Data analysis

Lecture 9Data Analysis

Ref Chapters 13 and 14

General Approach to Hypothesis Testing

Malhotra et al. 2005, 293

3

What is a hypothesis?

• A hypothesis is an unproven proposition or supposition that tentatively explains certain facts or phenomena.– Empirically testable

• Null hypothesis is a statement about status quo.– Any change from what has been thought to be

true will be due entirely to random sampling error.

• Alternative hypothesis is a statement indicating the opposite of the null.

4

Null and alternative hypotheses• Example: highly dogmatic consumers will

be less likely to try a new product than less dogmatic consumers.

• The null hypothesis (H0)is where there is no difference between high dogmatics and low dogmatics in their willingness to try an innovation.

• The alternative hypothesis (H1) is where there is a difference between high and low dogmatics.

5

Hypothesis testing

• The purpose of hypothesis testing is to determine which of the two hypotheses is correct.

6

The hypothesis–testing procedure

• The process is as follows:– Determine a statistical hypothesis.– Imagine what sampling distribution would be if

this is a true statement.– Take an actual sample and calculate sample

mean.– Determine if the deviation between the

obtained value and expected value of sample mean could have occurred by chance alone.

– Set a standard for determining if we reject the null or accept the alternative.

7

The hypothesis–testing procedure

• Significance level is the critical probability in choosing between the null and alternative hypotheses.– The probability level that is too low to warrant

support of the null hypothesis.• Assuming the null is true, if the probability

of occurrence of the observed data is smaller than the significance level, then the data suggest that the null should be rejected.– Evidence supporting contradiction of null.

8

Type I and type II errors• Since hypothesis testing is based on

probability theory, the researcher cannot be completely certain and runs the risk of committing two types of errors.– Type I error is an error caused by rejecting the

null hypothesis when it is true.• Probability of alpha ()

– Type II error is an error caused by failing to reject the null hypothesis when the alternative hypothesis is true.

• Probability of beta ()

9

The t–distribution• The t-distribution is a symmetrical, bell–

shaped distribution that is contingent on sample size.– It has a mean of zero and a standard deviation

equal to one.• The shape of the t–distribution is

influenced by its degrees of freedom. – Number of observations minus number of

constraints or assumptions.– For sample sizes over 30, t–distribution closely

approximates Z–distribution.

10

The t–distribution

11

Univariate hypothesis test using the t–distribution

• Store manager believes that average number of customers who return or exchange merchandise is 20.– H0: 20 and H1: 20

• Store records returns and exchanges for 25 days (n) and sample mean is 22 and standard deviation is 5.

• Confidence level of 95% or significance level of 5%.• Referring to Table B.3 in Appendix B, we find that for 24 degrees

of freedom (n–1), the t–value is 2.064. Thus:Lower limit = - tc.l. S = 20 – (2.064)5 = 17.936

n 25Upper limit = + tc.l. S = 20 + (2.064)5 = 22.064

n 25

• Since the sample mean lies within the critical limits, the null hypothesis cannot be rejected.

12

Univariate hypothesis test using the t–distribution

• Alternatively, we may test a hypothesis by calculating the observed t–value and comparing it to the critical t–value.

• To calculate the observed t:tobs = X - = 22 – 20 = 2

Sx 5 25

• Referring to Table B.3 in Appendix B, we find that for 24 degrees of freedom (n–1), the t–value is 2.064.

• Since the observed t–value is less than the critical t–value, the sample mean lies within the critical limits.

• Thus, the null hypothesis cannot be rejected.

13

The Chi-square test for goodness of fit

• The Chi-square (c2) test allows for investigation of statistical significance in the analysis of a frequency distribution.

• Categorical data on variables like sex, education, etc.,• Allows us to compare the observed frequencies with the

expected frequencies based on theoretical ideas about the population distribution.

• Tests whether the data came from a certain probability distribution.

14


• The process is as follows:• Formulate the null hypothesis and determine the

expected frequency of each answer.• Determine the appropriate significance level.• Calculate the c2 value, using the observed frequencies

from the sample and the expected frequencies.• Make the statistical decision by comparing the

calculated c2 value with the critical c2 value.

15


16

The Chi-square test for goodness of fit• To calculate the Chi-square statistic:

2 = (Oi – Ei)2

Ei

– where c2 is the Chi–square statistic, Oi is the observed frequency in the ith cell, and Ei is the expected frequency in the ith cell.

• Table 12.15 shows that calculated Chi–square value is 4.• The degrees of freedom is the number of cells associated with

column or row data minus one (k–1).– k is 2 since there are only two categorical responses.

• Referring to Table B.4 in Appendix B, we find that for 1 degree of freedom (k–1), the Chi-square value is 3.84.

• Since the calculated value is larger than the critical value, the null hypothesis is rejected.

17

Choosing the appropriate statistical technique

• The choice of statistical analysis depends on:– The type of question to be answered

• Example: researcher concerned with comparing average monthly sales central value would use t–test.

– The number of variables• Example: researcher concerned with one variable at a time

would use univariate statistical analysis.

– The scale of measurement• Example: testing a hypothesis about a mean requires interval

scaled or ratio scaled data.• Parametric versus nonparametric statistics.

18

Parametric versus hypothesis tests

• Parametric statistics are used when the data are interval or ratio scaled, when the sample size is large, and when the data are drawn from a population with a normal distribution.

• Nonparametric statistics are used when data are either nominal or ordinal.

Using the Normal tables

Using the p-value

One-Sample Statistics

448 1150.5960 2705.08330 127.80317

How much have youspent, in total, onInternet shopping overthe past 12 months?

N Mean Std. DeviationStd. Error

Mean

One-Sample Test

2.743 447 .006 350.5960 99.4263 601.7657


t df Sig. (2-tailed)Mean

Difference Lower Upper

95% ConfidenceInterval of the

Difference

Test Value = 800

If p-value 0.05, Reject H0

Conclude that the average amount spent on the internet is more than $800 per year800:

800:

1

0

H

H

Types of Hypothesis Tests

Parametric Tests

One sample t test

We are testing the hypothesis that the mean satisfaction rating exceeds 4.0, the neutral value on a 7-point scale.

4:

4:

1

0

H

H

Parametric Tests cont.

One-Sample Statistics

443 5.19 1.079 .051Shopping at thiswebsite is usually asatisfying experience


Mean

One-Sample Test

23.112 442 .000 1.19 1.08 1.29Shopping at thiswebsite is usually asatisfying experience


Difference Lower Upper


Difference

Test Value = 4

The p-value < 0.05, hence reject H0 and conclude that the satisfaction rating for the website is greater than 4 (generally agree)


Two Independent samples (Means)We are testing the hypothesis that mean amount spent on shopping on the Internet is different for males and females

H0: 1 = 2

H1: 1 2

Group Statistics

236 1283.4237 3502.02542 227.96244

212 1002.7311 1342.04673 92.17215

GenderMale

Female



Mean


Since p-value > 0.05, t test assuming equal variances should be used

Independent Samples Test

2.166 .142 1.097 446 .273 280.6926 255.91581 -222.258 783.64323

1.142 308.922 .255 280.6926 245.89139 -203.141 764.52641

Equal variancesassumed

Equal variancesnot assumed


F Sig.

Levene's Test forEquality of Variances


DifferenceStd. ErrorDifference Lower Upper


Difference

t-test for Equality of Means

Since p-value > 0.05, we do not reject H0 and conclude that there is no difference between men and women on the

amount they spend on internet shopping


Two Independent samples (Proportions)

We are testing the hypothesis that the proportion of heavy internet users is the same for male and females. H0: 1 = 2

H1: 1 2

Internet usage * Gender Crosstabulation

Count

39 58 97

199 154 353

238 212 450

Light

Heavy

Internetusage

Total

Male Female

Gender

Total

Sample data


2

22

1

11

2121

)1()1(

)()(.

nn

ppcalcZ

212)27)(.73(.

238)16)(.84(.

0)73.84(.

75.2


If Zcrit = 1.645 (using the normal tables where =0.05)

We reject H0 and conclude that there is a difference in the percentage (proportion) of heavy user of the internet between males and females.

Non-Parametric Tests

Chi-square

H0: There is no association between Internet usage and age of respondents

H1: There is an association between Internet usage and age of respondents

Non-Parametric Tests cont.

Internet usage * Age of respondents Crosstabulation

Count

22 17 44 14 97

164 107 71 11 353

186 124 115 25 450

Light

Heavy

Internetusage

Total

18 - 24 25 - 39 40-5960 yearsor over

Age of respondents

Total

Chi-Square Tests

51.444a 3 .000

47.450 3 .000

43.858 1 .000

450

Pearson Chi-Square

Likelihood Ratio

Linear-by-LinearAssociation

N of Valid Cases

Value dfAsymp. Sig.

(2-sided)

0 cells (.0%) have expected count less than 5. Theminimum expected count is 5.39.

a.

P-value < 0.05 hence reject H0

and conclude that there is an

association between internet

usage and age of respondents


Chi-square (Another one!)VU’s Open Day organisers are investigating whether visitors’ overall rating of Open Day is independent of the age of the visitor. Test at .05 level of significance.

H0: Overall rating of Open Day and age are independent [no association]

H1: Overall rating of Open Day and age are not independent [association]


P-value < 0.05 hence reject H0

and conclude that there is an

association between ratings of open day and

age of respondents

Overall rating of Open Day * Age of respondent Crosstabulation

Count

1 2 3

6 2 1 9

24 15 4 43

100 25 12 137

66 29 18 113

197 71 37 305

Poor

Fair

Good

Very Good

Excellent

Overallrating ofOpenDay

Total

18 or under 19 - 29 Over 29

Age of respondent

Total

Chi-Square Tests

18.369a 8 .019

15.161 8 .056

.023 1 .879

305

Pearson Chi-Square

Likelihood Ratio

Linear-by-LinearAssociation

N of Valid Cases

Value dfAsymp. Sig.

(2-sided)

5 cells (33.3%) have expected count less than 5. Theminimum expected count is .36.

a.

33

Chapter 13Bivariate statistical analysis: Tests of differences

34

What is the appropriate test of difference?

• Do two groups differ with respect to some behaviour, characteristic, or attitude?– For example, are there differences in muscle gain

between subjects in the experimental exposed to muscle supplements and control group?

– Two different variables.• Comparisons between two independent groups is

independent samples t–test for difference of means.

35

The independent samples t–test for differences of means

• Used to test a hypothesis stating that the means scores on a variable will be significantly different from two independent samples.

• Can only take the mean of an interval or ratio–scaled variable.– The other variable is nominal scaled with two

groups.• Example: frequency of purchase between gender

groups.

36


• The null hypothesis about differences between groups is as follows:

m1 = m2 or m1 – m2 = 0

• The t–value is a ratio with information about the difference between means (provided by the sample) in the numerator and the standard error in the denominator.

t = Mean 1 – Mean 2 = X1 – X2

Variability of random means Sx1-x2

where Sx1-x2 is the pooled estimate of the standard error.

37


• From Table 13.1, we can calculate the pooled estimate of the standard error (1.304) and the t–statistic (2.341).

• Referring to Table B.3 in Appendix B, the critical t–value is 2.021.

• Since the calculated t–value exceeds the critical t–value, the null is rejected.

38

Conducting an independent samples t-test in SPSS

• Running an independent samples t–test in SPSS would produce the results shown below.

39

Conducting an independent samples t-test in SPSS

• Mean is displayed in Table 13.2.– But is this a real difference or did it occur by

chance?• Equality of variances is displayed in Table 13.3.

– If significance value is more than 0.05, then we can assume equal variances.

• Equality of means and significance value are displayed in Table 13.3.– If significance value is less than 0.05, then the two

means are significantly different.

40

Analysis of variance (ANOVA)• Used when comparing means of more than two

groups or populations.– For example, comparing women working full–time

outside home, part–time outside home, and full time inside home on willingness to purchase life insurance.

• Using variances to allow us to compare means.• Null hypothesis about differences between groups is

as follows:m1 = m2 = m3

41

Analysis of variance (ANOVA)• If the grouping variable (i.e., working status) is

responsible for differences in purchase intention for life insurance, then the variation in responses between each of the three groups will be comparatively larger than the variation in responses within each of the groups.

• The F–test is used to compare one sample variance with another.– F–ratio: the larger sample variance is divided by

the smaller sample variance.

42

Analysis of variance (ANOVA)• A larger ratio of variance between

groups to variance within groups implies a greater F–ratio.

• If the F–ratio is large, the more likely the differences in means has occurred as a result of the grouping variable.

• A calculated F–ratio that exceeds the critical F–ratio indicates that results are statistically significant.

• Thus, the null has to be rejected.

43

Conducting an ANOVA in SPSS

• Running an ANOVA in SPSS would produce the results shown below.

44

Conducting an ANOVA in SPSS• Mean is displayed in Table 13.7.

– But is this a real difference or did it occur by chance?

• Variances between and within groups are displayed in Table 13.8.

• F–ratio and significance value are displayed in Table 13.8.– If significance value is more than 0.05,

then the means are not significantly different.

• A result of sampling error or chance.

45

Nonparametric statistics for tests of differences

• So far, it has been necessary to assume that population is normally distributed.– If it is normal, the error associated with making

inferences can be estimated.– If it is not normal, the error may be large and cannot

be estimated.• Nonparametric tests overcome these limitations but will

have a greater probability of Type II error and require a larger sample size compared to a parametric test.

46

Statistical and practical significance for test of differences

• There is a distinction between statistical significance and practical significance.

• Even if there is a statistical difference, the practical difference might be very small.– This is substantive significance.

• Researchers need to keep this in mind when interpreting statistical output.

ANOVA

• Analysis of variance (ANOVA) examines the differences in the mean values of the dependent variable (interval scale) associated with the effect of the controlled independent variable (nominal scale), after taking into account the influence of the uncontrolled independent variables.

e.g. Do the brand evaluation of groups exposed to different commercials vary?

How do consumers’ intentions to buy the brand vary with different price levels?

One-way ANOVAWe are testing to determine the effect of in-store promotion (X) on sales (Y).

H0: 1 = 2 = 3

H1: 1 2 3

Descriptives

Sales

10 8.3000 1.33749 .42295 7.3432 9.2568 6.00 10.00

10 6.2000 1.75119 .55377 4.9473 7.4527 4.00 9.00

10 3.7000 2.00278 .63333 2.2673 5.1327 1.00 7.00

30 6.0667 2.53164 .46221 5.1213 7.0120 1.00 10.00

high

medium

low

Total

N Mean Std. Deviation Std. Error Lower Bound Upper Bound

95% Confidence Interval forMean

Minimum Maximum

The DataStore number Coupon level In-store promotion Sales Clientele rating

123456789

101112131415161718192021222324252627282930

111111111111111222222222222222

111112222233333111112222233333

109

10898879657645897764556423212

91084684

1069896

104

10684968

104946

1098

Source: Malhotra et al. (2004) Table 11.1 page 317 for the data set

One-way ANOVA cont.ANOVA

Sales

106.067 2 53.033 17.944 .000

79.800 27 2.956

185.867 29

Between Groups

Within Groups

Total

Sum ofSquares df Mean Square F Sig.

Multiple Comparisons

Dependent Variable: Sales

Tukey HSD

2.1000* .76884 .029 .1937 4.0063

4.6000* .76884 .000 2.6937 6.5063

-2.1000* .76884 .029 -4.0063 -.1937

2.5000* .76884 .008 .5937 4.4063

-4.6000* .76884 .000 -6.5063 -2.6937

-2.5000* .76884 .008 -4.4063 -.5937

(J) In-store promotionmedium

low

high

low

high

medium

(I) In-store promotionhigh

medium

low

MeanDifference

(I-J) Std. Error Sig. Lower Bound Upper Bound

95% Confidence Interval

The mean difference is significant at the .05 level.*.

Main effect (in-store promotion)

Residuals

Reject H0, the means are not

equal

One-way ANOVA cont.

Interpretation• 57.1% (ie. 2 = 106.067/ 185.856) of the variation

in sales is accounted for by in-store promotion, indicating a modest effect.

• The mean sales figures are different, that is at least one pair of means is statistically different.

• All combination of means are statistically different, therefore the different levels of in-store promotion will impact sales.

Data analysis

Business

Transcript of Data analysis