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Inference Chapter About A Population
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Copyright 2005 Brooks/Cole, a division of Thomson Learning, Inc.
12.1 Introduction
In this chapter we utilize the approach developed before to describe a population.
Identify the parameter to be estimated or tested. Specify the parameter's estimator and its sampling distribution. Construct a confidence interval estimator or perform a hypothesis test.
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12.1 Introduction
We shall develop techniques to estimate and test three population parameters.
Population mean Population variance 2 Population proportion p
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ecall that when is known we use the following tatistic to estimate and test a population mean
z= x-
12.2 Inference About a Population Mean When the Population Standard Deviation Is Unknown
n
hen is unknown, we use its point estimator s,
nd the z-statistic is replaced then by the t-statistic
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The t - Statistic
Z tt Z Z-ttt Z Z= x ttt t t t = x - Z s n ssss n sss s s
hen the sampled population is normally distributed e t statistic is Student t distributed.
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The t - Statistic
t=
x-
Using the t-table
s n
The t distribution is mound-shaped, The "degrees of freedom", (a function of the sample size) and symmetrical around zero.
d.f. = v2 v1 < v2 d.f. = v1
0
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determine how spread the distribution is (compared to the normal distribution)
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Testing when is unknown
Example 12.1 - Productivity of newly hired Trainees
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Testing when is unknown
Example 12.1
In order to determine the number of workers required to meet demand, the productivity of newly hired trainees is studied. It is believed that trainees can process and distribute more than 450 packages per hour within one week of hiring. Can we conclude that this belief is correct, based on productivity observation of 50 8
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Testing when is unknown
Example 12.1 Solution
The problem objective is to describe the population of the number of packages processed in one hour. The data are interval.
H0: = 450 We wa reach 90% nt to pro v prod H1: > 450 uctiv e that th e ity of expe trainees x- rienc The t statistic ed w t= orker s s n
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Testing when is unknown
Solution continued (solving by hand)
Fromthe datawe have The rejection region is x = 23019 x2 = 10671 , thus , ,357 i , i t > t,n 1 23019 , x= = 46038 and . , t,n - 1 = t.05,49 50
t.05,50 = 1.676.
s2
n- 1 s = 1507 = 3883 .55 .
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=
( x ) x -
2 i i
2
n = 1507 . .55
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Testing when is unknown
The test statistic is Rejection region 1.676 1.89
t=
x- s
46038- 450 . = = 1.89 n 3883 50 .
Since 1.89 > 1.676 we reject the null hypothesis in favor of the alternative. There is sufficient evidence to infer that the mean productivity of trainees one week after being hired is greater than 450 packages at . Copyright 2005 Brooks/Cole, a significance Inc. 05 division of Thomson Learning, level.
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Testing when is unknown t-Test: Mean
Pack ages Mean 460.38 Standard Deviation 38.83 Hypothesized Mean 450 df 49 t Stat 1.89 P(T<=t) one-tail 0.0323 t Critical one-tail 1.6766 P(T<=t) two-tail 0.0646 t Critical two-tail 2.0096
.05 .0323
Since .0323 < .05, we reject the null hypothesis in favor of the alternative. There is sufficient evidence to infer that the mean productivity of trainees one week after being hired is greater than 450 packages at .05 significance level. Copyright 2005 Brooks/Cole, a division of Thomson Learning, Inc.
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Estimating when is unknown
Confidence interval estimator of when is unknown
x t 2 s n d.f. = n -1
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Estimating when is unknown
Example 12.2
An investor is trying to estimate the return on investment in companies that won quality awards last year. A random sample of 83 such companies is selected, and the return on investment is calculated had he invested in them. Construct a 95% confidence interval for the mean return.
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Estimating when is unknown
Solution (solving by hand)
The problem objective is to describe the population of annual returns from buying shares of quality award-winners. The data are interval. x = 15.02 s 2 = 68.98 s = 68.98 = 8.31 Solving by hand
From the Xm12-02 we determine
x t
2, n -1
s n
15.02 1.990
8.31 83
= [13.19,16.85]
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t.025,82 t.025,80
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Estimating when is unknown
t-Estimate: Mean Returns 15.02 8.31 13.20 16.83
Mean Standard Deviation LCL UCL
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Checking the required conditions
We need to check that the population is normally distributed, or at least not extremely nonnormal. There are statistical methods to test for normality (one to be introduced later in the book). From the sample histograms we see...
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14 12 10 8 6 4 2 0 400
A Histogram for Xm12- 01
425
30 25 20 15 10 5 0 -4
Packages A Histogram for Xm12- 02
450
475
500
525
550
575
More
2
8
Returns
14
22
30
More
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12.3 Inference About a Population Variance
Sometimes we are interested in making inference about the variability of processes. Examples:
The consistency of a production process for quality control purposes. Investors use variance as a measure of risk.
To draw inference about variability, the parameter of interest is 2.
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12.3 Inference About a Population Variance 2
The sample variance s is an unbiased, consistent and efficient point estimator for (n -1 s2 ) 2. 2 The statistic has a distribution called Chi-squared, if the population is (n - 1 s2 ) 2 normally distributed. = d.f. = n - 1 2 d.f. = 5
d.f. = 10
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Testing and Estimating a Population Variance
From the following probability statement P(21-/2 < 2 < 2/2) = 1- we have (by substituting 2 = [(n 1)s2]/ 2.) 2 2
(n - 1 s ) 2 /2 < <
2
(n - 1 s )
2 1- / 2
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Testing the Population Variance
Example 12.3 (operation management
application)
A container-filling machine is believed to fill 1 liter containers so consistently, that the variance of the filling will be less than 1 cc (.001 liter). To test this belief a random sample of 25 1-liter fills was taken, the and results recorded (Xm12-03) Do these data support the belief that the 22
Copyright 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Testing the Population Variance
Solution
The problem objective is to describe the population of 1-liter fills from a filling machine. The data are interval, and we are interested in the variability of the fills. tent The complete test is: is consis cess 2 r the pro H0: = 1 ethe know wh H1: <1
2
t to We wan
(n- 1)s The test statistic is = . 2 2 2 The rejection region is < 1- ,n- 1
2 2
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Testing the Population Variance Solving by hand
Note that (n - 1)s2 = (xi - x)2 = xi2 ( xi)2/n
From the sample (Xm12-03) we can calculate xi = 24,996.4, and xi2 = 24,992,821.3 Then (n - 1)s2 = 24,992,821.3-(24,996.4)2/25 =20.781)s2 2078 (n- . 2 = = 2 = 2078 There is insufficient evidence . , 2 1 to reject the hypothesis that the variance is less than 1. 12- ,n-1 = .295,25-1 = 138484 . .
Since 138484 2078 do not reject . < . , the null hypothesis .
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Testing the Population Variance
= .05
Rejection region
13.8484 20.8
1- = .95
2 < 138484 .
2
.295,25-1
Do not reject the null hypothesis
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Estimating the Population Variance
Example 12.4
Estimate the variance of fills in Example 12.3 with 99% confidence.
Solution
We have (n-1)s2 = 20.78. From the Chi-squared table we have 2/2,n-1 = 2.005, 24 = 45.5585 21 -/2,n-1 2.995, 24 = 9.88623
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Estimating the Population Variance
The confidence interval estimate is
(n- 1 s ) (n- 1 s ) 2 < < 2 2 /2 1- / 2
2
2
2078 . 2078 . 2 < < 455585 . 9.88623 .46< < 2.10
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12.4 Inference About a Population Proportion
When the population consists of nominal data, the only inference we can make is about the proportion of occurrence of a certain value. The parameter p was used before to calculate these probabilities under the binomial distribution.
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12.4 Inference About a Population Proportion
Statistic and sampling distribution
the statistic used when making inference about p is:
^ p= x where n x - the number successes of .
n - samplesize .
Under certain conditions, [np > 5 and n(1-p) > 5], ^ p is approximately normally distributed, with = p and 2 = p(1 - p)/n. 29
Copyright 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Testing and Estimating the Proportion
Test statistic for p ^ p- p
Z= p(1- p) / n
wherenp> 5 and n(1- p) > 5
Interval estimator for p (1- confidence level)
^ ^ ^ p z /2 p(1- p) / n ^ providedp > 5 andn(1- p) > 5 n^
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Additional example
Testing the Proportion
Example 12.5 (Predicting the winner in election day)
Voters are asked by a certain network to participate in an exit poll in order to predict the winner on election day. Based on the data presented in Xm12-05 where 1=Democrat, and 2=Republican), can the network conclude that the republican candidate will win the state college vote?
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Testing the Proportion
Solution
The problem objective is to describe the population of votes in the state. The data are nominal. The parameter to be tested is `p'. Success is defined as "Vote republican". The hypotheses are:
H0: p = .5 H1: p > .5
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More than 50% vote Republican
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Testing the Proportion
Solving by hand
The rejection region is z > z = z.05 = 1.645. From file we count 407 success. Number of voters participating is 765. ^ p is The sample proportion = 407765= .532 The value of the test statistic is ^ p- p .532- .5 Z= = = 1.77 p(1- p) / n .5(1- .5) / 765
The p-value is = P(Z>1.77) = .0382
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Testing the Proportion
z-Test : Proportion Sample Proportion Observations Hypothesized Proportion z Stat P(Z<=z) one-tail z Critical one-tail P(Z<=z) two-tail z Critical two-tail 0.532 765 0.5 1.77 0.0382 1.6449 0.0764 1.96
There is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. At 5% significance level we can conclude that more than 50% voted Republican.
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Estimating the Proportion
Nielsen Ratings
In a survey of 2000 TV viewers at 11.40 p.m. on a certain night, 226 indicated they watched "The Tonight Show". Estimate the number of TVs tuned to the Tonight Show in a typical night, if there are 100 million potential television sets. Use a 95% confidence level. ^ z / 2 p (1 ^ ^ p Solution - p) / n = .113 1.96 .113(1 - .113) / 2000
.113 .014
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Estimating the Proportion
Solution
z - Estimate: Proportion Viewers Sample Proportion Observations LCL UCL 0.113 2000 0.099 0.127
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A confidence interval estimate of the number of viewers who watched the Tonight Show: LCL = .099(100 million)= 9.9 million UCL = .127(100 million)=12.7 million
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Selecting the Sample Size to Estimate the Proportion
Recall: The confidence interval for the proportion is
^ ^ ^ p z / 2 p (1 - p ) / n
Thus, to estimate the proportion to within W, we can write ^ ^ W = z / 2 p (1 - p ) / n
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Selecting the Sample Size to Estimate the Proportion
The required sample size is
z / 2 p(1 - p ) ^ ^ n= W
2
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Sample Size to Estimate the Proportion
Example
Suppose we want to estimate the proportion of customers who prefer our company's brand to within .03 with 95% confidence. 1.96 p(1- p) Find the sample size. ^ ^ n= Solution .03 W = .03; 1 - = .95,
therefore /2 = .025, so z.025 = 1.96
Since the sample has not yet been taken, the sample proportion is still unknown.
2
We proceed using either one of the following two methods:
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Method 1:
Sample Size to Estimate the Proportion
^ There is no knowledge about the value of p ^ Let p = .5 . This results in the largest possible n needed ^ for a 1- confidence .03 p interval of the form . If the sample proportion does not equal .5, the actual W will be Method 2: narrower than .03 with the n obtained by the formula below. ^ p There is some idea about the value of
^ Use the value p of
to calculate the sample size
1.96 .5(1- .5) n= .03
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, = 1068 1.96 .2(1- .2) n= .03
2
= 683 40
2