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Topics:
- Identifying and Estimating the Target Parameter
- Confidence Interval for a Population Mean: Normal ( z ) Statistic
- Confidence Interval for a Population Mean: Student’s t - Statistic
- Large-Sample Confidence Interval for a Population Proportion
- Determining the Sample Size Learning Objectives:
- Estimate a population parameter (means, proportion, or variance) based on a large sample selected from the population
- Use the sampling distribution of a statistic to form a confidence interval for the population parameter
- Show how to select the proper sample size for estimating a population parameter 6 .1 Identifying and Estimating the Target Parameter
Target parameters - NOTATION:
- population mean
^2 - population variance
p - population proportion
Introductory concepts (review)
Parameter – a numerical feature of a population
Target Parameter : population mean, population proportion, population variance
- any parameter we are interested in estimating
Statistic is any numerical measure calculated from data: the proportion, mean,
median, range, variance, standard deviation, etc.
Statistical inference: a method that converts the information from random
samples into reliable estimates of the population parameters.
A point estimate: a single number calculated from a sample that can be
regarded as an educated guess for an unknown population parameter.
A point estimator of a population parameter is a rule or formula that tells us
how to use the sample data to calculate a single number that can be used as an
estimate of the target parameter
Goal: Use the sampling distribution of a statistic to estimate the value of a
population parameter with a known degree of certainty.
Determining the Target Parameter
Parameter Key Words or Phrases Type of Data
μ Mean; average Quantitative
p Proportion; percentage; fraction; rate Qualitative
Notation: Parameter Estimator
Proportion p p ˆ
Mean x
Variance ^2 s^2
When using a sample statistic to estimate a population parameter, some
statistics are good in the sense that they target the population parameter and
are therefore likely to yield good results. Such statistics are called unbiased
estimators. Sample mean, sample variance and sample proportion are the
examples of unbiased estimators.
Example 1: The sample mean X is an estimator of the population mean . The
observed (computed) value is called a point estimate of
An interval estimator (or confidence interval ) is a range of numbers that
contain the target parameter with a high degree of confidence.
Recall from the last chapter
By the rule of thumb, n=30 is “large enough” to justify the normality of the
distribution of sample means.
6 .2 Confidence Interval for a Population Mean: Normal ( z ) Statistic
For an approximately normal distribution we expect 95% of all data to stay within
2 standard deviations from the mean.
x 4
The confidence level is the confidence coefficient expressed as a percentage:
If our confidence level is 95%, then in the long run, 95% of our confidence
intervals will contain μ and 5% will not.
We can select any confidence level we like. Typical sizes are below.
Notation: The confidence level is expressed in percent and marked
100(1-α)%, so α = (1 – confidence coefficient)
Example: Confidence Level Confidence Coefficient α tail size α/
Critical Value
A critical value is a number of standard deviations separating likely location of
the population parameter from the unlikely locations on a number line
representing all data of standard distribution. For instance, z=1.96 is the number
of standard deviations separating 2.5% of the highest data under normal
distribution.
We will deal with two standard distributions: normal and t-distribution.
For critical value in confidence intervals we will use the notation zα/2 representing
standard normal distribution. (The tα/2 represents t-distribution which will be
introduced later)
zα/2 = critical value for the standard normal distribution (z-distribution)
Example: Find critical value zα/2 separating 95% of most likely scores from the
remaining 5% of the least likely scores under Standard Normal curve.
Solution: On the illustration below we see that 95% of the most likely scores are
represented by green area under the curve. Five percent of least likely scores
are represented by two red areas, “the tails”.
- z (^) α/2=-1.96 z (^) α/2=+1. Critical values for z distribution can be found in the last row (marked ∞) of Table Crit. Values of t: (in the end of the text). Exercise: Finding Commonly Used Critical Values zα/2 for (1-α) Confidence Level: Confidence level 90%, α=10%, α/2=5%, and zα/2=.............. Confidence level 80%, α=20%, α/2=10%, zα/2=..............
Compare the sizes of the intervals. What can be concluded?
Exercise : for sample mean = 5, standard deviation = 2 and confidence level
Find:
a) CI for sample size = 100
b) CI for sample size = 200
c) CI for sample size = 1000
Compare the sizes of the intervals. What can be concluded?
Classwork:
6.2 What is the confidence level of each of the following confidence intervals for μ? 6.3 A random sample of n measurements was selected from a population with unknown mean μ and known standard deviation σ. Calculate a 95% confidence interval for μ for each of the following situations:
Exercise:
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large
airline wants to estimate its average number of unoccupied seats per flight over
the past year. To accomplish this, the records of 225 flights are randomly
selected, and the number of unoccupied seats is noted for each of the sampled
flights.
Use enclosed printout to estimate μ, the mean number of unoccupied seats per
flight during the past year with a 90% confidence. Interpret the result.
6.13 Budget lapsing at army hospitals. Budget lapsing occurs when unspent funds do not carry over from one budgeting period to the next. Refer to the Journal of Management
Accounting Research (Vol. 19, 2007) study on budget lapsing at U.S. Army hospitals. Because budget lapsing often leads to a spike in expenditures at the end of the fiscal year, the researchers recorded expenses per full-time equivalent employee for each in a sample of 1,751 army hospitals. The sample yielded the following summary statistics: x¯=$6,563 and s=$2,484. Estimate the mean expenses per full-time equivalent employee of all U.S. Army hospitals using a 90% confidence interval. Interpret the result.
Chapter 6 .3 Confidence Interval for a Population Mean: Student t-Statistic
Key concepts: t-statistic, t-distribution, degrees of freedom (df) Derivation of confidence interval for large sample was based on the fact, that if the sample size n is large and if we replace σ by s, then both statistics 𝑥̅ −𝜇 𝑠 √𝑛 and 𝑥̅ −𝜇 𝜎 √𝑛 have approximately the same distribution ( z-distribution ). This is not true for small n. t-Distribution. If we are sampling from normal population, then the sampling distribution of sample means for small samples is not exactly normal. The shape is also bell shape, but the thickness of the tails varies with sample size n. the statistics ( t - statistic) 𝑡 =
has so called t - distribution with df = n - 1 degrees of freedom. Properties:
3.6, 4.2, 4.0, 3.5, 3.8, 3.1. What is the 90% confidence interval estimate of the population mean task time? Assume normality of distribution of the times. a) Solve by hand b) Check with the calculator Class Exercises
6.25 p. 317
Let t 0 be a particular value of t. Use Table to find t 0 values such that the following statements are true. a. P( - t 0 < t < t 0 ) = .90 where n= b. P( t ≤ t 0 ) = .05 where n=16. Exercise 1 [ 6.28, p. 318 ] The following sample of 16 measurements was selected from a population that is approximately normally distributed: 91 80 99 110 95 106 78 121 106 100 97 82 100 83 115 104
a. Compute the sample mean and the sample standard deviation
b. Construct an 80% confidence interval for the population mean by hand, and
then repeat using TI- 83
c. Construct a 95% confidence interval for the population mean and compare it
with that of part b.
d. Carefully interpret each of the confidence intervals and explain why the 80%
confidence interval is narrower.
e. What assumption is necessary to ensure the validity of this confidence interval?
Exercise To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The DA tests on this cigarette yielded mean nicotine content of 26.7 milligrams and standard deviation of 2.4 milligrams for a sample of 9 cigarettes. Construct a 98% confidence interval for the mean nicotine content of this brand of cigarette. What assumption do you have to make to solve the problem?
6.4 Large Sample Confidence Interval for the Population Proportion p Suppose that p is an unknown population proportion of elements of certain type S. The estimator of p is the sample proportion where x is the number of elements of type S in the sample. In Chapter 5 we studied sampling distribution of sample proportions P ˆ By CLT, for large random samples (np≥15 and nq≥15), the distribution is approximately normal with the mean p and standard deviation Large Sample (1-α)100% Confidence Interval for the Population Proportion p Estimated parameter: population proportion: p Assumptions: large sample size: ( n p ˆ ≥ 15 and ) n q ˆ ≥ 15 Random sample (1-α)*100% Confidence Interval: where q = 1 – p, q̂ = 1 - p̂, and zα/2 is the critical value for the standard normal distribution Example Solution: First, check the assumptions: Random sample, and
np ̂ = 34 > 15, nq ̂ = 66 > 15 assumptions are satisfied
p ̂ = 34/100 = 0.34, q ̂ = 1-.0.34 = 0.
1 - α = 0.95, α = 0.05, α/2 = 0.025, zα/2 = z0.025 = 1.960 (the Table)
pq n
n x P ˆ n pq p z ˆ ˆ ˆ (^) / 2
Class Exercises: 6.42 A random sample of size n=121 yielded p^=.88. a. Is the sample size large enough to use the methods of this section to construct a confidence interval for p? Explain. b. Construct a 90% confidence interval for p. c. What assumption is necessary to ensure the validity of this confidence interval? 6.50 Nannies who work for celebrities. The International Nanny Association reports that in a sample of 528 in-home child care providers (nannies), 20 work for either a nationally known, locally known, or internationally known celebrity ( 2011 International Nanny Association Salary and Benefits Survey ). Use Wilson's adjustment to find a 95% confidence interval for the true proportion of all nannies who work for a celebrity. Interpret the resulting interval. 6.54 Interviewing candidates for a job. The costs associated with conducting interviews for a job opening have skyrocketed over the years. According to a Harris Interactive survey, 211 of 502 senior human resources executives at U.S. companies believe that their hiring managers are interviewing too many people to find qualified candidates for the job ( Business Wire , June 8, 2006). a. Describe the population of interest in this study. b. Identify the population parameter of interest, p. c. Is the sample size large enough to provide a reliable estimate of p? d. Find and interpret an interval estimate for the true proportion of senior human resources executives who believe that their hiring managers interview too many candidates during a job search. Use a confidence level of 98%. e. If you had constructed a 90% confidence interval, would it be wider or narrower?
6.5 Determining the Sample Size Recall that a confidence interval is of the form point estimator ± margin of error, called by the author Sampling Error (SE) Example If a confidence interval is [33.9, 35.1] or 34.5 ± 0.6, then margin of error = 0. (a half of the width of the interval) Sampling Errors:
Sample Size Determination for 100(1 – ) % Confidence Interval for μ
To estimate population mean with given sampling error, confidence level and known standard deviation, the required sample size can be found by the formula derived from the equation above for SE by isolating n: 𝑛 =
𝑧𝛼^2 / 2 𝜎^2
𝑆𝐸^2
Round always UP! Example: The manufacturer wishes to estimate the mean inflation pressure to within .025 pound of its true value with a 99% confidence interval. The standard deviation of inflation pressure is about 0.1 (pound). What sample size should be used? Note: If sigma is unknown, some researchers use sample standard deviation s, or even a quarter of the range instead.
One More Exercise The countries of Europe report that 46% of the labor force is female. The United Nations wonders if the percentage of females in the labor force is the same in the United States. Representatives from the United States Department of Labor plan to check a random sample of over 10,000 employment records on file to estimate a percentage of females in the United States labor force. a). The representatives from the Department of Labor want to estimate a percentage of females in the United States labor force to within ±5%, with 90% confidence. How many employment records should they sample? b) They actually select a random sample of 525 employment records, and find that 229 of the people are females. Create the confidence interval. Show steps: find standard error and margin of error, then write the interval in the interval notation) Find Standard Error, Critical value and Margin of error, then Confidence Interval. c) Should the representatives from the Department of Labor conclude that the percentage of females in their labor force is lower than Europe’s rate of 46%? Explain.
