research hypothesis
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Research hypothesis marketingTRANSCRIPT
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Statistics for Business and Economics
Chapter 7Inferences Based on a Single Sample:
Tests of Hypotheses
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Content
1. The Elements of a Test of Hypothesis
2. Formulating Hypotheses and Setting Up the Rejection Region
3. Observed Significance Levels: p-Values4. Test of Hypothesis about a Population Mean:
Normal (z) Statistic5. Test of Hypothesis about a Population Mean:
Students t-Statistic
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Content
6. Large-Sample Test of Hypothesis about a Population Proportion
7. Test of Hypothesis about a Population Variance
8. Calculating Type II Error Probabilities: More about b*
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Learning Objectives
1. Test a specific value of a population parameter (mean or proportion), called a test of hypothesis
2. Provide a measure of reliability for the hypothesis test, called the significance level of the test
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Learning Objectives
3. Test a specific value of a population parameter (mean, proportion or variance) called a test of hypothesis
4. Show how to estimate the reliability of a test
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7.1
The Elements ofa Test of Hypothesis
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Hypothesis Testing
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Population
JJ
J
J
J
JJ
I believe the population mean age is 50 (hypothesis).
Mean`X = 20
Random sample
J
J J
Reject hypothesis! Not close.
J
Whats a Hypothesis?
A statistical hypothesis is a statement about the numerical value of a population parameter.
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I believe the mean GPA of this class is 3.5!
1984-1994 T/Maker Co.
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Null Hypothesis
The null hypothesis, denoted H0, represents the hypothesis that will be accepted unless the data provide convincing evidence that it is false. This usually represents the status quo or some claim about the population parameter that the researcher wants to test.
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Alternative Hypothesis
The alternative (research) hypothesis, denoted Ha, represents the hypothesis that will be accepted only if the data provide convincing evidence of its truth. This usually represents the values of a population parameter for which the researcher wants to gather evidence to support.
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Alternative Hypothesis1. Opposite of null hypothesis2. The hypothesis that will be accepted only if
the data provide convincing evidence of its truth
3. Designated Ha4. Stated in one of the following forms
Ha: m (some value)Ha: m < (some value)Ha: m > (some value)
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Identifying Hypotheses
Example problem: Test that the population mean is not 3Steps:
State the question statistically (m 3) State the opposite statistically (m = 3)
Must be mutually exclusive & exhaustive Select the alternative hypothesis (m 3)
Has the , sign State the null hypothesis (m = 3)
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What Are the Hypotheses?
State the question statistically: m = 12
State the opposite statistically: m 12
Select the alternative hypothesis: Ha: m 12
State the null hypothesis: H0: m = 12
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Is the population average amount of TV viewing 12 hours?
What Are the Hypotheses?
State the question statistically: m 12
State the opposite statistically: m = 12
Select the alternative hypothesis: Ha: m 12
State the null hypothesis: H0: m = 12
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Is the population average amount of TV viewing different from 12 hours?
What Are the Hypotheses?
State the question statistically: m 20
State the opposite statistically: m > 20
Select the alternative hypothesis: Ha: m > 20
State the null hypothesis: H0: m = 20
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Is the average cost per hat less than or equal to $20?
What Are the Hypotheses?
State the question statistically: m > 25
State the opposite statistically: m 25
Select the alternative hypothesis: Ha: m > 25
State the null hypothesis: H0: m 25
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Is the average amount spent in the bookstore greater than $25?
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Test Statistic
The test statistic is a sample statistic, computed from information provided in the sample, that the researcher uses to decide between the null and alternative hypotheses.
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Test Statistic - ExampleThe sampling distribution of assuming = 2,400. The chance of observing more than 1.645 standard deviations above 2,400 is only .05 if in fact the true mean is 2,400.
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Type I Error
A Type I error occurs if the researcher rejects the null hypothesis in favor of the alternative hypothesis when, in fact, H0 is true.
The probability of committing a Type I error is denoted by a.
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Rejection Region
The rejection region of a statistical test is the set of possible values of the test statistic for which the researcher will reject H0 in favor of Ha.
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Type II Error
A Type II error occurs if the researcher accepts the null hypothesis when, in fact, H0 is false.
The probability of committing a Type II error is denoted by b.
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Conclusions and Consequences for a Test of Hypothesis
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True State of NatureConclusion H0 True Ha TrueAccept H0(Assume H0 True)
Correct decision Type II error (probability b)
Reject H0(Assume Ha True)
Type I error (probability a)
Correct decision
Elements of a Test of Hypothesis
1. Null hypothesis (H0): A theory about the specific values of one or more
population parameters. The theory generally represents the status quo, which
we adopt until it is proven false.
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2. Alternative (research) hypothesis (Ha): - A theory that contradicts the null hypothesis. - The theory generally represents that which we will
adopt only when sufficient evidence exists to establish its truth.
Elements of a Test of Hypothesis
3. Test statistic: A sample statistic used to decide whether to reject the null hypothesis.
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4. Rejection region: The numerical values of the test statistic for which the null
hypothesis will be rejected. The rejection region is chosen so that the probability is a
that it will contain the test statistic when the null hypothesis is true, thereby leading to a Type I error.
The value of a is usually chosen to be small (e.g., .01, .05, or .10) and is referred to as the level of significance of the test.
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Elements of a Test of Hypothesis
5. Assumptions: Clear statement (s) of any assumptions made about the population (s) being sampled.
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6. Experiment and calculation of test statistic: Performance of the sampling experiment and determination of the numerical value of the test statistic.
Elements of a Test of Hypothesis
7. Conclusion:
a. If the numerical value of the test statistic falls in the rejection region, we reject the null hypothesis and conclude that the alternative hypothesis is true. We know that the hypothesis-testing process will lead to this conclusion incorrectly (Type I error) only 100a% of the time when H0 is true.
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Elements of a Test of Hypothesis
7. Conclusion:
b. If the test statistic does not fall in the rejection region, we do not reject H0. Thus, we reserve judgment about which hypothesis is true. We do not conclude that the null hypothesis is true because we do not (in general) know the probability b that our test procedure will lead to an incorrect acceptance of H0 (Type II error).
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Determining theTarget Parameter
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Parameter Key Words or Phrases Type of Data Mean; average Quantitativep Proportion; percentage;
fraction; rateQualitative
s2 Variance; variability; spread
Quantitative
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7.2
Formulating Hypotheses and Setting Up the Rejection Region
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Steps for Selecting the Null and Alternative Hypotheses
1. Select the alternative hypothesis as that which the sampling experiment is intended to establish. The alternative hypothesis will assume one of three forms:a. One-tailed, upper-tailed (e.g., Ha: > 2,400) b. One-tailed, lower-tailed (e.g., Ha: < 2,400) c. Two-tailed (e.g., Ha: 2,400)
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Steps for Selecting the Null and Alternative Hypotheses
2. Select the null hypothesis as the status quo, that which will be presumed true unless the sampling experiment conclusively establishes the alternative hypothesis. The null hypothesis will be specified as that parameter value closest to the alternative in one-tailed tests and as the complementary (or only unspecified) value in two-tailed tests.
(e.g., H0: = 2,400)
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One-Tailed Test
A one-tailed test of hypothesis is one in which the alternative hypothesis is directional and includes the symbol < or >.
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Two-Tailed Test
A two-tailed test of hypothesis is one in which the alternative hypothesis does not specify departure from H0 in a particular direction and is written with the symbol .
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Basic Idea
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Sample Meansm = 50H0
Sampling DistributionIt is unlikely that we would get a sample mean of this value ...
20
... if in fact this werethe population mean
... therefore, we reject the hypothesis that m = 50.
H0: = 50
Ha: < 50
Rejection Region (One-Tail Test)
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HoValueCritical
Value
a
Sample Statistic
RejectionRegion
Fail to RejectRegion
Sampling Distribution
1 a
Level of Confidence
H0: = 0Ha: < 0
Rejection Regions (Two-Tailed Test)
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HoValue Critical
ValueCriticalValue
1/2 a1/2 a
Sample Statistic
RejectionRegion
RejectionRegion
Fail to RejectRegion
Sampling Distribution
1 a
Level of ConfidenceH0: = 0Ha: 0
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Rejection Regions
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Alternative Hypotheses
Lower-Tailed
Upper-Tailed
Two-Tailed
a = .10 z < 1.28 z > 1.28 z < 1.645 or z > 1.645
a = .05 z < 1.645 z > 1.645 z < 1.96 or z > 1.96
a = .01 z < 2.33 z > 2.33 z < 2.575 or z > 2.575
7.3
Observed Significance Levels:p-Values
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p-ValueThe observed significance level, or p-value, for a specific statistical test is the probability (assuming H0 is true) of observing a value of the test statistic that is at least as contradictory to the null hypothesis, and supportive of the alternative hypothesis, as the actual one computed from the sample data.
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p-Value Probability of obtaining a test statistic more
extreme ( or ) than actual sample value, given H0 is true
Called observed level of significance Smallest value of a for which H0 can be
rejected
Used to make rejection decision If p-value a, do not reject H0 If p-value < a, reject H0
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Steps for Calculating the p-Value for a Test of Hypothesis
1. Determine the value of the test statistic zcorresponding to the result of the sampling experiment.
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Steps for Calculating the p-Value for a Test of Hypothesis
2a. If the test is one-tailed, the p-value is equal to the tail area beyond z in the same direction as the alternative hypothesis. Thus, if the alternative hypothesis is of the form > , the p-value is the area to the right of, or above, the observed z-value. Conversely, if the alternative is of the form < , the p-value is the area to the left of, or below, the observed z-value.
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Steps for Calculating the p-Value for a Test of Hypothesis
2b. If the test is two-tailed, the p-value is equal to twice the tail area beyond the observed z-value in the direction of the sign of z that is, if z is positive, the p-value is twice the area to the right of, or above, the observed z-value. Conversely, if z is negative, the p-value is twice the area to the left of, or below, the observed z-value.
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Reporting Test Results asp-Values: How to Decide Whether
to Reject H0
1. Choose the maximum value of a that you are willing to tolerate.
2. If the observed significance level (p-value) of the test is less than the chosen value of a, reject the null hypothesis. Otherwise, do not reject the null hypothesis.
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Two-Tailed z Test p-Value Example
Does an average box of cereal contain 368 grams of cereal? A random sample of 64 boxes showed x = 372.5. The company has specified s to be 25 grams. Find the p-Value.
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368 gm.
Two-Tailed z Test p-Value Solution
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z0 1.44z value of sample statistic (observed)
Two-Tailed Z Test p-Value Solution
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1/2 p-Value1/2 p-Value
z value of sample statistic (observed)
p-value is P(z 1.44 or z 1.44)
z0 1.441.44
From z table: lookup 1.44
.4251
.5000 .4251
.0749
Two-Tailed z Test p-Value Solution
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1/2 p-Value.0749
1/2 p-Value.0749
p-value is P(z 1.50 or z 1.50) = .1498
z0 1.441.44
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Two-Tailed z Test p-Value Solution
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0 1.441.44 z
Reject H0Reject H0
1/2 p-Value = .07491/2 p-Value = .0749
1/2 a = .0251/2 a = .025
p-Value = .1498 a = .05 Do not reject H0.
Test statistic is in Do not reject region
One-Tailed z Test p-Value Example
Does an average box of cereal contain more than 368 grams of cereal? A random sample of 64 boxes showed x = 372.5. The company has specified sto be 25 grams. Find the p-Value. How does it compare to a = .05?
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368 gm.
One-Tailed z Test p-Value Solution
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z0 1.44
z value of sample statistic
One-Tailed z Test p-Value Solution
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Use alternative hypothesis to find direction
p-Value is P(z 1.44)
z value of sample statistic
p-Value
z0 1.44From z table: lookup 1.44
.4251
.5000 .4251
.0749
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One-Tailed z Test p-Value Solution
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p-Value.0749
z value of sample statistic
From z table: lookup 1.44
Use alternative hypothesis to find direction
.5000 .4251
.0749
p-Value is P(z 1.44) = .0749
z0 1.44
.4251
One-Tailed z Test p-Value Solution
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a = .05
0 1.44 z
Reject H0
p-Value = .0749
(p-Value = .0749) (a = .05). Do not reject H0.
Test statistic is in Do not reject region
p-Value Thinking Challenge
Youre an analyst for Ford. You want to find out if the average miles per gallon of Escorts is less than 32 mpg. Similar models have a standard deviation of 3.8mpg. You take a sample of 60Escorts & compute a sample mean of 30.7 mpg. What is the value of the observed level of significance (p-Value)?
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p-Value Solution*
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Use alternative hypothesis to find direction
z02.65
z value of sample statistic
From z table: lookup 2.65
.4960
p-Value.004
.5000 .4960
.0040
p-Value is P(z -2.65) = .004.p-Value < (a = .01). Reject H0.
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Converting a Two-Tailedp-Value from a Printout to a One-
Tailed p-Value
if Ha is of the form > and z is positiveor Ha is of the form < and z is negative
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if Ha is of the form > and z is negativeHa is of the form < and z is positive
7.4
Test of Hypotheses about a Population Mean:Normal (z) Statistic
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Large-Sample Test of Hypothesis about
One-Tailed Test Two-Tailed TestH0: = 0 H0: = 0Ha: < 0 Ha: 0
(or Ha: > 0)
Test Statistic: Test Statistic:
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Large-Sample Test of Hypothesis about
One-Tailed TestRejection region:
z < za(or z > za when Ha: > 0)
where za is chosen so thatP(z > za) = a
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Large-Sample Test of Hypothesis about
Two-Tailed TestRejection region:
|z| > za/2where za/2 is chosen so that
P(|z| > za/2) = a/2
Note: 0 is the symbol for the numerical value assigned to under the null hypothesis.
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Conditions Required for a Valid Large-Sample Hypothesis Test for
1. A random sample is selected from the target population.
2. The sample size n is large (i.e., n 30). (Due to the Central Limit Theorem, this condition guarantees that the test statistic will be approximately normal regardless of the shape of the underlying probability distribution of the population.)
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Possible Conclusions for a Test of Hypothesis
1. If the calculated test statistic falls in the rejection region, reject H0 and conclude that the alternative hypothesis Ha is true. State that you are rejecting H0 at the a level of significance. Remember that the confidence is in the testing process, not the particular result of a single test.
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Possible Conclusions for a Test of Hypothesis
2. If the test statistic does not fall in the rejection region, conclude that the sampling experiment does not provide sufficient evidence to reject H0 at the a level of significance. [Generally, we will not acceptthe null hypothesis unless the probability b of a Type II error has been calculated.]
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Two-Tailed z Test Example
Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes had x = 372.5. The company has specified s to be 25 grams. Test at the .05level of significance.
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368 gm.
Two-Tailed z Test Solution
H0: Ha: a = n = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
m = 368m 368.0525
z0 1.961.96
.025
Reject H 0 Reject H0.025
Do not reject at a = .05
No evidence average is not 368
Two-Tailed z Test Thinking Challenge
Youre a Q/C inspector. You want to find out if a new machine is making electrical cords to customer specification: average breaking strength of 70 lb. with s = 3.5 lb. You take a sample of 36 cords & compute a sample mean of 69.7 lb. At the .05 level of significance, is there evidence that the machine is notmeeting the average breaking strength?
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Two-Tailed z Test Solution*
H0: Ha: a = n = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
m = 70m 70.05
36
z0 1.961.96
.025Reject H 0 Reject H0
.025Since -1.96 < z = -0.51
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One-Tailed z TestExample
Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified s to be 25 grams. Test at the .05level of significance.
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368 gm.
One-Tailed z Test Solution
H0: Ha: a = n = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
m = 368m > 368.0525
z0 1.645
.05
Reject Do not reject at a = .05
No evidence average is more than 368
One-Tailed z Test Thinking Challenge
Youre an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7mpg. At the .01 level of significance, is there evidence that the miles per gallon is less than 32?
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One-Tailed z Test Solution*
H0: Ha: a = n = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
m = 32m < 32.0160
z0-2.33
.01
Reject Reject at a = .01
There is evidence average is less than 32
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7.5
Test of Hypothesis about a Population Mean:
Students t-Statistic
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Small-Sample Test of Hypothesis about
One-Tailed TestH0: = 0Ha: < 0 (or Ha: > 0)
Test statistic:
Rejection region: t < ta(or t > ta when Ha: > 0)where ta and ta/2 are based on (n 1) degrees of
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Small-Sample Test of Hypothesis about
Two-Tailed TestH0: = 0Ha: 0
Test statistic:
Rejection region: |t| > ta/2
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Conditions Required for a Valid Small-Sample Hypothesis Test for
1. A random sample is selected from the target population.
2. The population from which the sample is selected has a distribution that is approximately normal.
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Two-Tailed t TestExample
Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of 372.5 and a standard deviation of 12 grams. Test at the .05 level of significance.
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368 gm.
Two-Tailed t Test Solution
H0: Ha: a = df = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
m = 368m 368.0536 1 = 35
t0 2.030-2.030
.025Reject H0 Reject H0
.025Reject at a = .05
There is evidence population average is not 368
Two-Tailed t TestThinking Challenge
You work for the FTC. A manufacturer of detergent claims that the mean weight of detergent is 3.25 lb. You take a random sample of 64containers. You calculate the sample average to be 3.238 lb. with a standard deviation of .117 lb. At the .01 level of significance, is the manufacturer correct?
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3.25 lb.
Two-Tailed t Test Solution*
H0: Ha: a = df = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
m = 3.25m 3.25.0164 1 = 63
t0 2.656-2.656
.005Reject H0 Reject H0
.005Do not reject at a = .01
There is no evidence average is not 3.25
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One-Tailed t TestExample
Is the average capacity of batteries less than 140ampere-hours? A random sample of 20 batteries had a mean of 138.47 and a standard deviation of 2.66. Assume a normal distribution. Test at the .05 level of significance.
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One-Tailed t Test Solution
H0: Ha: a = df = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
m = 140m < 140.0520 1 = 19
t0-1.729
.05
Reject H0 Since t = -2.57 < -1.729, reject H0 at a = .05
There is evidence population average is less than 140
One-Tailed t TestThinking Challenge
Youre a marketing analyst for Wal-Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was:8 11 0 4 7 8 10 5 8 3At the .05 level of significance, is there evidence that the average bear sales per store is more than 5 ($ 00)?
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One-Tailed t Test Solution*
H0: Ha: a = df = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
m = 5m > 5.0510 1 = 9
t0 1.833
.05
Reject H0 Do not reject at a = .05
There is no evidence average is more than 5
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7.6
Large-Sample Test of Hypothesis about a Population Proportion
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Large-Sample Test of Hypothesis about p
One-Tailed TestH0: p = p0Ha: p < p0 (or Ha: p > p0)
Test statistic:
Rejection region:z < za (or z > za when Ha: p > p0)
Note: p0 is the symbol for the numerical value of passigned in the null hypothesis
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Large-Sample Test of Hypothesis about p
Two-Tailed TestH0: p = p0Ha: p p0
Test statistic:
Rejection region: |z| > za /2
Note: p0 is the symbol for the numerical value of passigned in the null hypothesis
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Conditions Required for a Valid Large-Sample Hypothesis Test for
p
1. A random sample is selected from a binomial population.
2. The sample size n is large. (This condition will be satisfied if both np0 15 and nq0 15.)
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One-Proportion z Test Example
The present packaging system produces 10% defective cereal boxes. Using a new system, a random sample of 200 boxes had 11 defects. Does the new system produce fewer defects? Test at the .05level of significance.
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One-Proportion z Test Solution
H0: Ha: a = n = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
p = .10p < .10.05200
z0-1.645
.05Reject H0 Since z = -2.12 < -1.645
Reject at a = .05
There is evidence new system < 10% defective
One-Proportion z Test Thinking Challenge
Youre an accounting manager. A year-end audit showed 4% of transactions had errors. You implement new procedures. A random sample of 500 transactions had 25 errors. Has the proportion of incorrect transactions changed at the .05 level of significance?
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One-Proportion z Test Solution*
H0: Ha: a = n = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
p = .04p .04.05500
z0 1.96-1.96
.025Reject H0 Reject H0
.025Do not reject at a = .05
There is evidence proportion is not 4%
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7.7
Test of Hypothesis about a Population Variance
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Variance
Although many practical problems involve inferences about a population mean (or proportion), it is sometimes of interest to make an inference about a population variance, s2.
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Test of a Hypothesis about s 2
One-Tailed TestH0: s 2 = s02Ha: s 2 < s02 (or Ha: s 2 > s02)
Test statistic:
Rejection region:(or c 2 > ca2 when Ha: s 2 > s02 )
where s02 is the hypothesized variance and the distribution of c2 is a chi-square distribution with (n 1) degrees of freedom.
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Test of a Hypothesis about s 2
Two-Tailed TestH0: s 2 = s02Ha: s 2 s02
Test statistic:
Rejection region:
where s02 is the hypothesized variance and the distribution of c2 is a chi-square distribution with (n 1) degrees of freedom.
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Conditions Required for a Valid Hypothesis Test for s 2
1. A random sample is selected from the target population.
2. The population from which the sample is selected has a distribution that is approximately normal.
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Several c2 probability Distributions
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Part of Table VI: Critical Values of Chi Square
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Finding Critical Value Example
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What is the critical c2 value given:Ha: s2 > 0.7n = 3a =.05?
c20
Upper Tail AreaDF .995 .95 .051 ... 0.004 3.8412 0.010 0.103 5.991
c2 Table (Portion)
df = n - 1 = 25.991
Reject
a = .05
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Finding Critical Value Example
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What is the critical c2 value given:Ha: s 2 < 0.7n = 3a =.05?
What do you do if the rejection region is on the left?
Finding Critical Value Example
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What is the critical c2 value given:Ha: s 2 < 0.7n = 3a =.05?
.103 c20
Upper Tail AreaDF .995 .95 .051 ... 0.004 3.8412 0.010 0.103 5.991
c2 Table (Portion)
Upper Tail Area for Lower Critical Value = 1.05 = .95a = .05
Reject H0
df = n - 1 = 2
Chi-Square (c2) Test Example
Is the variation in boxes of cereal, measured by the variance, equal to 15 grams? A random sample of 25boxes had a variance of17.7 grams. Test at the .05level of significance.
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Chi-Square (c2) Test Solution
H0: Ha: a = df = Critical Value(s):
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Test Statistic:
Decision:
Conclusion:
s 2 = 15s 2 15.0525 1 = 24
c20
a/2 = .025
39.36412.401
Since 12.401
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7.8
Calculating Type II Error Probabilities: More about b
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Type II Error The Type II error probability b is calculated assuming
that the null hypothesis is false because it is defined as the probability of accepting H0 when it is false.
The situation corresponding to accepting the null hypothesis, and thereby risking a Type II error, is not generally as controllable.
For that reason, we adopted a policy of nonrejection of H0 when the test statistic does not fall in the rejection region, rather than risking an error of unknown magnitude.
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Steps for Calculating b for a Large-Sample Test about
1. Calculate the value(s) of x corresponding to the border(s) of the rejection region. There will be one border value for a one-tailed test and two for a two-tailed test. The formula is one of the following, corresponding to a test with level of significance a:
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Steps for Calculating b for a Large-Sample Test about
Upper-tailed test:
Lower-tailed test:
Two-tailed test:
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Steps for Calculating b for a Large-Sample Test about
2. Specify the value of a in the alternative hypothesis for which the value of b is to be calculated. Then convert the border value(s) of to z-value(s) using the alternative distribution with mean a. The general formula for the z-value is
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Steps for Calculating b for a Large-Sample Test about
Sketch the alternative distribution (centered at a) and shade the area in the acceptance (nonrejection) region. Use the z-statistic(s) and Table IV in Appendix B to find the shaded area, which is b.
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Power of Test
Probability of rejecting false H0 Correct decision
Equal to 1 b Used in determining test adequacy Affected by
True value of population parameter Significance level a Standard deviation & sample size n
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Two-Tailed z Test Example
Does an average box of cereal contain less than 368grams of cereal? A random sample of 25 boxes had x = 372.5. The company has specified s to be 15 grams. Test at the .05 level of significance.
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368 gm.
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Finding PowerStep 1
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`xm0 = 368
Reject H0Do NotReject H0
Hypothesis:H0: m 368Ha: m < 368 a = .05
Draw
Finding PowerSteps 2 & 3
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True Situation:ma = 360 (Ha)
Draw
Specify
b1b
`xm0 = 368
Reject H0Do NotReject H0
Hypothesis:H0: m 368Ha: m < 368 a = .05
Draw
Finding PowerStep 4
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`xm a = 360
True Situation:ma = 360 (Ha)
Draw
Specify1b
`xm0 = 368
Reject H0Do NotReject H0
Hypothesis:H0: m 368Ha: m < 368 a = .05
Draw
Finding PowerStep 5
2011 Pearson Education, Inc 116363.065 `xma = 360
True Situation:ma = 360 (Ha)
Draw
Specify
= 368m `x0
Reject H0Do NotReject H0
Hypothesis:H0: m 368Ha: m < 368 a = .05
Draw
b = .1541b =.846z Table
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Properties ofb and Power
1. For fixed n and a, the value of bdecreases (the power increases) as the distance between the specified null value 0 and the specified alternative value aincreases.
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Propertiesof b andPower
2. For fixed n and values of 0 and a, the value of bincreases (the power decreases) as the value of ais decreased.
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Properties of b and Power
3. For fixed a and values of 0 and a, the value of bdecreases (the power increases) as the sample size n is increased.
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Key Ideas
Key Words for Identifying the Target Parameter
m Mean, Averagep Proportion, Fraction, Percentage, Rate, Probabilitys 2 Variance, Variability, Spread
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Key Ideas
Elements of a Hypothesis Test
1. Null hypothesis (H0)
2. Alternative hypothesis (Ha)
3. Test statistic (z, t, or c2)
4. Significance level (a)
5. p-value
6. Conclusion
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Key Ideas
Errors in Hypothesis Testing
Type I Error = Reject H0 when H0 is true(occurs with probability a)
Type II Error = Accept H0 when H0 is false(occurs with probability b )
Power of Test = P(Reject H0 when H0 is false)= 1 b
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Key Ideas
Forms of Alternative Hypothesis
Lower-tailed : Ha : m < 50
Upper-tailed : Ha : m > 50
Two-tailed : Ha : m 50
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Key Ideas
Using p-values to Decide
1. Choose significance level (a )
2. Obtain p-value of the test
3. If a > p-value, reject H0
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Assignments
#97,#121,#130, #132, #133, #142
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