# Statistical Measurements, Analysis

Lecture 4 FieldworkThis week we want you, as the Marketing Analyst, to consider these questions from our clients.
Client 1:
For one of our survey results, Consider the following:
Ho : µ -5
A random sample of 25 observations yields a sample mean of -8. The pop sd = 10. Calculate the
p-value. What is the conclusion if α = .05?
In order to test if the population mean differs from 16, you draw a random sample of 32
observations and compute the sample mean and sample standard deviation as 15.2 and 0.6
Use the p-value approach and the critical value approach at 1% level of significance.
Client 2:
The manager of the small store MomandPops Coffee does not want the customers to stand in
line for too long. She is willing to hire someone if the wait is more than 5 minutes. She observes
the following data:
3.5 5.8 7.2 1.9 6.8 8.1 5.4
a. Set up null and alternative hypotheses to determine if the manager needs to hire someone.
b. Calculate the value of the test statistic. What assumption is needed to do this?
Client 3:
In order to deal with financial hardships, people have been raiding their retirement accounts. It
is reported that between 1998 and 2004, about 12% of families with 401k plans borrowed from
those plans. An economist is worried that this amount is now more than 20%. He randomly
surveys 190 households with these plans and finds that 50 are borrowing from them.
a. Set up the null and the alternative hypotheses to test the economist’s concern
b. Compute the value of the test statistic.
c. Use the p-value approach to test if the economist’s concern is justifiable at α= .05
Client 4:
1. A movie production company is releasing a movie with the hopes that people will come a
second time. Their target is to have 30 million viewers and they want more than 30% of the
viewers to return to see the movie again. They show this to 200 people, 68 of which said they
will watch again.
a. At 5% level of significance, test if more than 30% of the viewers will return.
Client 5:
1. Import the Franchises file into SPSS
2. Sort the dataset by Ascending order on Sales
3. I want some basic summary statistics. Also, check for normality.
4.
a. Select the sales and sqft variables from the table that you imported into SPSS
b. Now Select all the sales information whose ages were greater than 40. What do you notice?
c. Find the probability (using z scores) all the franchises where advertising spent was more
than \$8,200. Do it again with the sales were greater than \$200,000. Do you think it was
worth that money?
d. Rename SQFT to “Square Feet”
e. Make a new variable where the owners ages are “very young” if it is between 20-30,
“young” if 31-40 and “old” is 41-55
f. Create a scatterplot matrix grouped by the Franchises ages.
5. At the end, interpret your findings.
FRANCHISE NUMBERSALES SQFT
INVENTORY ADVERTISING FAMILIES STORES OWNER AGE OF FRANCHISE
1
231
3,00
294
8,20
8,20
11
20
2
156
2,20
232
6,90
4,10
12
50
3
10
0,50
149
3,00
4,30
15
35
4
519
5,50
600
12,00
16,10
1
44
5
437
4,40
567
10,60
14,10
5
25
6
487
4,80
571
11,80
12,70
4
35
7
299
3,10
512
8,10
10,10
10
55
8
195
2,50
347
7,70
8,40
12
35
9
20
1,20
212
3,30
2,10
15
43
10
68
0,60
102
4,90
4,70
8
46
11
570
5,40
788
17,40
12,30
1
35
12
428
4,20
577
10,50
14,00
7
33
13
464
4,70
535
11,30
15,00
3
33
14
15
0,60
163
2,50
2,50
14
53
15
65
1,20
168
4,70
3,30
11
25
16
98
1,60
151
4,60
2,70
10
45
17
398
4,30
342
5,50
16,00
4
46
18
161
2,60
196
7,20
6,30
13
34
19
397
3,80
453
10,40
13,90
7
35
20
497
5,30
518
11,50
16,30
1
53
21
528
5,60
615
12,30
16,00
0
45
22
99
0,80
278
2,80
6,50
14
33
23
0,5
1,10
142
3,10
1,60
12
24
24
347
3,60
461
9,60
11,30
6
46
25
341
3,50
382
9,80
11,50
5
64
26
507
5,10
590
12,00
15,70
0
34
27
400
8,60
517
7,00
12,00
8
36
The data are for each franchise store.
SALES = annual net sales/\$1000
SQFT = number sq. ft./1000
INVENTORY = inventory/\$1000
FAMILIES = size of sales district/1000 families
STORES = number of competing stores in distric
OWNER AGE OF FRANCHISE
STATISTICAL MEASUREMENTS,
ANALYSIS AND RESEARCH
Instructor: Amreeta Choudhury
Lecture 4
M2.4: Hypothesis Testing – One mean and Two mean test
Section Objectives
1. Test on one mean and Test on two means
2. Tests on proportions and variances
3. Use in Marketing
Chapter 6
Sampling and inference
1)
2)
3)
4)
5)
3
Exploit the probabilistic nature of sampling to show how
statements on the population parameters can be based on
sample estimates
Consider the degree of uncertainty (level of confidence) in such
an operation
Build a confidence interval, which is a range of values around
the sample mean which is expected to include the true
population mean at a given probability (confidence) level
Compute the probability associated with a given statement on
the population parameters (or parameters from different
population)
Decide whether the statement is false depending on its
probability level (hypothesis testing)
Chapter 6
Direct and indirect problems
• Direct problem – If we knew the population mean and the
standard error then we would know the exact probability of any
sample mean
• Indirect problem –
• Confidence interval: Given the sample mean, we can exploit the
normal probability curve to find a range of value which will
contain the true population mean at a given confidence level
• Hypothesis testing: Given the sample mean, we can exploit the
normal probability curve to check the probability of an hypothesis
on the true mean – then we can decide whether to discard or
retain that hypothesis
4
Chapter 6
The normal distribution of the sample means
• Once more the normal distribution…
• The central point is the
true population mean and
is the most likely sample
mean
• The larger the standard
error the flatter the curve
95% of
probability
2.5% of
probability
2.5% of
probability
−
5

+
• 95% of all possible
sample means fall within a
standard errors from the
mean (1.96 to be precise)
Chapter 6
Indirect problem
• We have only one sample mean (x-bar)
• 95% of sample means fall within the range
[  − 1.96  ;  + 1.96  ]
• If 100% of the sample means were in that range we could
state with certainty that the true mean falls in the range
[ x − 1.96  ; x + 1.96  ]
• However, we can affirm that only for 95% of the sample
means, thus we refer to a 95% confidence level
6
Chapter 6
Confidence level
95% of
probability
x − 1.96 
x + 1.96 
2.5% of
probability
2.5% of
probability
−

+
HOWEVER, ONLY 95% OF THE SAMPLE MEANS FALL IN THIS RANGE – THUS WE CAN
STATE THAT THE TRUE POPULATION MEAN IS INCLUDED IN THE INTERVAL BETWEEN
AND
. WITH A 95% CONFIDENCE LEVEL
x + 1.96 
x − 1.96 
If the single sample mean we extract falls in this range, then the true population mean is also included in the
interval between
and
. In fact, even if we get one of the extremes, it will still be at a
distance of 1.96 from 
Chapter 6
Confidence levels and critical values
• The critical value 1.96 corresponds to a probability of 95%, but it is
possible to know exact critical values for any confidence level
• For example, if we want a 99% confidence level, the critical value
based on the normal curve is 2.58
• Most packages including Microsoft Excel allow computation of critical
values
8
Chapter 6
A further complication
• The standard error of the mean  is unknown (sample
estimate)
• This adds some further uncertainty on the top of the
sampling error
• Hence, we use an approximation of the normal
distribution which is more conservative, is flatter and
assigns higher probabilities to the tails (extreme values)
compared to the normal distribution
• This distribution is the so-called Student-t distribution and
its critical values are different from those of the normal
distribution
9
Chapter 6
The Student-t distribution
0.4
Normal distribution
t(20)
Dotted lines show the
Student t distribution with
different degrees of
freedom (1, 5 and 20)
t(5)
0.35
0.3
t(1)
The bold line represents
the standard normal
distribution (with mean zero
and standard deviation
equal to one).
0.25
0.2
0.15
The degrees of freedom
are equal to the sample
size minus one.
0.1
0.05
0
-4
10
-3
-2
-1
0
1
2
3
4
Chapter 6
Critical values and sample size
Level of
confidence
99%
95%
90%
11
Student t values
Normal
value
t according to sample
size
z
10
20
30
40
3.17
2.23
1.81
2.85
2.09
1.72
2.75
2.04
1.70
2.70
2.02
1.68
2.58
1.96
1.64
Chapter 6
How to build a confidence interval from a sample
1. Compute the sample mean x-bar and standard
deviation s
2. Estimate the standard error of the mean sx
3. Choose a confidence level a
4. Choose the appropriate coefficient for critical values
(see previous slide), using the Student t approximation
instead of the Normal (z) value if the sample size is
below fifty
5. Compute the lower and upper confidence limits as:
 x − za / 2 sx ; x + za / 2 sx 
 x − ta / 2 sx ; x + ta / 2 sx 
12
for the Normal distribution or
for the Student t approximation
Chapter 6
SPSS confidence intervals
13
Chapter 6
Confidence interval
Descriptives
In a typical week how
much do you spend
on fresh or frozen
chicken (Euro)?
Mean
95% Confidence
Interval for Mean
5% Trimmed Mean
Median
Variance
Std. Deviation
Minimum
Maximum
Range
Interquartile Range
Skewness
Kurtosis
14
Lower Bound
Upper Bound
Statistic
5.6677
5.2817
6.0537
5.2958
5.0000
17.089
4.13383
.00
30.00
30.00
4.50
2.084
8.005
Std. Error
.19640
Boundaries of
the c.i.
.116
.231
Chapter 6
Hypothesis testing (HT)
• HT is a form of inference
• HT is a statistical tool to decide whether to reject or not a statement
about the target population on the basis of statistics computed on
sample
• This is only possible if the sampling distribution is known!
15
Example (Trust data-set)
• The sample mean expenditure for chicken is £ 5.67
• Considering only the sub-sample of those belonging to consumer
organizations the mean is £ 5.04
• May we safely conclude that those who belong to consumer
organizations spend less?
• This depends on the precision of the estimates.
• the upper limit of the 99% confidence interval of the sub-sample of
respondents associated to consumer organizations is 6.84
• the upper limit of the confidence interval for the overall sample is 6.18
• There is a chance that those who belong to consumer organization actually
spend more than other people
• Statistical tests based on the sample help in deciding whether the
hypothesis should be rejected; for example, that there is no
difference between respondents who belong to a consumer
organization and those who don’t.
16
Chapter 6
Hypothesis testing
• The difference between the two means follows a known statistical
distribution
• If the hypothesis of mean equality is true, the difference between the
two means should be zero
• The actual difference is it is £ 0.63 (£ 5.67 minus £ 5.04), but this
difference may be generated by sampling error
• When no difference exists at the population level, what is the
probability that a sample with a difference of £ 0.63 is extracted?
• If the probability is very high (say 90%) then it is wise not to reject the hypothesis
to equality
• Instead, a very low probability (e.g. 0.001%) means that it is very
unlikely that the difference is due sampling error, so one should
choose to reject the null hypothesis and conclude that belonging to a
consumer organization actually makes a difference on chicken
expenditure.
17
Chapter 6
The hypotheses
• The initial hypothesis is called the null hypothesis,
denoted by H0
• Contextually, the researcher sets an alternative
hypothesis (H1) which is complementary to H0 and
remains valid if H0 is rejected.
• Two-tailed tests are those where the alternative
hypothesis can go in either direction
• When the alternative hypothesis is formulated in a
unique direction (for example >6), the test is onetailed
18
Chapter 6
Chapter 6
Significance and confidence
• The threshold probability level below which the null
hypothesis is rejected is the significance level arbitrarily set by
the researcher (usually at the 5% or 1% level).
• It is denoted by a and its complementary 1-a is the
confidence level of a test
• The smaller the significance level the smaller the rejection
region, (which is the set of values that leads to rejection of the
hypothesis) and the larger is the acceptance region, (which is
the set of values that lead to non-rejection of the hypothesis)
19
Errors confidence and power
Chapter 6
The statistical power of a test is the probability of correctly
rejecting the null hypothesis when it is false and it is equal to
1-b. It can be estimated on sample data and depends on
• sample size
• significance level a
• effect size (a measure of “how wrong” is the null hypothesis)
When the null hypothesis is not rejected and power is above
80% then the conclusion is usually regarded as robust.
Reject H0
Non-reject H0
H0 is true
Type I Error
prob. =significance level (a)
Correct
prob. = confidence level (1-a)
H0 is false, H1 is true
Correct
prob. = power level (1-b)
Type II Error
prob. = b
20
Chapter 6
Hypothesis testing
1. Formulate the null hypothesis (H0) and the alternative hypothesis (H1)
2. Determine the distribution of the sample test statistic under H0
3. Choose a significance level a (i.e. a confidence level 1-a)
4. Compute the sample test statistic and its probability level (p-value)
5. When p-value 1.15.
• Critical values are obtained like for confidence intervals:
-za/2 defines the left rejection region (negative values < 2.5% probability) +za/2 defines the right rejection region (positive values < 2.5% probability) • For a 5% significance level (or a 95% confidence level): –z0.025=-1.96 and +z0.025=+1.96 25 Chapter 6 Chapter 6 Acceptance and rejection areas The test statistic lies in the rejection area 26 Chapter 6 Hypothesis testing in SPSS Test value here 27 Chapter 6 SPSS output One-Sample Test Test Value = 1.15 t In a typical week how much fresh or frozen chicken do you buy for your household consumption (Kg.)? 2.183 df 91 Sig. (2-tailed) Mean Difference .032 .60152 95% Confidence Interval of the Difference Lower Upper .0541 1.1490 p value The null hypothesis is rejected (as the p-value is smaller than 0.05) • We reject the null hypothesis that the average weekly consumption is Kg. 1.15 28 Chapter 6 One sided hypothesis (one tailed test) • We want to test whether average consumer evaluation of animal welfare is larger than 4.9 (on a 7-point scale) • It is convenient to formulate the hypothesis as follows: H 0 :   4.9 H1 :   4.9 • This is an one tailed test, as the alternative hypothesis is expressed directionally: the rejection area lies on the right of the critical value (all values on the left are consistent with the null hypothesis) Sample mean: 5.01 5.01 − 4.90 Standard deviation: 1.65 z= = 1.49 0.074 Standard error of the mean: 0.074 • instead of two critical values with a/2=0.025 as in the two-tailed test, we only require a single critical value for a=0.05 (which corresponds to the a=0.10 two-tailed critical values), which is 1.64 29 Chapter 6 One-tailed test The test statistic lies in the acceptance area 30 Chapter 6 One mean one-tailed test in SPSS • Exactly like two-tailed tests, but when interpreting the output: - One should only consider values larger than the positive critical value for rejection (or smaller of the negative critical value if the null direction is >)
– To get the correct critical value one should consider the a one instead of a/2 (for example, z0.05
rather than z0.25)
– Thus, the critical value for a 5% significance level in a one-tailed test corresponds to the critical
value for a 10% two-tailed test. For example z0.05=1.64
– Similarly, when looking at the two-tailed test p-value in SPSS, one should consider a “double”
threshold, i.e. reject the null at the 95% level when p>0.10 (since SPSS always assumes twotailed tests)
31
Chapter 6
SPSS output and one-sided test
One-Sample Test
Test Value = 4.9
Animal welfare
t
1.545
df
496
Sig. (2-tailed)
.123
Mean
Difference
.114
95% Confidence
Interval of the
Difference
Lower
Upper
-.03
.26
• The null hypothesis is not rejected at the 5% significance level
because Sig>0.10
• Note that differently from two two-tailed test here the
threshold is twice the chosen significance level.
32
Chapter 6
Test on two means
• Tests of equality on two means follow directly from the singlemean test
• The difference of two normally distributed sample means is still a
normally distributed, provided that the samples are unrelated or
paired
• The mean of the difference distribution is zero under the null
hypothesis of mean equality
33
Unrelated samples
• The sampled units belong to different populations and
are randomly extracted from each of the population. The
key condition is that the sampled units are randomly
assigned to the two groups
• This excludes the case where:
a) a single sample is drawn
b) the units are subdivided into two groups according to some
variable (gender, living in urban versus rural areas, etc.)
c) the sampling process might have some selection bias creating
dependency between the two groups
• Most social studies consider the samples to be unrelated
if the units are randomly extracted and the groups are
mutually exclusive.
34
Chapter 6
Chapter 6
Related and paired samples
• In related samples he same subjects may belong to both groups.
• For example, if the same individual is interviewed in two waves, the
two samples are said to be related.
• Two sub-sets from the same sample are generally related (not for
stratified sampling)
• Paired samples are a special case where exactly the same units
appear in both samples
• In this case it is possible to compute the difference for each of the
sampled units and the result corresponds to a single sample
• Matched samples are artificially paired samples, where the two
samples are matched according to some characteristics.
35
Chapter 6
Example (Trust data-set)
• 71% of respondents are female
• the targeted population are people in charge of food purchases;
males and females in the Trust data-set are associated by the fact
that they are responsible for food purchases
• This excludes those females and males who are not; any gender
comparison is conditional on this external factor
• We could not be conclusive in testing – for example – whether
males like chicken more than females
36
Chapter 6
Test for unrelated samples
• Example (Trust data-set)
• Italian versus UK respondents (extracted independently)
• Does the attitude towards chicken (question q9, “In my household we like
chicken”) differ between the two countries?
H0: UK = ITA vs.
37
H1: UK ≠ ITA
Chapter 6
Unrelated samples
• The two means are normally distributed
• Under the null hypothesis their difference is also normally distributed
• Consider the difference variable D = UK – ITA
• The test becomes identical to the one mean test for D = 0
• However, a measure of the standard error for D is necessary
38
Chapter 6
Standard error for
mean comparison testing
• In the (unlikely) event that the true standard errors are known the
joint standard error is:
  − =
1
2
 12
n1
+
 22
n2
• Everything proceeds as for the one mean test, thus the test statistic
is:
t=
x1 − x2
  −
1
39
2
Test statistics (unknown standard errors)
Chapter 6
• With unknown but equal standard errors, the test statistics is
t=
x1 − x2
1
1
sx
+
n1 n2
• Given that the standard error is estimated, additional uncertainty requires
the use of the t distribution with
n1+ n2-2 degrees of freedom
• With different standard errors the test statistic is
z=
x1 − x2
sx21
+
sx22
n1
n2
• This statistic can only be applied to large samples and the standard normal
distribution is the reference
40
Chapter 6
How to decide whether the standard errors are equal
or different?
• There are appropriate hypothesis tests for the equality of two variances
(discussed later)
• SPSS shows the p-value for the Levene’s test (Brown and Forsythe, 1974)
• SPSS provides the outcomes of both test, with and without assuming equality of
standard errors.
• In most cases these two tests provide consistent outcomes.
41
Chapter 6
Mean comparison with
unrelated samples
• The mean for the 100 UK respondents is 6.12 (standard error 0.15)
• the mean for the 100 Italian respondents is 5.62 (standard error 0.11)
• Is a mean difference of 0.5 significantly different from 0?
• Are the standard error equal?
42
Chapter 6
SPSS example
Target variable
here
Sub-groups defined here
through a grouping variable
43
Chapter 6
SPSS output
The
null hypothesis of equality of variance (Levene’s test) is not
CONCLUSION:
rejected
thea5%
s.l. (as p-value
is larger thandifference
0.05), thus the two
Thereatis
statistically
significant
standard errors could be regarded as equal
between the UK and Italy in terms of attitudes
towards chicken (as measured by q9)
Independent Samples Test
Levene’s Test for
Equality of Variances
F
In my household
we like chicken
Equal variances
assumed
Equal variances
not assumed
Sig.
.243
.622
t-test for Equality of Means
t
df
Sig. (2-tailed)
Mean
Difference
Std. Error
Difference
2.682
198
.008
.500
.186
.132
.868
2.682
183.426
.008
.500
.186
.132
.868
At any rate, the null hypothesis is rejected in both cases
(as the p-value is smaller than 0.05)
44
95% Confidence
Interval of the
Difference
Lower
Upper
Chapter 6
Paired samples
• As a second case study, consider the situation where two measures
are taken on the same respondents.
• For example, all respondents were asked a second question on
their general attitude towards chicken, “A good diet should always
include chicken” (q10)
• Do the two questions measure the same item (general attitude
towards chicken)?
• Can we assume that the results are – on average – equal?
• In this case the samples are paired and it is possible to compute a
difference between the response to q9 and q10 for each of the
sampled household.
• Everything goes back to one mean test discussed earlier
45
Chapter 6
Paired samples
• One may compute a third variable for each of the respondent as the difference
between q9 and q10, then compute the mean and the standard error for this
variable
• SPSS does this automatically
– mean of 5.73 for q9 versus a mean of 5.50 for q10
– average difference 0.23
– estimated standard error 0.06.
– The t test statistic is 3.84, largely above the two-tailed 99% critical value,
• the null hypothesis of mean equality should be rejected.
• It is not safe to assume that the two questions are measuring the same construct
46
Chapter 6
Paired samples
Select two variables from the
same data-set
47
Chapter 6
Output
Paired Samples Test
Paired Differences
Mean
Pair
1

In my household we like
chicken – A good diet
should include chicken
.231
Std. Deviation
Std. Error
Mean
1.348
.060
95% Confidence
Interval of the
Difference
Lower
Upper
.112
.350
t
3.824
df
Sig. (2-tailed)
497
Note how mean comparison tests can be used in the post-editing phase to
check consistency (see lecture 4):
1) Try again with q9 and q12e, the two questions are targeted to measure
the same construct with slightly different wording
2) If the null hypothesis is not rejected, one can compute the difference
between q9 and q12 and look for outliers
3) Cases with outliers show an inconsistency
48
.000
Chapter 6
Other type of tests
• Independent samples
• Standard t-test (as seen)
• Paired samples
– It becomes a one-sample t-test
• Proportion test
• Related samples
• Non-parametric tests
49
Chapter 9
Jaggia
Hypothesis test for population proportion
• The population proportion is useful for descriptive data that we want
proportions of.
• For normality, we use the rule: np>=5 and np(1-p) >=5
• Our test statistic: Z =
𝑃ℎ𝑎𝑡 −𝑝
𝑝(1−𝑝)/𝑛
• The hypothesis test works the same way for proportions
Communicating with
Numbers Jaggia
50
Chapter 9
Jaggia
Example
• A popular weekly magazine asserts that fewer than 40% of
households have changed their lifestyles due to gas prices. A survey
of 180 households find that 67 households have made lifestyle
changes.
• Specify and conduct a hypothesis test to test the magazine’s claim. Do
this for a 10% level of significance.
• Z=
𝑃ℎ𝑎𝑡 −𝑝
𝑝(1−𝑝)/𝑛
Communicating with
Numbers Jaggia
51
Chapter 9
Jaggia
Example
• A popular weekly magazine asserts that fewer than 40% of
households have changed their lifestyles due to gas prices. A survey
of 180 households find that 67 households have made lifestyle
changes.
• Step 1:
• Ho : p >= .40
• Ha : p < .40 Business Statistics Communicating with Numbers Jaggia 52 Chapter 9 Jaggia Example • A popular weekly magazine asserts that fewer than 40% of households have changed their lifestyles due to gas prices. A survey of 180 households find that 67 households have made lifestyle changes. • Step 2/3: 67 • The significance level is 10%. Phat = = .3722 𝑃ℎ𝑎𝑡 −𝑝 .3722 −.4180 • Test statistic: Z = = = -.76 𝑝(1−𝑝)/𝑛 .4 1−.4 − 18 • If we do P(Z

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