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Chapter 6 Review

Chapter 6 Review. 6-1 Preparation 6-2 The Standard Normal Distribution 6-3 Applications of Normal Distributions 6-4 Sampling Distributions and Estimators 6-5 The Central Limit Theorem 6-6 Assessing Normality 6-7 Normal as Approximation to Binomial. Example.

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Chapter 6 Review

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  1. Chapter 6Review 6-1 Preparation 6-2 The Standard Normal Distribution 6-3 Applications of Normal Distributions 6-4 Sampling Distributions and Estimators 6-5 The Central Limit Theorem 6-6 Assessing Normality 6-7 Normal as Approximation to Binomial

  2. Example Bone mineral density test scores are normally distributed with a mean of 0 and a standard deviation of 1. . These scores can be helpful in identifying the presence of osteoporosis. The result of the test is can be measured as z scores. If a randomly selected adult undergoes a bone density test, find the probability that the result is a reading less than 1.27.

  3. Look at Table A-2The are from the left hand side of the curve to 1.27 is .8980note that the hundredths are on top.

  4. The probability of random adult having a bone density less than 1.27 is 0.8980. Here is a picture that describes this

  5. Example 2 Using the same bone density test, find the probability that a randomly selected person has a result above –1.00 (which is considered to be in the “normal” range of bone density readings.

  6. Since we are looking for the area to the right we have to subtract. The probability of a randomly selected adult having a bone density above –1 is 0.8413. Example 2 – continued

  7. Example 3 Find this probability of getting a bone density reading between –1.00 and –2.50. This indicates the subject has osteopenia.

  8. 1. Find the area to the left of z = –2.50 0.0062.2. Find the area to the left of z = –1.00 0.1587.3. The area between z = –2.50 and z = –1.00 is the difference between these areas. Example 3

  9. Example 4 Find the 95th Percentile. This is the z score which separates the lower 95%. 5% or 0.05 This is an example of finding a z score from an area.

  10. Solution 5% or 0.05 1.645

  11. Example Find the value of z0.025. This means find the 97.5 percentile.

  12. Solution The notation z0.025 is used to represent the z score with an area of 0.025 to its right. Referring back to the bone density example, z0.025 = 1.96.

  13. Chapter 6Normal Probability Distributions 6-1 Preparation 6-2 The Standard Normal Distribution 6-5 The Central Limit Theorem 6-6 Assessing Normality 6-7 Normal as Approximation to Binomial

  14. Tall Clubs International has a requirement that women must be at least 70 inches tall. Given that women have normally distributed heights with a mean of 63.8 inches and a standard deviation of 2.6 inches, find the percentage of women who satisfy that height requirement. 6-3 Applications of Normal DistributionsExample

  15. Convert to a z score and use Table A-2 or technology to find the shaded area. Solution

  16. Draw the normal distribution and shade the region. The area to the right of 2.38 is 0.008656, and so about 0.87% of all women meet the requirement. Solution

  17. Example Men’s heights are normally distributed with a mean of 69.5 inches and a standard deviation of 2.4 inches. When designing aircraft cabins, what ceiling height will allow 95% of men to stand without bumping their heads? This is an example of finding a z score from an area.

  18. Solution With z = 1.645, μ = 69.5, andσ= 2.4. we can solve for x.

  19. 6-4 Sampling Distributions and Estimators Example Roll a die 5 times. Find the mean , variance , and the proportion of odd numbers of the results. What will happen if you continue this process? What will the mean of all of your sample means approach?

  20. Assume male weights follow a normal distribution with a mean of 182.9 lb and a standard deviation of 40.8 lb. Suppose an elevator has a maximum capacity of 16 passengers with a total weight of 2500 lb.Assuming a worst case scenario in which the passengers are all male, what are the chances the elevator is overloaded?a. Find the probability that 1 randomly selected male has a weight greater than 156.25 lb.b. Find the probability that a sample of 16 males have a mean weight greater than 156.25 lb (which puts the total weight at 2500 lb, exceeding the maximum capacity). Example – Elevators

  21. a. Find the probability that 1 randomly selected male has a weight greater than 156.25 lb.Use the methods presented in Section 6.3. We can convert to a z score and use Table A-2.Using Table A-2, the area to the right is 0.7422. Solution

  22. b. Find the probability that a sample of 16 males have a mean weight greater than 156.25 lb. Since the distribution of male weights is assumed to be normal, the sample mean will also be normal.Converting to z: Solution

  23. b. Find the probability that a sample of 16 males have a mean weight greater than 156.25 lb. While there is 0.7432 probability that any given male will weigh more than 156.25 lb, there is a 0.9955 probability that the sample of 16 males will have a mean weight of 156.25 lb or greater. Solution

  24. Review Binomial Probability Distribution 1. The procedure must have a fixed number of trials. 2. The trials must be independent. 3. Each trial must have all outcomes classified into two categories (commonly, success and failure). 4. The probability of success remains the same in all trials. Solve by binomial probability formula, Table A-1, or technology.

  25. Approximation of a Binomial Distributionwith a Normal Distribution a distribution. (normal) then , and the random variable has

  26. Example – NFL Coin Toss In 431 NFL football games that went to over time, the teams that won the coin toss went on to win 235 of those games. What is the probability of such an event occurring assuming there is a 0.5 probability of winning a game after winning the coin toss?

  27. Solution If the coin-toss method is fair, teams winning the toss would win about 50% of the games (we’d expect 215.5 wins in 431 overtime games). The given problem involves a binomial distribution with n = 431 trials and an assumed probability of success of p = 0.5. Use the normal approximation to the binomial distribution. Step 1: The conditions check:

  28. Example – NFL Coin Toss Step 2: Find the mean and standard deviation of the normal distribution: Step 3: We want the probability of at least 235 wins, so x = 235. Step 4: The vertical strip will go from 234.5 to 235.5.

  29. Example – NFL Coin Toss Step 5: We will shade the area to the right of 234.5.

  30. Example – NFL Coin Toss Step 6: Find the z score and use technology or Table A-2 to determine the probability. The probability is 0.0336 for the coin flip winning team to win at least 235 games. This probability is low enough to suggest the team winning coin flip has an unfair advantage.

  31. Don’t Forget When we use the normal distribution (which is a continuous probability distribution) as an approximation to the binomial distribution (which is discrete), a continuity correction is made to a discrete whole number x in the binomial distribution by representing the discrete whole number x by the interval from x– 0.5 to x + 0.5 (that is, adding and subtracting 0.5).

  32. Example a. Find the margin of error E that corresponds to a 95% confidence level. b. Find the 95% confidence interval estimate of the population proportion p. c. Based on the results, can we safely conclude that more than 75% of adults know what Twitter is? d. Assuming that you are a newspaper reporter, write a brief statement that accurately describes the results and includes all of the relevant information. In theChapter Problem we noted that a Pew Research Center poll of 1007 randomly selected adults showed that 85% of respondents know what Twitter is. The sample results are n = 1007 and

  33. Requirement check: simple random sample; fixed number of trials, 1007; trials are independent; two outcomes per trial; probability remains constant. Note: number of successes and failures are both at least 5. Example - Continued a) Use the formula to find the margin of error.

  34. b) The 95% confidence interval: Example - Continued

  35. Based on the confidence interval obtained in part (b), it does appear that more than 75% of adults know what Twitter is. Because the limits of 0.828 and 0.872 are likely to contain the true population proportion, it appears that the population proportion is a value greater than 0.75. Example - Continued

  36. Here is one statement that summarizes the results: 85% of U.S. adults know what Twitter is. That percentage is based on a Pew Research Center poll of 1007 randomly selected adults. In theory, in 95% of such polls, the percentage should differ by no more than 2.2 percentage points in either direction from the percentage that would be found by interviewing all adults in the United States. Example - Continued

  37. Example Many companies are interested in knowing the percentage of adults who buy clothing online. How many adults must be surveyed in order to be 95% confident that the sample percentage is in error by no more than three percentage points? a. Use a recent result from the Census Bureau: 66% of adults buy clothing online. b. Assume that we have no prior information suggesting a possible value of the proportion.

  38. a) Use Example - Continued To be 95% confident that our sample percentage is within three percentage points of the true percentage for all adults, we should obtain a simple random sample of 958 adults.

  39. b) Use Example - Continued To be 95% confident that our sample percentage is within three percentage points of the true percentage for all adults, we should obtain a simple random sample of 1068 adults.

  40. Example A common claim is that garlic lowers cholesterol levels. In a test of the effectiveness of garlic, 49 subjects were treated with doses of raw garlic, and their cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 0.4 and a standard deviation of 21.0. Use the sample statistics of n = 49, = 0.4, and s = 21.0 to construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?

  41. Example - Continued Requirements are satisfied: simple random sample and n = 49 (i.e., n > 30). 95% implies α= 0.05.With n = 49, the df = 49 – 1 = 48 Closest df is 50, two tails, so = 2.009 Using = 2.009, s = 21.0 and n = 49 the margin of error is:

  42. Example - Continued Construct the confidence interval: We are 95%confident that the limits of –5.6 and 6.4 actually do contain the value of μ, themean of the changes in LDL cholesterol for the population. Because the confidence interval limits contain the value of 0, it is very possiblethat the mean of the changes in LDL cholesterol is equal to 0, suggesting that thegarlic treatment did not affect the LDL cholesterol levels. It does not appear thatthe garlic treatment is effective in lowering LDL cholesterol.

  43. Important Properties of the Student t Distribution 1. The Student t distribution is different for different sample sizes. (See the following slide for the cases n = 3 and n = 12.) 2. The Student t distribution has the same general symmetric bell shape as the standard normal distribution but it reflects the greater variability (with wider distributions) that is expected with small samples. 3. The Student t distribution has a mean of t = 0 (just as the standard normal distribution has a mean of z = 0). 4. The standard deviation of the Student t distribution varies with the sample size and is greater than 1 (unlike the standard normal distribution, which has σ= 1). 5. As the sample size n gets larger, the Student t distribution gets closer to the normal distribution.

  44. Student t Distributions for n = 3 and n = 12

  45. Finding the Point Estimate and E from a Confidence Interval Margin of Error: = (upper confidence limit) – (lower confidence limit) 2 Point estimate of μ: =(upper confidence limit) + (lower confidence limit) 2

  46. Finding a Sample Size for Estimating a Population Mean = population mean = population standard deviation = sample mean = desired margin of error = z score separating an area of in the right tail of the standard normal distribution

  47. Round-Off Rule for Sample Size n If the computed sample size n is not a whole number, round the value of n up to the next larger whole number.

  48. Finding the Sample Size nWhen σis Unknown Use the range rule of thumb (see Section 3-3) to estimate the standard deviation as follows: Start the sample collection process without knowing σ and, using the first several values, calculate the sample standard deviation s and use it in place of σ. The estimated value of σ can then be improved as more sample data are obtained, and the sample size can be refined accordingly. Estimate the value of σby using the results of some other earlier study. .

  49. Example Assume that we want to estimate the mean IQ score for the population of statistics students. How many statistics students must be randomly selected for IQ tests if we want 95% confidence that the sample mean is within 3 IQ points of the population mean? α = 0.05 α/2 = 0.025 zα/2 = 1.96 E = 3 σ= 15 With a simple random sample of only 97 statistics students, we will be 95% confident that the sample mean is within 3 IQ points of the true population mean μ.

  50. Part 2: Key Concept This section presents methods for estimating a population mean. In addition to knowing the values of the sample data or statistics, we must also know the value of the population standard deviation, σ.

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