**Unformatted text preview: **Business Statistics
Fifth Edition Ken Black Chapter 9:
Statistical Inference:
Hypothesis Testing for Single Populations Not to be reproduced without the written permission of PB. Alt 9.1 Introduction to Hypothesis Testing [Black, page 292]
9.2 Hypothesis Testing for a Population Mean (0' Known ) [Black, page 301] 0 Two Basic Types of Inference
0 Estimation (Chapter 8)
' Point estimation (a single number)
I Interval estimation (an interval of probable values)
0 Statistical Hypothesis Testing (Chapter 9)
' Null Hypothesis (H0): Parameter = Some Value I Alternative or Research Hypothesis (Ha):
Parameter i ( or < ; or >) Some Value Example:
The quality control engineer in a bottling mantras to determine if the average content of the containers is 16 02. (Or, is the p 5 mean at the target value of
16 02.?) Past studies indicate that the content is normally distributed and that the
process standard deviation is 0.25 02. A random sample of size 25 yielded a
sample mean of 16.15 oz. Using a signiﬁcance level of 5%, determine if the
process mean is at the target value. By the way, the quality control engineer is
interested in detecting deviation in either direction from 16 ounces. State the null and alternative hypotheses: H0: E vs. Ha: ( — Sided Alternative Hypothesis)
<16 >16
H8 H0 Ha
“—37—“ Possible Values of ll
0 Ha is one of 3 possibilities
- Ha: 16 or Ha: 16 or Ha: 16 Not to be reproduced without the written permission of F .8. Alt 2 o HTAB System of Hypothesis Testing [Black, page 297] N. -m M WWWWMWW44 «m WWW fhzk WHW «M4 MW,» We M 4%.“..me .M Wm M Hi.“ wigewo .M .m‘ WM.» Mywaw N, This 3: ii} pg, ﬁliesim‘ >
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_ WWNWWWWWWW Figures 9.2 and 9.3 Not to be reproduced without the written permission of PB. Alt 0 Type I and Type II Errors [Black, page 300]
0 Recall H0: [1 =16 vs. Ha: ##16 . Base decision about the parameter ,1! on the statistic ; . ; is "close" to 16, conclude that H0 is true (do not reject it). If
- If 1: is not "close" to 16, reject HO. 0 Since decision is based on and ,U is unknown, the decision to reject or not reject could be incorrect. 0 Two possible types of errors. [E Null Hypothesis True = ? F (115E.1
a Do Not Reject
Decision Based on H0 E] Sample Mean X
Reject H0 [E] E Not to be reproduced without the written permission of RB. Alt 4 o If H0 is true and we incorrectly reject it, then a Type I error occurs. I The probability of making a Type I error is symbolized by a
aka the “level of signiﬁcance.” or = P( Type I Error) = P( Reject H0 | True) 0 If H0 is false and we incorrectly fail to reject it, a Type II error occurs. I The probability of making a Type 11 error is symbolized by ,6 . ,3 = P( Type II Error) = P( Not Rejecting H0 l False) 0 Decision will be: I Reject HO if X’ is not close to 16 01‘ I Do not reject HO if A7 is close to 16. 0 Never say “Accept H0” o What is "close"? Not to be reproduced without the written permission of F.B. Alt 5 o Rejection and Nonrejection Regions [Black, page 299] 0 Fix (1 at 5%. Given that the sample size is 25 and o = .25 oz. 0 a: P T e I Error aka the sini zcance level of the test 0 .05=P( Rejecting HOIHO is true)
0 =P(/X7>[_7E_]u=16)+P(A7<k1|u=16) o .025 = P(X> k2 I u: 16) ' Splitting up or evenly o .025 = P(Z > 1.96) (From Table A5) Therefore, _
LE 2 1.96 .2 5
or, k2=16+1%5/Jz—5)=16098=1610
kI =15.90
0 Distribution of I? if H0: ,1! = 16 is true. E1 .025 _025 k1=15 90 16 k2=16.10
Critical values 0 The critical values determine the rejection region (RR) 0 RR. in terms of X :
‘ 15.90 16.10 <—-———{ |——-—---—>
E 16'
0 Recall n=25 and/l7: 16.1 I Decision: H0: # = 16 at the 5% level of signiﬁcance. Not to be reproduced without the written permission of F3. Alt 6 o In general, in testing H0: # = #0 vs. Ha: ,11 ¢ #0 with 0' known, reject H0 if [Black, page 305- 308]: 55 > #0 +Za/2(0/\/n—) =Upper_Xc or if g X<,u0 —Za/2(0'/\/n—)=L0wer_X C o Equivalently, calculate the standardized test statistic Z [Black, page 302]: z: X410 z .t.
0/6 { statls 1c I Reject H0 if I Z l > Zn ,2 } Rejection Region Example (Bottling Plant): The sample of size 25 yielded 55 = 16.15 _-M)X __1——___6 15- 16 :3 16.15 = estimate of population mean
—U/\/—— .25/5 16 = hypothesized value for pop. mean
R-R- -1.96 1.96 R-R-
——-———| l |———' R.R in terms of Z-statistic
0 i 0 Note: I differs from #0 by 3 standard errors. Decision: H0: # = 16 at the 5% level of signiﬁcance. 0 Summary of the rejection regions for each type of research hypothesis
is on page 13. Not to be reproduced without the written permission of EB. Alt 7 9.6 Solving for Type II Errors and Power of a Test [Black, page 326]
0 Probability of Type 11 error ([3) o ,3 = P( Not rejecting H0 I H0 is false ) Example (Bottling Plant): Suppose the true process mean is 16.2.
What’s the chance of sayiEIIe process is on target? ﬂ=P(15.90S)?316.10|,ua 216.2) P(1590-162 S231610—162) .05 .05
= P(-6 s z s —2) =0.0228 = .5 —0.4772 About 2.3% of the time, we conclude the process is on target (/1 = 16 oz.)
when in fact ,1! = 16.2 oz. E a/2 = .025 ﬂ= .0228 16.2 0 Graph showing the relationship between 06 and ,3 . Not to be reproduced without the written permission of F8. Alt 8 o B is not a single number but a collection of probabilities. Examgle (Bottling Plant): What’s the chance of saying the process is on target for other values of H a ? #1.. 3 16.05 0.830 =.4985 + .3315
16.10 0.484 =.5000 - .0160
16.20 0.0228 Conclusion: LV—J o The graph of :8 VS. #0 is called the Operating Characteristic curve. 0 General comments on a and ,6
I As P( Type 1 Error) decreases, P( Type 11 Error) E] . . Given any two of the three (a, ﬂ, n), we can determine the other. 0 The power of a statistical test is the probability of rejecting the null hypothesis when the null hypothesis is false. Examgle (Bottling Plant): Find the power of the test if #a = 16.2. Power=l~ﬂ=1- = Not to be reproduced without the written permission of F.B. Alt 9 0 Using p-Values to Test Hypotheses [Black, page 305] o The p-value is the chance of observing the sample result, or a more extreme result, if the null hypothesis is true. Example (Is a coin fair?):
H0: Coin is fair (p=0.5) Ha; Coin is not fair
I Toss coin 10 times. I Observe 10 heads.
' If coin is fair, P(10 heads in 10 tosses) = (1/2)10 = .000976 = -Value
Since p—value is small, hypothesis of a fair coin must be E . 16.15—16 P(X >16.15|,u=l6)=P[Z> .05 )=P(Z>3)=.00135 Example (Bottling Plant): Find the p—Value of the test. We had X = 16.15. We could have gotten a comparable value of X to the left of 16.
o p—value = 2(.00135) = .0027
I Use the factor “2” only for two—sided Ha I and when calculating the p-Value manually. o Reaching a conclusion: If p-value < a, reject H0 Example (Bottling Plant):
Since p-Value = .0027 < 0.05, H0 Not to be reproduced without the written permission of F.B. Alt 10 0 Procedure for using Minitab to do Z—Test:
0 Click on Stat > Basic Statistics > l—Sample Z. 0 Enter column Where data is stored, unless data is summarized. 0 Click on “Perform hypothesis test” 0 Enter “16” for hypothesized mean. Click on “Options” 0 Enter Conﬁdence level “95.0”. [Corresponds to signiﬁcance level of 5%]
0 Select “not equal” for alternative. 0 Suppose the 25 data values are: Ounces
15.67 16.16
15.78 16.17
15.82 16.18
15.86 16.20
15.91 16.22
15.93 16.25
15.97 16.28
16.03 16.32
16.05 16.36
16.06 16.47
16.12 16.52
16.14 16.57 16.71 0 The Minitab output is: One-Sample Z: Ounces
Test ofmu =16 vs not = 16
The assumed standard deviation = 0.25 E] Variable N Mean StDev SE Mean 95% Cl Z P
Ounces 25 16.1500 0.2558 0.0500 (16.0520,16.2480) 3.00 0.003 0 Note: 0.003 is the p—value for a two—sided alternative hypothesis. Not to be reproduced without the written permission of PB. Alt 11 o Assumptions for Validity of Z—test
0 Population standard deviation (0') is known.
0 Independence between the observations within the sample. 0 Data comes from a normally distributed population. I We really require X be normally distributed. I Consequences of nonnormality depend on the severity of skewness. Example (Bottling Plant): Were the assumptions met to use the Z-test? o The population standard deviation(c) known ?
0 Independence between the observations within the sample. E 0 Does the data come from a normally distributed population ? . At a minimum, is X normally distributed ? Normal Probabiltiy Plot
Normal 8 I I I l I i I Mean 16.15
I i I i I I : swev 0-2558
4 ~ 25 I AD 0.142 P—Value 0.968 a I
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25 88883 8 8 "-'T'-‘-F__“1"——T"“r‘
~ I I I I I
‘-T—”——T"‘"F“‘1‘--'T—"‘T— I I I I I
T— r 7 —"T—“_T‘_ F '—W——-'T‘—__F— l l I I I I I I I I I I I I I I I I I I I 15.75 16.00 16.25 16.50 16.75
Ounces Conclusion: E Not to be reproduced without the written permission of PB. Alt 12 Tests of Hypotheses on a Mean( ,1! )
When Sampling from a Normal Distribution with Known Variance (0'2) Null hxgothesis Value of test statistic under H0
I? w
0 0 0/6
Alternative hypothesis Rejection region
Ha: ,u i ,u0 Reject H0 when Z < “Za/g or when Z > Za/z Ha: it > #0 Reject H0 when Z > 20, E
Ha: I! < #0 Reject H0 when Z < “'20, o A 100(1 - a)% conﬁdence interval for ,U is: XiZa/Z .9:- J; Not to be reproduced without the written permission of F8. Alt 13 9.3 Hypothesis Testing About I” with 0' Unknown [Black, page 310] 0 An easy extension of the Z-statistic: X- a/x/E
[E] . 0 Replace the Z-percentile by the t—percentile. Z—statistic- —— 0 0' is unknown, replace it by o The t—test statistic: t—statistic = 0 Summary of the rejection regions for each type of research hypothesis is on page 18. Exam le: A bank ms 5 a “snake system” waiting line at its counters to try t-the averagew witin ng time. The mean waiting time under speciﬁc conditions
with the previous system was 6.1 minutes. A sample of 15 waiting times is taken;
the times are measured at widely separated intervals to eliminate the possibility of dependent observations. The resulting sample statistics are shown below. Descriptive Statistics: Wait'lEL
Variable N Mean SE Mean StDev WaitTimes 15 5.732 0.608 2.356 Test the null hypothesis of no change against an appropriate alternative hypothesis, using a 20.10. Assume that the population distribution of waiting times is approximately normal. Not to be reproduced without the written permission of F.B. Alt 14 Let H denote the true average waiting time with the snake system. Conclusion: (Reject, H0: /1 =6.1 at the 10% level of signiﬁcance. Business Decision: Not to be reproduced without the written permission of PB. Alt 15 0 Procedure for using Minitab to do t-Test:
0 Click on Stat > Basic Statistics > 1—Sample t.
0 Enter column Where data is stored, unless data is summarized. 0 Click on “Perform hypothesis test”
0 Enter “6.1” for hypothesized mean. Click on “Options” 0 Enter Conﬁdence level “90.0”. [Corresponds to signiﬁcance level of 10%] 0 Select “less than” for alternative. 0 Suppose the 15 data values are: WaitTimes
1.95
3.79
5.64
6.36
9.50
7.11
0.83
6.86
6.74
6.30
4.10
8.45
4.76
5.52
8.06 o The Minitab output is: One-Sample T: WaitTimes Testofmu=6.1vs<6.l E 90% Upper
Variable N Mean StDev SE Mean Bound T P [—75—]
WaitTimes 15 5.732 2.356 0.608 6.551 -0.60 0.278 0 Note: 0.278 is the p-value for a one—sided alternative hypothesis. Not to be reproduced without the written permission of PB. Alt 16 o Assumptions for Validity of t-test 0 Independence between the observations within the sample. 0 Data comes from a normally distributed population. . The consequences of nonnormality depend on the type of nonnormality. I If the population distribution is skewed, a and the p-value may be in error, particularly for one-tailed tests. I If the population distribution is symmetric, but heavy—tailed, a and the p—Value are reasonably accurate. Example : Were the assumptions met to use the t-test? 0 Independence between the observations within the sample. 0 Does the data come from a normally distributed population ? Normal Probabiltiy Plot
Normal 68 I I I l I I I I I I I I ____I___.J----L_-_L__._
I I I 95 I
I I I I I
SO —-I——--I—-—-+-—-I--——I—---I--——I——--I--——
I I I I I I I I
I I I I I I I I
so -I----I—--‘I—-“I—--—~I--~-I—---I—-—-f—--
I I I I I I I I.
70 4--...I.--_I_-_.I._-_I.__-I.-__I___
‘5 I I I I I I
o 60 I I I I I I
U 50 -I——-—I-———I————I————I—-——I--
8- I I I I I o I I I I I I I
I? 4° “""""'""'"“r"“. "I"T'"T'"T"'T"’T"l
30 ‘ I'“‘1""'1"‘”‘T"‘T““T"'“T"“‘I
I I I I I I I I
20 —-— I“°‘P-‘-P‘-—+--—+-—-+-—-+---+----§ .I I I I I I I I I H
O U! Conclusion: Not to be reproduced without the written permission of PB. Alt 17 Tests of Hypotheses on a Mean( ,1! )
When Sampling from a Normal Distribution with Unknown Variance (02) Null hygothesis Value of test statistic under H0
Horﬂ=ﬂo tZX‘ﬂo s / J1;
Alternative hygothesis Rejection region
Ha 3!! 75 #0 Reject H0 when t < —ta/2,n_1 or when t > ta/Zm—l
Ha: ,u > #0 Reject H0 when t > tam—1 Ha: ,u < #0 Reject Ho when t < ‘tagﬂ
0 A 100(1 - a)% conﬁdence interval for ,U is: fit S
a/2,n—l Tn- Not to be reproduced without the written permission of F .8. Alt 18 ...

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