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...

...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%...

...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...

...malfunction. The problem is to compute 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...

...Solve the problem.
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%...

...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...

...
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...

...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...

...
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...