7 hypothesis testing1
TRANSCRIPT
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Chapter
7
Elementary Statistics Larson Farber
Hypothesis Testing
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A Statistical Hypothesis
Alternativehypothesis H a contains a statement
of inequality , such as.
Null hypothesis H 0 tatistical hypothesis
hat contains a
tatement of equality ,uch as , = or .
If I am false,you are true If I am false,
you are true
H 0 H a
Complementary Statements
A claim about a population.
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Write the claim about the population. Then,write its complement. Either hypothesis, the
ull or the alternative, can represent the claim.
A hospital claims its ambulance responsetime is less than 10 minutes .
H0 : 10 min
Ha : 10 min claim
Ha : 60.0 p
H0 : 60.0 p claim
Writing Hypotheses
A consumer magazine claims theproportion of cell phone calls made duringevenings and weekends is at most 60%.
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A type I error: Null hypothesis is actuallyrue but the decision is to reject it.
Level of significance, a Maximum probability of committinga type I error.
D e c
i s i o n
Actual Truth of H 0
Errors and Level of Significance
H0 True H0 False
Do not
reject H 0
Reject H 0
Correct
Decision
CorrectDecision
Type II
Error
Type IError
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Sampling distribution for x
The rejection region is the range of values for which the null hypothesis is not
probable . It is always in the direction of the alternative hypothesis. Its area is equalto a .
Rejection Region
0 z z0
A critical value separates the rejectionregion from the non-rejection region
Critical Value z0
Rejection Regions
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z00
-z0
0 z0
Right- tail test Ha: >valu Reject H 0 if z > z 0 otherwise fail to reject H 0.
Two- tail test Ha: value
Reject H 0 if z > z 0 or z
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Claim is H 0
There is notenoughevidence toreject the claim
There isenoughevidence toreject the claim
Interpreting the Decision
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Claim is H a
There is notenoughevidence tosupport theclaim
There isenoughevidence tosupport theclaim
Interpreting the Decision
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1 . Write the null and alternative hypothesis
2 . State the level of significance
3 . Identify the sampling distribution
Write H 0 and H a as mathematical statements.Remember H 0 always contains the = symbol.
This is the maximum probability of rejecting thenull hypothesis when it is actually true. (Makinga type I error.)
The sampling distribution is the distribution for the test statistic assuming that H 0 is true andthat the experiment is repeated an infinitenumber of times.
8 Steps in a Hypothesis Test
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7 . Make your decision
6. Find the test statistic
5. Find therejection region
4 . Find thecritical value
8 . Interpret your decision
The critical value separates the rejection regionof the sampling distribution from the non-rejection region.
Perform the calculations to standardize your sample statistic.
If the test statistic falls in the critical region,reject H 0. Otherwise, fail to reject H 0.
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he critical value z 0 separates the rejection regionrom the non-rejection region. The area of theejection region is equal to a .
z0 0
ejectiongion
z00
Rejectionregion
-z0 0 z 0
Reject ionregion
Rejectionregion
ind z 0 for a left-tailest with a =.01
Find z 0 for a right-tailtest with a =.05
Find - z0
and z0
for a two-tail test with a =.01
z0=-2.33
-z0=-2.575 and z 0 =2.575
z0=1.645
Critical Values
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A cereal company claims the meansodium content in one serving of its cereal isno more than 230 mg. You work for anational health service and are asked to test
this claim. You find that a random sample of 52 servings has a mean sodium content of 232 milligrams and a standard deviation of 10 mg. At a = 0.05, do you have enough
evidence to reject the companys claim? 1. Write the null and alternative hypothesis
H0: 230 mg.(claim) Ha: > 230 mg.
2. State the level of significancea = 0.05
3. Determine the sampling distribution
Since the sample size is at least 30, the samplingdistribution is normal.
The z-test for a Mean
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7. Make your decision
6. Find the test statistic
8. Interpret your decision
5. Find the rejectionregion
Rejectionregion
Since H a contains the > symbol, this is a right tail test.
n=52x = 232s=10
44.1387.1
2
5210
230232 z
= 1.44 does not fall in the rejection region, soail to reject H 0
here is not enough evidence to reject the
ompanys claim that there is at most 230mg of odium in one serving of its cereal.
z00
1.645
4. Find the criticalvalue
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The P-value is the probability of obtaininga sample statistic with a value as extremeor more extreme than the one determinedby the sample data. Reject H 0 if P > a .
z 0
Area inleft tail
For a left tail test
z 0
Area in right tail
For a right tail test
z 0 z
If z is
negative,twice thearea inthe lefttail
If z is positivetwice the
area in theright tail
For a two-tail test
P-Values
P-value = indica ted area
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The standardized z -score for the hypothesistest on the sodium content of cereal was z =1.44. Find the P-value and make your decisionat the 0.05 level of significance.
The standardized z -score for the hypothesisest on the sodium content of cereal was z =.44. Find the P-value and make your decisiont the 0.05 level of significance.
ince this was a right-tail test, the P-value is therea to the right of the z score. The area to theght of z = 1.44 is 0.0749 so the P-value is .0749.ince 0.0749 > 0.05, fail to reject H 0.
If P = 0.0246, what is your decision if 1) a =0.052) a =0.01
If P = 0.0246, what is your decision if 1) a =0.052) a =0.01
1) Since 0.0246 < 0.05, reject H 0.
2) Since 0.0246 > 0.01, fail to reject H 0 .
Using P-Values
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For a P-value decision, compare areas.If P < , reject H 0. P , fail to reject H 0.
0
For = 0.05, z 0 = -1.645. If z = -1.23
the P-value is 0.1093. The actual area to theleft of the z -score is 0.1093. With astandardized z -score of -1.23, fail to rejectthe null hypothesis at the 0.05 level of
significance.
z0
Rejection area
0.05
z
If = 0.05 and z = -1.23
Using the P-value of a testcompare areas
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Find the critical value t 0 for a left-tailedtest given a = 0.01 and n = 18.
Find the critical value t 0 for a left-tailedtest given a = 0.01 and n = 18.
0
Area inleft tail
Find the critical values - t 0 and t 0 for atwo-tailed test given a = 0.05 and n = 11.
Find the critical values - t 0 and t 0 for atwo-tailed test given a = 0.05 and n = 11.
d.f.= 18-1 = 17
to
t0 = - 2.567
0
d.f.= 11-1 = 10
-t0
t0
-t0 =-2.228 and t 0=2.228
The t SamplingDistribution
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A university says the mean number of classroomhours per week for full-time faculty is 11.0. Aandom sample of the number of classroom hoursor full-time faculty for one week is listed below.
You work for a student organization and are askedo test this claim At a = 0.01, do you have enough
evidence to reject the university s claim?11.8 8.6 12.6 7.9 6.4 10.4 13.6 9.1
A university says the mean number of classroomours per week for full-time faculty is 11.0. Aandom sample of the number of classroom hoursor full-time faculty for one week is listed below.
You work for a student organization and are askedo test this claim At a = 0.01, do you have enoughvidence to reject the universitys claim? 1.8 8.6 12.6 7.9 6.4 10.4 13.6 9.1
. Write the null and alternative hypothesis
H0: = 11.0 (claim) Ha: 11.0
. State the level of significance
a = 0.01
3. Determine the sampling distribution
ince the sample size is 8, the samplingistribution is a t-distribution with 8 - 1 = 7 d.f.
Testing -small Sample
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7. Make your decision
6. Find the test statistic
8. Interpret your decision
n=8x = 10.050s=2.485
08.1878.0
95.0
8485.2
0.11050.10t
= -1.08 does not fall in the rejection region, soail to reject H 0 at the 0.01 level of significance.
here is not enough evidence to reject the
niversitys claim that faculty spend a mean of 11lassroom hours.
5. Find the rejectionregion
Rejectionregion
Rejectionregion
ince H a contains the symbol, this is a two- tail test.
-t0 0 t0
4. Find the criticalvalues
-3.499 3.499
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T-Test of the Mean
Test of mu = 11.000 vs mu not = 11.000
Variable N Mean StDev SE Mean T PHours 8 0.050 2.485 0.879 -1.08 0.32
Enter the data in C1, Hours.Choose t-test in the STAT menu .
Minitab reports the t-statistic and the P-value.
Since the P-value is greater than the level of significance (0.32 > 0.01), reject the nullhypothesis at the 0.01 level of significance.
Minitab Solution
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Test for Proportions
A communications industry spokespersonlaims that over 40% of Americans either owncellular phone or have a family member whooes. In a random survey of 1036 Americans,56 said they or a family member owned aellular phone. Test the spokespersons claim
t a = 0.05. What can you conclude?
. Write the null and alternative hypothesis
H0: p 0.40 Ha: p > 0.40 (claim)
. State the level of significance a = 0.05
The standardized test statistic is:n
pq
p p z
is the population proportion of successes.The test statistic isf np 5 and nq 5 the sampling distribution for
normal.
n x p
p
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3. Determine the sampling distribution
7. Make your decision
6. Find the test statistic
8. Interpret your decision
6.201522.0
04.0
1036)60)(.40(.
40.44. z
z = 2.63 falls in the rejection region, so reject H 0
here is enough evidence to support the claim that
ver 40% of Americans own a cell phone or have aamily member who does.
0
036(.40) > 5 and 1036(.60)>5 . The s amplingistribution is normal.
n =1036x = 456
4. Find the criticalvalue
1.645
5. Find the rejection
region
Rejectionregion
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Find a c 20 critical value for a left-tail test when=17 and a = 0.05.
2 is the test statistic for the population variance. Itsampling distribution is a c 2 distribution with n-1 d.f.
Find critical values c 20 for a two-tailed test when= 12 and a = 0.01.
he standardized test statistic is2
22 )1(
c
sn
c 20 =7.962
c 2l =2.603 and c 2R =26.757
0 1 0 2 0 3 0 4 0
Critical Values for 2
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A state school administrator says that the standarddeviation of test scores for 8th grade students whotook a life-science assessment test is less than 30.You work for the administrator and are asked to test
this claim. You find that a random sample of 10 testshas a standard deviation of 28.8. At a = 0.01, do youhave enough evidence to support the administratorsclaim? Assume test scores are normally distributed.
. Write the null and alternative hypothesis
H0: 30 Ha: < 30 (claim)
. State the level of significance a = 0.01
3. Determine the sampling distribution
he sampling distribution is c 2 with 10 - 1 = 9 d.f.
Test for
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7. Make your decision
6. Find the test statistic
8. Interpret your decision
n=10s = 28.8
c 2 = 8.2944 does not fall in the rejection region,so fail to reject H 0
here is not enough evidence to support the
2944.8
30
8.28)110()1(2
2
2
22
c sn
0 1 0 2 0 3 0 4 0
5. Find the rejectionregion
4. Find the criticalvalue
2.088