1 sample t test

Problem Statement - Measuring depressive system of the sample of the people taken.

H0: The depression level of the sample is same as that of the population

Ha: The depression level of the sample is significantly different as that of the population

Interpretation:

As we do not know the population variance so we are using 1 sample test. We have to find whether the mean of the sample is different from the normal population average of 1.5.

Two tables are shown. The first is the one sample statistics table which shows the amount of depression in the people. N is the no of people participated in the study. Mean is the average of the depression present of the sample of the people, stand deviation and standard error.

In the second table we measure the amount of depression from the first 3 values of the table. In this t absorbed, df is degrees of freedom and the significant value which is same as p value. If significance value is smaller than the alpha which is equal to .05 , then we will reject the hypothesis . Here the significance value is .02 that means more depressive symptoms are there than the population.

Independent sample t test

Problem statement: To find out if there is a sifnificant difference between IQ scores of males and females

H0 : There is no difference between the IQ scores of males and females

Ha : There is significant difference between IQ scores of males and females

The genders alog with the IQ scores are shown below:

Alpha = .05

Interpretation

In Levene’s test the value uf “Sig” > .05 , hence we cannot decide at this point and we will have to check “sig(2tailed)” value for “Equal variances assumes” i.e. .000 . Since this value is less than .05 , we reject null hyouthesis and conclude that there is a significant difference between IQ score of males and females

Paired sample t test

Problem statement

Marketing research analytics of company want to compare the price of watch of company A and B .He has taken sample of 8 retail stores in India with the 0.01 level. Verify whether company A watch sell for less than company B watch. Sample taken from the retail shop are as follows:

Price A($) Price B($)10 11

8 117 109 12

11 1110 13

9 128 10

H0: ud=0 (uA-uB=0)

Ha: ud<0

α= 0.01

Paired Samples Statistics

Mean N Std. Deviation Std. Error Mean

Pair 1PriceA 9.00 8 1.309 .463

PriceB 11.25 8 1.035 .366

Interpretation:

As we can see from the test output , p value =.001 is less than significant level (0.01).So we reject the null hypothesis and conclude that company A watch sells at lesser price than the company B watch .

Problem statement: A marketing firm want to know whether the income level of people influence the purchase of their for brands which are BrandA, BrandB, BrandC and BrandD respectively .A sample of 600 consumers were taken and following data emerged

H0: There is no association between the brand preference and income level (These two attributes are independent).

Ha: There is association between brand preference and income level (These two attributes are dependent).

 Level of Significance = .05

Case Processing Summary

Paired Samples Test

Paired Differences t df Sig. (2-

tailed)Mean Std.

Deviati

on

Std.

Error

Mean

99% Confidence

Interval of the

Difference

Lower Upper

Pair 1 PriceA - PriceB -2.250 1.165 .412 -3.691 -.809 -5.463 7 .001

  Brands  Brand1 Brand2 Brand3 Brand4 TotalIncome          Lower 25 15 55 65 160Middle 30 25 35 30 120Upper Middle 50 55 20 22 147Upper 60 80 15 18 173Total 165 175 125 135 600

Cases

Valid Missing Total

N Percent N Percent N Percent

Income * Brand 600 100.0% 0 0.0% 600 100.0%

Income * Brand Crosstabulation

Count

Brand Total

BrandA BrandB BrandC BrnadD

Income

Low 25 15 55 65 160

Middle 30 25 35 30 120

Uppermiddle 50 55 20 22 147

High 60 80 15 18 173

Total 165 175 125 135 600

Chi-Square Tests

Value df Asymp. Sig. (2-

sided)

Pearson Chi-Square 131.766a 9 .000

Likelihood Ratio 138.105 9 .000

Linear-by-Linear Association 84.357 1 .000

N of Valid Cases 600

Interpretation: chi square

As we can see from the SPSS output, value of p <0.05 which proves that null hypothesis is rejected

and there is relation between brand and income level.

Problem statement: X Company recorded the sale of Product A for the first 20 days of the month Verify

whether data is normally distributed or not. Sales details are provided below –

Day 1 2 3 4 5 6 7 8 910

11

12

13

14

15

16

17

18

19 20

Sale16

20

22

23

23

25

24

20

17

17

25

22

20

19

17

15

17

18

23 23

In this example, the null hypothesis is that the data is normally distributed and the alternative

hypothesis is that the data is not normally distributed.

Statistic Std. Error

Sale

Mean 20.20 .694

95% Confidence Interval for

Mean

Lower Bound 18.75

Upper Bound 21.65

5% Trimmed Mean 20.22

Median 20.00

Variance 9.642

Std. Deviation 3.105

Minimum 15

Maximum 25

Range 10

Interquartile Range 6

Skewness -.008 .512

Kurtosis -1.237 .992

Tests of Normality

Kolmogorov-Smirnova Shapiro-Wilk

Statistic df Sig. Statistic df Sig.

Sale .149 20 .200* .943 20 .274

*. This is a lower bound of the true significance.

a. Lilliefors Significance Correction

Interpretation:

 For dataset small than 2000 elements, we use the Shapiro-Wilk test; otherwise, the Kolmogorov-Smirnov test is

used. In our case, since we have only 20 elements, the Shapiro-Wilk test is used. From test of normality, the p-value

is 0.274 which is greater than 0.05. We can reject the alternative hypothesis and conclude that the data comes from

a normal distribution.

Problem statement : A company has fifteen male and fifteen as his employee .Salary structure of each employee

is different which has been provided below .Identify the outliers using box plot.

Gender Salary (Rs)

Male2000

03000

04000

01200

01500

0 70001400

01260

01800

01230

01450

01700

06000

02700

01300

0

Female12000

18000

10000 8000 5000 6000 7000 9000 4000 6000

17000

30000 3200

23000

14000

Case Processing Summary

Cases

Valid Missing Total

N Percent N Percent N Percent

Salary 15 75.0% 5 25.0% 20 100.0%

Interpretation:

Dark line in the middle of the box is median of the salary .We also observed that salary for third and

thirteen male employer are outliers as they have value more than three times the height of the box

while twelfth female employer is outlier .We can also see that average salary of female is less than the

male .