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Hypothesis and Hypothesis Testing
HYPOTHESIS A statement about the value of a population parameter developed for the purpose of testing.
HYPOTHESIS TESTING A procedure based on sample evidence and probability theory to determine whether the hypothesis is a reasonable statement.
TEST STATISTIC A value, determined from sample information, used to determine whether to reject the null hypothesis.
CRITICAL VALUE The dividing point between the region where the null hypothesis is rejected and the region where it is not rejected.
Important Things to Remember about H0 and H1
H0: null hypothesis and H1: alternate hypothesis
H0 and H1 are mutually exclusive and collectively exhaustive
H0 is always presumed to be true H1 is the research hypothesis A random sample (n) is used to “reject H0” If we conclude 'do not reject H0', this does
not necessarily mean that the null hypothesis is true, it only suggests that there is not sufficient evidence to reject H0; rejecting the null hypothesis then, suggests that the alternative hypothesis may be true.
Equality is always part of H0 (e.g. “=” , “≥” , “≤”).
“≠” “<” and “>” always part of H1 In actual practice, the status quo is set up
as H0
In problem solving, look for key words and convert them into symbols. Some key words include: “improved, better than, as effective as, different from, has changed, etc.”
KeywordsInequalitySymbol
Part of:
Larger (or more) than > H1
Smaller (or less) < H1
No more than H0
At least ≥ H0
Has increased > H1
Is there difference? ≠ H1
Has not changed = H0
Has “improved”, “is better than”. “is more effective”
See left text
H1
Signs in the Tails of a Test
Rejection Region
Rejection RegionAcceptance
Region
RejectionRegion Acceptance
Region
Two-tailed Test
One-tailed Test
Two-tailed tests - the rejection region is in both tails of the distribution
One-tailed tests - the rejection region is in only on one tail of the distribution
0H is true 0H is false
0HReject
Do not reject0H
Type I errorP(Type I)= Correct
Decision
CorrectDecision
Type II errorP(Type II)=
Types of Errors
Type I Error - Defined as the probability of rejecting the null hypothesis when it is actually true.This is denoted by the Greek letter “”Also known as the significance level of a test
Type II Error: Defined as the probability of “accepting” the null hypothesis when it is actually false.This is denoted by the Greek letter “β”
Hypothesis Setups for Testing a Mean () or a Proportion ()
MEAN
PROPORTION
Steps in hypothesis testing
- Define Null hypothesis- Define Alternative hypothesis- Calculate Test statistic- Determine Rejection region- Compare Value of the test statistic with Critical Value- Conclusion
Testing for a Population Mean with aKnown Population Standard Deviation- Example
EXAMPLEJamestown Steel Company manufactures and
assembles desks and other office equipment . The weekly production of the Model A325 desk at the Fredonia Plant follows the normal probability distribution with a mean of 200 and a standard deviation of 16. Recently, new production methods have been introduced and new employees hired. The mean number of desks produced during last 50 weeks was 203.5. The VP of manufacturing would like to investigate whether there has been a change in the weekly production of the Model A325 desk, at 1% level of significance.
Step 1: State the null hypothesis and the alternate hypothesis.
H0: = 200
H1: ≠ 200
(note: This is a 2-tail test, as the keyword in the problem “has changed”)
Step 2: Select the level of significance.
α = 0.01 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since σ is known
Step 4: Formulate the decision rule.Reject H0 if |Z| > Z/2
Step 5: Make a decision and interpret the result.Because 1.55 does not fall in the rejection region, H0 is not
rejected. We conclude that the population mean is not different from 200. So we would report to the vice president of manufacturing that the sample evidence does not show that the production rate at the plant has
changed from 200 per week.
58.2not is 55.1
50/16
2005.203
/
2/01.
2/
2/
Z
Zn
X
ZZ
Suppose in the previous problem the vice president wants to know whether there has been an increase in the number of units assembled. To put it another way, can we conclude, because of the improved production methods, that the mean number of desks assembled in the last 50 weeks was more than 200?
Recall: σ=16, =200, α=.01
Step 1: State the null hypothesis and the alternate hypothesis.
H0: ≤ 200
H1: > 200
(note: This is a 1-tail test as the keyword in the problem “an increase”)
Step 2: Select the level of significance.
α = 0.01 as stated in the problem
Step 3: Select the test statistic.
Use Z-distribution since σ is known
Testing for a Population Mean with a Known Population Standard Deviation- Another Example
Step 4: Formulate the decision rule.
Reject H0 if Z > Z
Step 5: Make a decision and interpret the result.
Because 1.55 does not fall in the rejection region, H0 is not rejected. We conclude that the average number of desks assembled in the last 50 weeks is not more than 200
EAMPLE p-ValueRecall the last problem where the hypothesis
and decision rules were set up as:
H0: ≤ 200
H1: > 200
Reject H0 if Z > Z
where Z = 1.55 and Z =2.33
Reject H0 if p-value <
0.0606 is not < 0.01
Conclude: Fail to reject H0
p-value in Hypothesis Testing
p-VALUE is the probability of observing a sample value as extreme as, or more extreme than, the value observed, given that the null hypothesis is true.
In testing a hypothesis, we can also compare the p-value to the significance level ().
Decision rule using the p-value:
Reject null hypothesis, if p< α
Describing the p-value– If the p-value is less than 1%, there is
overwhelming evidence that supports the alternative hypothesis.
– If the p-value is between 1% and 5%, there is a strong evidence that supports the alternative hypothesis.
– If the p-value is between 5% and 10% there is a weak evidence that supports the alternative hypothesis.
– If the p-value exceeds 10%, there is no evidence that supports the alternative hypothesis.
Interpreting the p-value
The Power of Statistical TestThe power of a statistical test, given as 1 –
= P (reject H0 when H0 is false), measures the
ability of the test to perform as required. This
1 – is called the power of the function. This
means that greater the power of the function
the better would be the decision rule. There are two types of tail test
1. One-tailed tests - the rejection region is in only one tail of the distribution
2. Two-tailed tests - the rejection region is in both tails of the distribution
Steps in Hypothesis Testing using SPSS State the null and alternative
hypotheses Define the level of significance (α) Calculate the actual significance :
p-value Make decision : Reject null hypothesis,
if p≤ α, for 2-tail test; and
if p*≤ α, for 1-tail test.(p* is p/2 when p is obtained from 2-tail test)
Conclusion
In practice, the population standard deviation will be unknown.Recall that when is known we use the following statistic to estimate and test a population mean
When is unknown or when the sample
size is small, we use its point estimator
s, and the z-statistic is replaced then
by the t-statistic
Inference About a Population Mean When the Population Standard Deviation Is Unknown or When the Sample Size is Small
n
xz
The t - Statistic
n
x
n
x
s
0
The t distribution is mound-shaped, and symmetrical around zero.
The “degrees of freedom”,(a function of the sample size)determine how spread thedistribution is (compared to the normal distribution)d.f. = v2
d.f. = v1v1 < v2
t
Example – 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 trainees (see file PROD.sav).
Testing when is unknown
Example – Solution– The problem objective is to describe the
population of the number of packages processed in one hour.
– H0: = 450 H1: > 450
– The t statistic
d.f. = n - 1 = 49
ns
xt
Testing when is unknown
Solution continued (solving by hand) – The rejection region is
t > t,n – 1
t,n - 1 = t.05,49
t.05,50 = 1.676.
83.3855.1507s
.55.15071n
nx
xs
and,38.46050019,23
x
thus,357,671,10x019,23x
havewedatatheFrom
2
i2i2
2ii
Testing when is unknown
• The test statistic is
89.15083.38
45038.460
ns
xt
• 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 .05 significance level.
1.676 1.89
Rejection region
Testing when is unknown
Solution using SPSS (use file PROD.sav)
One-Sample Statistics
N Mean Std. Deviation Std. Error MeanPackages 50 460.38 38.827 5.491
One-Sample Test
Test Value = 450
t dfSig. (2-tailed)
Mean Difference
95% Confidence Interval of the Difference
Lower UpperPackages
1.890 49 .065 10.380 -.65 21.41
.sizesamplen.successesofnumberthex
wherenx
p̂
.sizesamplen.successesofnumberthex
wherenx
p̂
Statistic and sampling distribution– the statistic used when making inference
about p is:
– Under certain conditions, [np > 5 and n(1-p) > 5], is approximately normally distributed, with = p and 2 = p(1 - p)/n.p̂
Inference About a Population Proportion
Testing and Estimating the Proportion
Test statistic for p
5)p1(nand5npwhere
n/)p1(ppp̂
Z
5)p1(nand5npwhere
n/)p1(ppp̂
Z
Example 12.6 – A pharmaceutical company claimed that its
medicine was 80% effective in relieving allergy. In a sample of 200 persons, who were given medicine only 150 persons had relief. Do you thank that the effectiveness is below 80%? Use 0.05 level of significance.
Testing the Proportion
Solution– The problem objective is to test the
effectiveness of medicine.– The data are nominal.– The parameter to be tested is ‘p’.– Success is defined as “having relief”.– The hypotheses are:
H0: p = .8
H1: p < .8
Testing the Proportion
– Solution• The rejection region is z < z = z.05 = -1.645.
• The sample proportion is• The value of the test statistic is
Since calculated z is less than critical value, we reject null hypothesis and conclude that the claim of the company that its medicine is 80% effective is not justified.
75.200150ˆ p
786.1200/)8.1(8.
8.75.
/)1(
ˆ
npp
ppZ
Testing the Proportion
T-Tests : When sample size is small (<30) or When the Population Standard Deviation Is Unknown
Variable : Normal Types of t-tests:
One-sample t-testPaired or dependent
sample t-test Independent samples t-test (Equal and
Unequal Variance)
One-sample t-test
01
01
01
00
:
:
:
:
H
H
H
H
Paired sample t-test
0:
0:
0:
0:
1
1
1
0
d
d
d
d
H
H
H
H
Matched pairs
The mean of the population differences is D
that is D 21
DD
DD
nsx
t
Test statistic:
Degree of freedom = 1Dn
Independent sample t-test
211
211
211
210
:
:
:
:
H
H
H
H
The sampling process.
Population 1 Population2
Parameters:2
11 andParameters:
2
22 and
Statistics: Statistics:2
11andsx 222andsx
Sample size: 1n Sample size: 2n
If the two population standard deviations areunknown, then we can estimate the standarderror of the difference between two means.
2
2
2
1
2
1
21 nˆ
nˆ
ˆ xx
2
22
1
21
21
ˆˆ
nn
xxz
Test statistic:
If population variance unknown and the sample sizeis small and the population variances are equal
Then we will use the weighted average called a “ pooled estimate” of 2
21
2
21
11nn
s pxx
2
11
21
2
22
2
112
nnsnsn
s p
Where:
Test statistic:
21
2
21
11nn
s
xxt
p
Degree of freedom = 221 nn
One way Analysis of Variance ( ANOVA )
ANOVA is a technique used to test a hypothesis concerning the means of three or more populations.
Comparing Means of Three or More PopulationsThe F distribution is used for testing whether two or more sample means came from
the same or equal populations. Assumptions:
– The sampled populations follow the normal distribution.– The populations have equal standard deviations.– The samples are randomly selected and are independent.
The Null Hypothesis is that the population means are the same. The Alternative Hypothesis is that at least one of the means is different.
H0: µ1 = µ2 =…= µk
H1: The means are not all equalReject H0 if F > F,k-1,n-k
The test statistic used to test the hypothesis is F statistic
Assumptions:
1. The random variable is normally distributed.
2. The population variances are equal.
same are means allNot :
........:
1
3210
H
H
EXAMPLERecently a group of four major carriers
joined in hiring Brunner Marketing Research, Inc., to survey recent passengers regarding their level of satisfaction with a recent flight. The survey included questions on ticketing, boarding, in-flight service, baggage handling, pilot communication, and so forth.
Twenty-five questions offered a range of possible answers: excellent, good, fair, or poor. A response of excellent was given a score of 4, good a 3, fair a 2, and poor a 1. These responses were then totaled, so the total score was an indication of the satisfaction with the flight. Brunner Marketing Research, Inc., randomly selected and surveyed passengers from the four airlines.
Is there a difference in the mean satisfaction level among the four airlines?
Use the .01 significance level.
ANOVA – Example (File Airlines.sav)
Step 1: State the null and alternate hypotheses.
H0: µE = µA = µT = µO
H1: The means are not all equalReject H0 if F > F,k-1,n-k
Step 2: State the level of significance. The .01 significance level is stated in the
problem.
ANOVA – Example
Step 3: Find the appropriate test statistic. Use the F statistic
Calculations: It is convenient to summarize the calculations of F statistic in an ANOVA Table.
Compute the value of F and make a decision
ANOVA – Example
We find deviation of each observation from the grand mean, square the deviations, and sum this result for all 22 observations.SS total = {(94-75.64)2 + (90-75.64)2 + ……+
(65-75.64)2 } = 1485.10
To compute SSE, find deviation between each observation and its treatment mean. Each of these values is squared and then summed for all 22 observations.SSE = {(94-87.25)2 + (90-87.25)2 + ……+ (80-87.25)2 } + {(75-78.20)2 + (68-78.20)2 + ……+ (88-78.20)2 } + {(70-72.86)2 + (73-72.86)2 + ……+ (65-72.86)2 } + {(68-69)2 + (70-69)2 + ……+ (65-69)2 } = 594.41 Finally, determine SST = SS total – SSE.SST = 1485.10 – 594.41 = 890.69
ANOVA – ExampleStep 3: Find the appropriate test statistic. Use the F statistic
Calculations: It is convenient to summarize the calculations of F statistic in an ANOVA Table.
Step 4: State the decision rule.
Reject H0 if: F > F,k-1,n-k
F > F.01,4-1,22-4
F > F.01,3,18
F > 5.09
Step 5: Make a decision.
The computed value of F is 8.99, which is greater than the critical value of 5.09, so the null hypothesis is rejected. Conclusion: The mean scores are not the same for the four airlines; at this point we can only conclude there is a difference in the treatment means. We cannot determine which treatment groups differ or how many treatment groups differ.
ANOVA Example – SPSS OutputTest of Homogeneity of Variances
SatisfactionLevene Statistic df1 df2 Sig.
.962 3 18 .432
ANOVA
SatisfactionSum of
Squares df Mean Square F Sig.Between Groups
890.684 3 296.895 8.991 .001
Within Groups 594.407 18 33.023
Total 1485.091 21
ANOVA Example – SPSS OutputMultiple Comparisons
SatisfactionTukey HSD
(I) Carrier (J) Carrier Mean Difference (I-
J) Std. Error Sig.
95% Confidence IntervalLower Bound
Upper Bound
EasternTWA 9.050 3.855 .124 -1.85 19.95
Allegheny 14.393* 3.602 .004 4.21 24.57Ozark 18.250* 3.709 .001 7.77 28.73
TWA
Eastern -9.050 3.855 .124 -19.95 1.85Allegheny 5.343 3.365 .410 -4.17 14.85
Ozark 9.200 3.480 .071 -.63 19.03
Allegheny
Eastern -14.393* 3.602 .004 -24.57 -4.21TWA -5.343 3.365 .410 -14.85 4.17Ozark 3.857 3.197 .631 -5.18 12.89
Ozark
Eastern -18.250* 3.709 .001 -28.73 -7.77TWA -9.200 3.480 .071 -19.03 .63
Allegheny -3.857 3.197 .631 -12.89 5.18*. The mean difference is significant at the 0.05 level.
ANOVA Example – SPSS Output
Satisfaction
Tukey HSDa,b
Carrier
N
Subset for alpha = 0.05
1 2Ozark 6 69.00
Allegheny 7 72.86
TWA 5 78.20 78.20
Eastern 4 87.25
Sig. .078 .085
Means for groups in homogeneous subsets are displayed.
a. Uses Harmonic Mean Sample Size = 5.266.
b. The group sizes are unequal. The harmonic mean of the group sizes is used. Type I error levels are not guaranteed.
Homogeneous Subsets
Chi-squared Test of a Contingency Table
Test of Independence : Test on association between two nominal variables regarding contingency tables.
Null Hypothesis : Two variables are independent
Alternative Hypothesis : The two variables are dependent
The Chi-square Distribution
At the outset, we should know that the chi-
square distribution has only one parameter
called the ‘degrees of freedom’ (df ) as is the
case with the t-distribution. The shape of a
particular chi-square distribution depends on
the number of degrees of freedom.
1. Chi-square is non-negative in value; it is either zero or positively valued.
2. It is not symmetrical; it is skewed to the right.
3. There are many chi-square distributions. As with the t-distribution, there is a different chi-square distribution for each degree-of-freedom value.
Properties of Chi-square Distribution
The chi-squared statistic measures the differencebetween the actual counts and the expected counts ( assuming validity of the null hypothesis)
The sum( Observed count - Expected count )2
Expected count
k
ii
ii
EEO
1
2
Contingency table 2 test – Example
– In an effort to better predict the demand for courses offered by a certain MBA program, it was hypothesized that students’ academic background affect their choice of MBA major, thus, their courses selection.
– A random sample of last year’s MBA students was selected. The data is given in the file Chi-Sq_MBA.sav. The following contingency table summarizes relevant data.
The file Chi_Sq_MBA_Table.sav gives the data as per the contingency table.
Contingency table 2 test – Example
Degree Accounting Finance MarketingBA 31 13 16 60
BENG 8 16 7 31BBA 12 10 17 60
Other 10 5 7 3961 44 47 152
The observed values
Solution– The hypotheses are:
H0: The two variables are independent
H1: The two variables are dependent
k is the number of cells in the contingency table.
– The test statistic
k
i i
ii
E
EO
1
22 )(
– The rejection region
2)1c)(1r(,
2
Contingency table 2 test – Example
Under the null hypothesis the two variables are independent:
P(Accounting and BA) = P(Accounting)*P(BA)
Undergraduate MBA MajorDegree Accounting Finance Marketing Probability
BA 60 60/152BENG 31 31/152BBA 39 39/152Other 22 22/152
61 44 47 152Probability 61/152 44/152 47/152
The number of students expected to fall in the cell “Accounting - BA” iseAcct-BA = n(pAcct-BA) = 152(61/152)(60/152) = [61*60]/152 = 24.08
= [61/152][60/152].
60
61 152
The number of students expected to fall in the cell “Finance - BBA” iseFinance-BBA = npFinance-BBA = 152(44/152)(39/152) = [44*39]/152 = 11.29
44
39
152
Estimating the expected frequencies
The expected frequencies for a contingency table
Eij = (Column j total)(Row i total)Sample size
• The expected frequency of cell of raw i and column j in the contingency table is calculated by
k
1i i
2ii2
e)ef(
Undergraduate MBA MajorDegree Accounting Finance Marketing
BA 31 (24.08) 13 (17.37) 16 (18.55) 60BENG 8 (12.44) 16 (8.97) 7 (9.58) 31BBA 12 (15.65) 10 (11.29) 17 (12.06) 39Other 10 (8.83) 5 (6.39) 7 (6.80) 22
61 44 47 152
The expected frequency
31 24.08
31 24.08
31 24.08
31 24.08
31 24.08
(31 - 24.08)2
24.08 +….+
5 6.39
5 6.39
5 6.395 6.39
(5 - 6.39)2
6.39 +….+
7 6.80
7 6.80
7 6.80
(7 - 6.80)2
6.80
7 6.80
2= = 14.70
k
i i
ii
E
EO
1
22 )(
Calculation of the 2 statistic• Solution – continued
Contingency table 2 test – Example
• Conclusion: Since 2 = 14.70 > 12.5916, there is sufficient evidence to infer at 5% significance level that students’ undergraduate degree and MBA students courses selection are dependent.
• Solution – continued– The critical value in our example is:
5916.122)13)(14(,05.
2)1c)(1r(,
SPSS OutputChi-Square Tests
Value dfAsymp. Sig. (2-
sided)Pearson Chi-Square
14.702a 6 .023
Likelihood Ratio13.781 6 .032
Linear-by-Linear Association2.003 1 .157
N of Valid Cases152
a. 0 cells (.0%) have expected count less than 5. The minimum expected count is 6.37.
Yates’ Correction for Continuity
Chi-square distribution is a continuous distribution. Whenever the degrees of freedom (in case of a 2x2 table), certain corrections for continuity can be made
Required conditions – the rule of five
The test statistic used to perform the test is only approximately Chi-squared distributed.
For the approximation to apply, the expected cell frequency has to be at least 5 for all the cells (np 5).
If the expected frequency in a cell is less than 5, combine it with other cells.
