Unit 1 Advanced Statistics

Which of the following statements are true features of the Central Limit Theorem?

-The distribution shape is approximately Normal. -It describes the sampling distribution of the sample mean. -For large sample size, the distribution of sample means is independent of the shape of the population.

Choose the statement that best defines the Sampling Distribution of the Sample Mean

A probability distribution of all possible sample means of a given sample size

Which one of the following statements is true about the dispersion of the distribution of sample means?

As the sample size increases, the variability in the sample means decreases.

Which of the following populations is a good candidate for Systematic random sampling?

Companies listed on the New York Stock Exchange

The average height of American women (in 2016) is 5 ft. 4 inches (64 inches) with a standard deviation of 3 inches. What is the probability that the average height of a random group of nine American women would be less than five feet three inches? Assume that the heights of American women are normally distributed.

Reason: z=(63−64)(39√)=−1z=63-6439=-1 The probability corresponding to z-value of −1 is 0.1587.

The average height of American males (in 2016) is 5 ft. 9 1/2 inches (69.5 inches) with a standard deviation of 3 inches. What is the probability that the average height of a random group of 16 American men would be over five feet ten inches? Assume that the heights of American men are normally distributed.

Reason: z=(70 − 69.5)(316√)=0.67z=70 - 69.5316=0.67 In Appendix B.3, locate the probability corresponding to a z-value of 0.67. it is 0.2486. The likelihood of a z-value greater than 0.67 is 0.2514, found by 0.5000−0.24860.5000-0.2486.

Choose the statement that best describes sampling error.

The difference between a sample statistic and its corresponding population parameter.

Which of the following statements correctly describe characteristics of the set of sample means?

The dispersion of the sample means is narrower than the population dispersion. The mean of the sample means equals the mean of the population.

Which statement correctly describes the relationship between the mean of the sample means and the population mean?

The mean of the sample means equals the population mean.

Which of the following statements about the sampling distribution of the sample means is not correct?

The standard deviation of sample means is the same as that of the population.

Pick the statement that describes the formula for the standard error of the mean in ordinary language.

The standard error is equal to the population standard deviation divided by the square root of the sample size.

There are two conditions under which we can assume that the sample means follow a normal distribution. What are they?

We know that the population is normally distributed. We don't know the population distribution, but the sample size is 30 or larger.

When is it inappropriate to use systematic random sampling?

When the order of items in the population is related to some particular characteristic.

Listed below are the 35 members of the Metro Toledo Automobile Dealers Association. We would like to estimate the mean revenue from dealer service departments. The members are identified by numbering them 00 through 34. a. We want to select a random sample of five dealers. The random numbers are 31, 42, 16, 53, 59, 18, 84, 27, 58, 92, 49, and 30. Which dealers would be included in the sample? (Enter the numbers as they appear. c. Using systematic random sampling, every fourth dealer is selected starting with the fourth dealer in the list. Which dealers are included in the sample?

a. 31, 16, 18, 27, 30 c. 3, 7, 11, 15, 19

Which of the following is an expression that represents sampling error?

x bar - mu

Identify the steps involved in taking a cluster sample. Select all that apply.

Randomly select a subset of clusters. Select a random sample from each sub group. Divide the population into groups using naturally occurring boundaries.

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $128,000. This distribution follows the normal distribution with a standard deviation of $33,000. a. If we select a random sample of 66 households, what is the standard error of the mean? b. What is the expected shape of the distribution of the sample mean? c. What is the likelihood of selecting a sample with a mean of at least $130,000? d. What is the likelihood of selecting a sample with a mean of more than $120,000? e. Find the likelihood of selecting a sample with a mean of more than $120,000 but less than $130,000.

Explanation a.The standard error of the mean is the standard deviation divided by the square root of the sample size. In symbols, s sub x bar= 33,000/√66=4,062. b.The population is normally distributed. So, the sampling distribution of the sample mean is normal. c.z=130,000 − 128,00033,000/√66=0.49 z=130,000 - 128,00033,000/√66=0.49 So, probability is 0.3121, found by 0.5000 − 0.1879. d.z=120,000 − 128,00033,000/√66=−1.97z=120,000 - 128,00033,000/√66=-1.97 So, probability is 0.9756, found by 0.5000 + 0.4756. e.0.6635, found by 0.4756 + 0.1879

Human Resource Consulting (HRC) surveyed a random sample of 78 Twin Cities construction companies to find information on the costs of their health care plans. One of the items being tracked is the annual deductible that employees must pay. The Minnesota Department of Labor reports that historically the mean deductible amount per employee is $504 with a standard deviation of $110. a. Compute the standard error of the sample mean for HRC. b. What is the chance HRC finds a sample mean between $477 and $527? c. Calculate the likelihood that the sample mean is between $492 and $512. d. What is the probability the sample mean is greater than $550?

Explanation a.The standard error of the mean is the standard deviation divided by the square root of the sample size.In symbols, σx¯=11078√=12.4550σx¯=11078=12.4550 b.The mean of the sample equals the mean of the population ($504) and the standard deviation of the sample mean equals the population standard deviation 110 divided by the square root of the sample size ($110/√78). The z-values corresponding to $477 and $527 are z1=477 − 504/(110/√78)=−2.17 and z2=527 − 504/(110/√78)=1.85 respectively. Using a normal table the probability is 0.9528, found by 0.4850 + 0.4678. c.The z-values corresponding to $492 and $512 are z1=492 − 504/(110/√78)=−0.96 and z2=512 − 504/(110/√78)=0.64 Using a normal table the probability is 0.5704, found by 0.3315 + 0.2389. d.The z-value corresponding to $550 is z1=550−504/(110/√78)=3.69 z1=550 - 504/(110/√78)=3.69 That is beyond the last value in the normal table. So probability is almost 0.

Scrapper Elevator Company has 20 sales representatives who sell its product throughout the United States and Canada. The number of units sold last month by each representative is listed below. Assume these sales figures to be the population values. 2 3 2 3 3 4 2 4 3 2 2 7 3 4 5 3 3 3 3 5 b. Compute the population mean. c. Compute the standard deviation. d. If you were able to list all possible samples of size five from this population of 20, how would the sample means be distributed? e. What would be the mean of the sample means? f. What would be the standard deviation of the sample means?

Explanation b.μ=3.3μ=3.3 c.σ=1.23 σ=1.23 d.The central limit theorem says that the sample means would tend to be normally distributed. e.The mean of the distribution of sample means would be the same as the population mean, 3.30. f.The standard deviation of the distribution of sample means would be: σx¯=σn√=1.235√=0.55σx¯=σn=1.235=0.55 .

Choose the two statements that are correct descriptions of the sampling distribution of the sample mean.

It is a probability distribution of all possible sample means. It is a distribution of means from samples of all one size.

Based on all student records at Camford University, students spend an average of 5.20 hours per week playing organized sports. The population's standard deviation is 3.40 hours per week. Based on a sample of 81 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates. a. Compute the standard error of the sampling distribution of sample means. b. What is the chance HLI will find a sample mean between 4.5 and 5.9 hours? c. Calculate the probability that the sample mean will be between 4.8 and 5.6 hours. d. How strange would it be to obtain a sample mean greater than 7.70 hours?

a.The standard error of the mean is the standard deviation divided by the square root of the sample size or 0.38, found by 3.4/81−−√which is 3.4/9.0.38, found by 3.4/sqrt(81) which is 3.4/9. b.The central limit theorem states that, for large random samples, the shape of the sampling distribution of the mean is close to the normal probability distribution. A sample of 30 or more is considered "large." The z-value for 5.90 is 1.84, found by (5.90 − 5.20)/0.38. Similarly, the z-value for 4.5 is −1.84 = (4.5 − 5.2)/0.38. Using the standard normal distribution, this probability is 0.9342, found by 2(0.4671). c.The z-value for 5.60 is 1.05, found by (5.60 − 5.20)/0.38. Similarly, the z-value for 4.8 is −1.05 = (4.8 − 5.20)/0.38. So this probability is 0.7062, found by 2(0.3531). d.The z-value for 7.70 is 6.58, found by (7.70 − 5.20)/0.38. This is very unlikely with a probability of 0.000 (rounded to three decimal places).