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...

...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) 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 confidence interval 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% confidence interval 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 < 0.370 D) 0.262 < p < 0.379
Determine whether the given conditions justify using the margin of error E = when
finding a confidence interval estimate of the...

...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% 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 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. 10
B. 9
C. 8
D. 7
5. Let t0 be a specific value of t. Find t0 such that the...

...Submit completed tests as word or pdf files via email to paul.kurose@seattlecolleges.edu
Due: Sunday, May 19 (by 8am).
1. a) In Chapter 6, 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 population proportion.
b) Is finding an interval estimate an example of inferential or descriptive statistics? Explain.
It is an interval estimate is an example of inferential statistics, as an estimate of the value of the population parameter is made based on sample statistics.
c) An interval estimate (23.8, 30.6) is determined for the mean age of NSCC students. Identify the point estimate and the margin of error of the interval estimate.
The point estimate=27.2 and the margin of error=3.4
d) The proportion of registered voters in Washington State in favor of a referendum to lower college tuitions is estimated with a 95% confidence level to be 45% with a margin of error ±3%. Is it possible the referendum will pass? Explain.
Yes, It is possible the referendum will pass. The margin of...

...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)
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
Confidence Interval for µ using
normal distribution
95% confidence interval
• Suppose we want to construct a 95%
confidence interval 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 new
clinic in KL. Before operation, he wants to know
the average charge for common illness (e.g....

...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...

...pounds. a. What is the estimated population mean? b. Determine a 95 percent 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 Interval using Student’s t distribution
Here, t = 2.0301 for 95% confidence with 34 d.f.
A 95 percent confidence interval for the population mean is given by
(xbar – t*SE, xbar + t*SE)
= (3.01 – 2.0301 *0.005, 3.01 + 2.0301 *0.005)
= (3.01 – 0.0102, 3.01 +0.0102)
= (2.9998, 3.0202)
Exercise 34. A recent survey of 50 executives who were laid off from their previous position revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. Construct a 95 percent confidence interval for the population mean. Is it reasonable that the population mean is 28 weeks? Justify your answer.
Here it is given that, xbar = 26, s = 6.2, n = 50.
The Standard error of the mean, SE = s/Sqrt(n) = 6.2/Sqrt(50) = 0.8768
The z- score for 95% confidence is z = 1.96
A 95 percent confidence...

...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 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% confidence interval. 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% confidence interval. 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% confidence interval. Thus, the sample mean gives us an approximation of the population mean. The same holds true for a 90%, a 99%, or...