confidence intervals
The confidence intervals represent upper and lower bounds of variation around each reference forecast. Values may occur outside the confidence intervals due to external shocks, such as extreme weather, structural changes to the economic system, geopolitical events, or technology development. The confidence intervals 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 confidence interval. The confidence interval 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 34.0%. If another group of home owners has taken and we found that they have a sample of proportion of 38.0%, So the sampling proportion will be the key to our ability to generalize from our sample to the population.

Now, we know that the sampling distribution model is centered at the true proportion, p, of all home owners who use electricity to heat their homes. But we don’t know p. it isn’t 34.0%. That’s the p̂ from our sample. What we do know is that the sampling distribution model of p̂ centered at p, and we know that the standard deviation of the sampling distribution is

...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%...

...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%...

...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...

...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...

...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...

...percent confidenceinterval for the population mean.
a) The estimated population mean = 3.01 pounds
b) Here it is given that, xbar = 3.01, s = 0.03, n = 36.
The Standard error of the mean, SE = s/Sqrt(n) = 0.03/Sqrt(36) = 0.005
The z- score for 95% confidence is z = 1.96
A 95 percent confidenceinterval for the population mean is given by
(xbar – z*SE, xbar + z*SE)
= (3.01 – 1.96*0.005, 3.01 + 1.96*0.005)
=...

...
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...

...
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...