Instructions: Use this page as your cover page, and attach your groupwork behind your work. Your assignment answers should be in complete and grammatically correct sentences.

1. According to the National Center for Health Statistics (2004), 22.4% of adults are smokers. A random sample of 300 adults is obtained.

(a) Describe the sampling distribution of phat, the sample proportion of adults who smoke. (b) In a random sample of 300 adults, what is the probability that at least 50 are smokers? (c) Would it be unusual if a random sample of 300 adults results in 18% or less being smokers? Explain your answer.

2. A machine at K&A Tube & Manufacturing Company produces a certain copper tubing component in a refrigeration unit. The tubing components produced by the manufacturer have a mean diameter of 0.75 inch with a standard deviation of 0.004 inch. The quality-control inspector takes a random sample of 30 components once each week and calculates the mean diameter of these components. If the mean is either less than 0.748 inch or greater than 0.752 inch, the inspector concludes that the machine needs an adjustment.

(a) Describe the sampling distribution of the sample mean diameter for a random sample of 30 such components.
(b) What is the probability that, based on a random sample of 30 such components, the inspector will conclude that the machine needs an adjustment when, in fact, the machine is correctly calibrated?

3. In a random sample of 678 adult males 20 to 34 years of age, it was determined that 58 of them have hypertension (high blood pressure). Source: The Centers for Disease Control.
(a) Obtain a point estimate for the proportion of adult males 20 to 34 years of age who have...

...corresponds to a degree of confidence of 91%.
A) 1.70 B) 1.34 C) 1.645 D) 1.75
2) The following confidence interval is obtained for a population proportion, p:0.817 < p < 0.855
Use these confidence interval 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% confidence interval used to estimate the population proportion.
3) n = 186, x = 103
A) 0.0643 B)...

...assumed that the standard deviation, σ, is 2.4. Construct the 95% confidence interval 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% confidence interval 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 confidence interval. Which of the following will result in a reduced interval...

...you learned to find interval estimates for two population parameters, a population mean and a population proportion. Explain the meaning of an interval estimate of a population parameter.
An interval estimate for a specified population parameter (such as a mean or proportion) is a range of values in which the parameter is estimated to lie. In Chapter 6, you were assigned to find interval estimates for a population mean and a...

...Sample sizes and confidence
intervals 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
• Confidence interval of proportion
• Hypothesis testing with proportion
Population and Sample
Samples
Populations
Sampling distribution of
sample means (SDSM)...

...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...

...confidence interval 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 confidence interval 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)
= (3.01 – 0.0098, 3.01 +0.0098)
= (3.0002, 3.0198)...

...Confidence Intervals
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...