(n=15\) and \(\alpha=0.01\) c. \(H_{o}: \sigma=20.3\) and \(H_{a}: \sigma \neq 20.3,\) with \(n=10\) and \(\alpha=0.10\) d. \(H_{o}: \sigma^{2}=0.05\) and \(H_{a}: \sigma^{2} \neq 0.05,\) with \(n=8\) and \(\alpha=0.02\) e. \(H_{o}: \sigma=0.5\) and \(H_{a}: \sigma<0.5,\) with \(n=12\) and \(\alpha=0.10\)

Use a computer or calculator to find the area (a) to the left, and (b) to the right of \(\chi^{2} \star=20.2\) with df \(=15\).
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