Q1.
A researcher wishing to estimate the proportion of X-ray machines that malfunction and produce excess radiation. A random sample of 40 machines is taken and 12 of the machines malfunction. The problem is to compute the 95% confidence interval 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% confidence interval is 1.96. The limits of the confidence interval are therefore: Lower limit = .30 - (1.96)(0.072) = .16
Upper limit = .30 + (1.96)(0.072) = .44.
The confidence interval 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% confidence interval for the mean time to process μ.

Solution:
For large n, a 90% confidence interval 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% confidence interval 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% confidence interval for the population mean μ, amount of PCBs in the soil.

Solution:
For large n, a 95% confidence interval for μ is given by X ±z0.025 ×S/√n Using z0.025 = 1.96, n = 40, S = .5ppm, and x = 3.56ppm, the 95% confidence interval for μ (true mean amount of PCBs in the soil) is given by ± 1.96 × .5/√ 40 = 3.56 ± 0.155 = (3.405, 3.715).

Q4.
Radiation of microwave ovens has normal distribution with standard deviation σ=0.6. A sample of 25 microwave ovens produced X = 0.11. Determine a 95% confidence interval...

...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)
=...

...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...

...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...

...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...

...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%...

...
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...