34. Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000. This distribution follows the normal distribution with a standard deviation of $40,000. a. If we select a random sample of 50 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 $112,000? d. What is the likelihood of selecting a sample with a mean of more than $100,000? e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000. a) The standard error of the mean, S.E. = /Sqrt(n) = 40,000/Sqrt(50) = 5656.8542 b) It will be normally distributed with mean $110,000 and standard deviation $5656.85. Note that Z = (Xbar – 110,000)/5656.85 follows a Standard Normal distribution. c) The likelihood of selecting a sample with a mean of at least $112,000 is given by,

P[Xbar 112,000] = P[(Xbar–110,000)/5656.85 (112,000-110,000)/5656.85]
= P[Z 0.3536] = 1- P[Z < 0.3536]
= 1 – 0.6382 = 0.3618
d) The likelihood of selecting a sample with a mean of more than $100,000 is given by,
P[Xbar > 100,000] = P[(Xbar–110,000)/5656.85 > (100,000-110,000)/5656.85]
= P[Z > -1.7678] = 1- P[Z < 0.3536]
= 1 – 0.0385 = 0.9615
e) The likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000 is given by, P[100,000 < Xbar < 112,000]
= P[(100,000-110,000)/5656.85 < (Xbar–110,000)/5656.85 < (112,000- 110,000)/5656.85]
= P[-1.7678 < Z < 0.3536]
= P[Z < 0.3536] - P[Z<-1.7678]
= 0.6382 - 0.0385 = 0.5997
Exercise 32. A state meat inspector in Iowa has been given the assignment of estimating the mean net weight of packages of ground chuck labeled “3 pounds.” Of course, he realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveals the mean weight...

...Chapter 9 Quiz
1 |2 |3 |4 |5 |6 |7 |8 |9 |10 |11 |12 |13 |14 |15 | | | | | | | | | | | | | | | | | |
1 Compute the critical value za/2 that corresponds to a 94% level of confidence.
A. 1.88
B. 1.66
C. 1.96
D. 2.33
2. In a sample of 10 randomly selected women, it was found that their mean height was 63.4 inches. Form previous studies, it is assumed that the standard deviation, σ, is 2.4. Construct the 95% confidenceinterval for the population mean.
A. (61.9, 64.9)
B. (58.1, 67.3)
C. (59.7, 66.5)
D. (60.8, 65.4)
3. Suppose a 95% confidenceinterval for µ turns out to be (120, 310). To make more useful inferences from the data, it is desired to reduce the width of the confidenceinterval. Which of the following will result in a reduced interval width?
A. Increase the sample size.
B. Decrease the confidence level.
C. Increase the sample size and decrease the confidence level.
D. All of the choices will result in a reduced interval width.
4. A nurse at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 90% confidence that the true mean is within 4 ounces of the sample mean? The standard deviation of the birth weights is known to be 6 ounces.
A....

...confidenceintervals
The confidenceintervals represent upper and lower bounds of variation around each reference forecast.
Values may occur outside the confidenceintervals due to external shocks, such as extreme weather,
structural changes to the economic system, geopolitical events, or technology development. The
confidenceintervals increase in width throughout the forecast period due to the increasing level of
uncertainty in each subsequent year. The upper and lower bounds were based on one to two standard
deviations of the historic values, indicating at least a 68 percent probability that future values would be
expected to fall within the confidenceinterval. The confidenceinterval for the first forecast year is based on one standard deviation and grows linearly until it reaches two standard deviations, or a 95 percent probability.
=====================================================================================
For example, if we have polled a number of respondents from the home owners let’s say 3500 respondents, and from those only 1190 are using electricity to heat their homes, this means that 34.0% are using electricity to heat their homes,
p̂ = 1190/3500 = 34.0%.
And we know that a second sample of 3500 home owners wouldn’t have a sample proportion of exactly...

...the 95% confidenceinterval on π, the proportion that malfunction in the population.
Solution:
The value of p is 12/40 = 0.30. The estimated value of σp is
= 0.072.
A z table can be used to determine that the z for a 95% confidenceinterval is 1.96. The limits of the confidenceinterval are therefore:
Lower limit = .30 - (1.96)(0.072) = .16
Upper limit = .30 + (1.96)(0.072) = .44.
Theconfidenceinterval is: 0.16 ≤ π ≤ .44.
Q2.
A manager at a power company monitored the employee time required to process high-efficiency lamp bulb rebates. A random sample of 40 applications gave a sample mean time of 3.8 minutes and a standard deviation of 1.2 minutes. Construct a 90% confidenceinterval for the mean time to process μ.
Solution:
For large n, a 90% confidenceinterval for μ is given by
√X±z0.05×S/ n
Using z0.05 = 1.645, n = 40, S = 1.2minutes, and x = 3.8minutes, the 90% confidenceinterval for μ (true mean processing time) is given by
3.8 ± 1.645 × 1.2/ √40 = 3.8 ± 0.31 = (3.49, 4.11) minutes.
Q.3
The amount of PCBs (polychlorinated biphenyls) was measured in 40 samples of soil that were treated with contaminated sludge. The following summary statistics were obtained. x = 3.56, s = .5ppm Obtain a 95% confidenceinterval for...

...1) Find the critical value that corresponds to a degree of confidence of 91%.
A) 1.70 B) 1.34 C) 1.645 D) 1.75
2) The following confidenceinterval is obtained for a population proportion, p:0.817 < p < 0.855
Use these confidenceinterval limits to find the point estimate,
A) 0.839 B) 0.836 C) 0.817 D) 0.833
Find the margin of error for the 95% confidenceinterval used to estimate the population proportion.
3) n = 186, x = 103
A) 0.0643 B) 0.125 C) 0.00260 D) 0.0714
Find the minimum sample size you should use to assure that your estimate of will be within the required margin of error around the population p.
4) Margin of error: 0.002; confidence level: 93%; and unknown
A) 204,757 B) 410 C) 204,750 D) 405
5) Margin of error: 0.07; confidence level: 95%; from a prior study, is estimated by the
decimal equivalent of 92%.
A) 58 B) 174 C) 51 D) 4
Use the given degree of confidence and sample data to construct a confidenceinterval for the
population proportion p.
6) When 343 college students are randomly selected and surveyed, it is found that 110 own
a car. Find a 99% confidenceinterval for the true proportion of all college students who own a car.
A) 0.256 < p < 0.386 B) 0.279 < p < 0.362 C) 0.271 < p <...

...Sample sizes and confidenceintervals for proportions
Chong Chun Wie
Ext: 2768
ChongChunWie@imu.edu.my
Content
• Sampling distribution of sample means (SDSM)
• Normality Test
• Estimating a population mean: σ known
• Estimating a population mean: σ unknown
• Standard deviation of proportion
• Confidenceinterval of proportion
• Hypothesis testing with proportion
Population and Sample
Samples
Populations
Sampling distribution of
sample means (SDSM)
Sampling distribution
• Example
– Select randomly from the following number (1, 1, 2, 5, 5,
6, 7, 10)
– Mean = 4.625
•
If we choose 3 points randomly and produced the mean
1, 1, 2 = 4, mean = 1.33
2, 5, 6 = 13, mean = 4.33
1, 5, 7 = 13, mean = 4.33
2, 6, 7 = 15, mean = 5
5, 1, 10 = 16, mean = 5.33
Sampling distributions for (a) normal, (b) reverse-Jshaped, and (c) uniform variables
Test for normality
• Shapiro-Wilk Test (SPSS)
• Probability Plot (Minitab)
Estimating a population
mean: σ known
ConfidenceInterval for µ using
normal distribution
95% confidenceinterval
• Suppose we want to construct a 95%
confidenceinterval for μ,
Another word: 95% of all sample means are in the interval
Margin of Error
Z table
95%
confidence
α = 0.05
99%
confidence
α = 0.01
Example 1
• A doctor graduated from IMU is setting up a...

...
A Study in Determining ConfidenceIntervals at 95%
Charlesatta Johnson
PH6014
October 9, 2013
Dr. Rodrick Frazier
A Study in Determining ConfidenceIntervals at 95%
As hypothesized, high cholesterol levels in children can lead to their children being affected with hyperlipidemia. A study is conducted to estimate the mean cholesterol in children between the ages of 2 - 6 years of age. It also attempted to establish a correlation as to the effect family history has on the onset of the disease. From data collected as shown in Table 1 of the spreadsheet attached, a sample size of 9 (n=9) participants enrolled in the study. Total cholesterol levels measured in children between ages 2 – 6 years was summarized at 1,765. The sample mean (X) and standard deviation (S) computed as (1765/90) =196.1 and square root summation (X-X) square / n-1 =29.0 respectively. Now to generate a 95% confidenceinterval for the true mean total cholesterol levels in children from data collected, we used the z value for 95% as (z= 1.96). From sample statistics the confidenceinterval for 95% computed from the formula is (196.1 +/- 1.96 X 29/3) we now have 196.1 +/-19.0. Now, by adding and subtracting the margin of error, we have (215.1, 177.1) respectively. A point estimate for the true mean cholesterol levels in the population is 196.1 and that we are 95% confident...

...ConfidenceIntervals
Consider the following question: someone takes a sample from a population and finds both the sample mean and the sample standard deviation. What can he learn from this sample mean about the population mean?
This is an important problem and is addressed by the Central Limit Theorem. For now, let us not bother about what this theorem states but we will look at how it could help us in answering our question.
The Central Limit Theorem tells us that if we take very many samples the means of all these samples will lie in an interval around the population mean. Some sample means will be larger than the population mean, some will be smaller. The Central Limit Theorem goes on to state that 95% of the sample means will lie in a certain interval around the population mean. That interval is called the 95% confidenceinterval. Practically spoken it means that whenever someone is taking a sample and calculates the mean of that sample, he can be 95% confident that the mean of the sample he just took is in the 95% confidenceinterval. More importantly, if someone takes a sample from a population and calculates the mean of that sample, he can be 95% confident that the population mean is also in the 95% confidenceinterval. Thus, the sample mean gives us an approximation of the population mean. The same...

...
Question 1 of 1:
Interpreting the ConfidenceInterval
Solve the following problems:
A simple random sample of 40 salaries of NCAA football coaches has a mean of $415,953 and a standard deviation of $463,364. Construct a 95% confidenceinterval estimate of the mean salary of an NCAA football coach.
In a study designed to test the effectiveness of acupuncture for treating migraine, 142 subjects were treated with acupuncture and 80 subjects were given a sham treatment. The numbers of migraine attacks for the acupuncture treatment group had a mean of 1.8 and a standard deviation of 1.4. The numbers of migraine attacks for the sham treatment group had a mean of 1.6 and a standard deviation of 1.2.
Construct the 95% confidenceinterval estimate of the mean number of migraine attacks for those treated with acupuncture.
Construct the 95% confidenceinterval estimate of the mean number of migraine attacks for those given a sham treatment.
Compare the two confidenceintervals. What do the results suggest about the effectiveness of acupuncture?
Submission Requirements:
Submit the assignment in a Microsoft Word or Excel document.
Show detailed steps and provide appropriate rationale with your answers.
Evaluation Criteria:
Correctly answered each question
Included appropriate steps or rationale to determine the answer to each question
...