# Hypothesis Testing

Updated on February 23, 2010

Continued from: Data Analysis and Interpretation

Statistics has two major categories: (i) descriptive statistics and (ii) inferential statistics.

Descriptive statistics describe characteristics or property of data. One can get an idea about the central tendency and its dispersion. Various indicators or ratios are worked out to find the central tendency or concentration like Mean, Median and Mode.

On the other hand, dispersion is covered by Range, Variance and Standard Deviation. For example, two countries may have the same per capita income but standard deviations of individual’s income may be small in one case and large in another. This would reflect rich-poor gap or inequality of income distribution.

This is known as quantitative analysis and can be strengthened with the use of graphs.

Inferential statistics helps the researcher to reach conclusions about the population on the basis of a sample. In other words, the inferential statistics try to infer from the sample data qualities or character of the population. A researcher may like to find out how many people in a country read newspaper. The researcher would not interview the whole population but by choosing an appropriate sample, the answer can be found.

Also, inferential statistics are used to make judgments of the probability that an observed difference between groups is a dependable one or it is mere a chance.

A variety of tests are used to make statistical decision using experimental data. A result is called statistically significant if it is unlikely to have occurred by chance. The phrase "test of significance" was coined by Ronald Fisher: "Critical tests of this kind may be called tests of significance, and when such tests are available we may discover whether a second sample is or is not significantly different from the first."

The following examples would show use of statistical tables to test the hypotheses.

## t - Test

Adamjee Insurance has worked out that it costs Rs.3,600 on the average to handle paper-work, pay investigators and settle an insurance claim. This was more than what other insurance companies were experiencing. So some cost-cutting methods were introduced. In order to measure the impact a sample of 25 claims was taken and average claim was found to be Rs.3,420 with a standard deviation of Rs.600.

At the 0.01 level, is there a reduction in the average cost or can the difference be mere chance?

## T - Test applied

Since sample is rather small i.e. less than 30, the researcher would use t-test. It may be noted that there may be 5,000 claims in a particular period. It would be cumbersome or expensive to include all in the study. So the researcher draws a random sample of 25 claims and include them in study. Taking a random in this case is quite easy. All claims are numbered. So samples can be taken manually or through computer command RAND().

Necessary formula and calculation are given in the side sheet. It would be seen that there was no significant reduction. (If more samples were taken, the average would move towards Rs.3,600.

## t - test

Click thumbnail to view full-size

## A question on the use of Z - Score

Haleeb Foods Ltd is worried about the shelf life of its skimmed milk. It is receiving complaints about the bad quality of milk from its various sales outlets in the city. Average shelf life of a milk pack is 180 days with a standard deviation of 16 days.

Given the above, find out probability of packs which may:

1. Get bad ( rotted ) by 160 days or before
2. Stay fresh beyond 200
3. Stay fresh Between 160 and 200 days
4. How many days of freshness with money-back guarantee, the company may advertise so that no more than 5% are lodged?

The formula for calculating z value for any observation is to substract it from mean and multiply it by the standard deviation. The z value shows distance from the mean in term of standard deviation.

In this case, the observation or X value is 160 days while mean is 180 days, giving a difference of -20 days. If one divides this by standard deviation of 16, the result is - 1.25. From the z table, area under the curve would be 10.56%,

Generally table give only positive value beyond ZERO. Such table would show an area of 0.39435, Since total area before zero or half of the curve is 0.50, one gets 0.89435 by adding the previous figure. Now the z value is -1.25 and not 1.25. Therefore by deducting it from one one get 0.1056 ( 1.00 - 0.89435 ) or 10.56 %. This area is shaded in the chart.

In MS Excel, there is a function NORMSDIST(), which converts the z-score it its probability.

Based on above, the company can conclude that 11.56% packs would get bad within first 160 days of storage.

Part 2 of the question says, how many packs would stay fresh beyond 200 days. Since absolute difference between observation "x" and Mean is the same, the z value would remain 1.25 which gives a reading of 89.44%. Since we are interested in area beyond this measure, we get 10.56%, same as before, by deducting 89.44% from 100%. In this way, 78.87% would be the area between -1.25 and +1.25.

In the last part of the question, we know z value (-1.645 corresponding to 100% - 5%) besides mean and standard deviation. All we have to do is find out value of "X" which in this case be 153 days. Thus the company can announce money-back guarantee if the pack gets bad within 153 days. The overall claims are expected to remain with 5% of total packs sold.

## Z- SCORE

Click thumbnail to view full-size

## Chi-Squared Test

This test evaluates whether observed frequencies for a qualitative variable (or variables) are in line with the expected frequencies. In the question given in the side sheet, 120 cups were sold in one day bearing pictures of 6 cricket players. If there was no difference between the popularity of the players, customers would have bought 20 cups of each player. These were expected frequencies. As against this, actual or observed frequencies were matched and some difference was found. Was this difference mere a chance and if more samples were taken the position would not change much thereby confirming that each player is equally popular as seen from the sale of cups with his picture.

In order verify that there is indeed more preference for Inzamam and Yousuf, we test the following hypothesis:

NULL: There is no preference for cup bearing pictures of different players.

ALTERNATE: There is more preference for Inzamam and Yousuf than others.

The calculations would be as follows:

Frequencies Observed
Frequencies Expected
Frequencies Observed minus Frequencies Expected
(Fo-Fe) x (Fo-Fe)

Danish
13
20
-7
49
2.45
Inzamam
33
20
13
169
8.45
Shahid
14
20
-6
36
1.8
Shoaib Akhtar
7
20
-13
169
8.45
Younis
17
20
-3
9
0.45
Yousuf
36
20
16
256
12.8

120
120

34.40

The calculated value, 34.40 must be compared with the table value of 11.07 being the chi-square value for k-1 = 5 and at 95% confidence.  Since the calculated figure exceeds the table-value, one can say with 95% confidence that there is certainly a preference for cups bearing pictures of two players.  In other words, more sales of the two types of cups were not due to chance except for 5% probability that it was only a chance.  With such low probability we can reject Null Hypothesis and accept the Alternate.

## Chi square

Click thumbnail to view full-size

## F - Test

In this test, we compare the two variations. More variation is indicator of more risk. If price of two shares is the same at Rs.20 but one share has a standard deviation of 10 while other only one, we can say that first share is volatile and hence risky while other is stable.

F-test determine whether there is real difference between the two variations or it was just a chance. In the side example, one route is shorter but there being more traffic signals, it may take between 44 to 68 minutes. While other is longer but certain as it takes 54 to 64 minutes. Those who travel on the first route take a chance but on the second route one can travel with certainty.

Using the appropriate formula, we find calculated value to be and table value to be . Since calculated value is more than the table value, we can conclude with 90% certainty that there is difference in variations of the both route. Hence, if one plans to go to an important meeting, one should take the route with lesser variation.

## F Distribution

Click thumbnail to view full-size

0

1

7

3

2

## Popular

10

56

• ### Ancient Egypt Lapbook

39

0 of 8192 characters used
• Shahid Bukhari

8 years ago from My Awareness in Being.

hafeezrm

I read your descriptive of Hypothesis Testing, and learned from it ... Thank you.

But, I noted, that the "Variable" Cause of contention ... the factor of Corporate Politics, is absent from your analysis ... which, you will agree, is the "Prime Factor" influencing, what has now become, Multinationals controlled Businesses today.

In, that your paper, relies purely on Frequencies and Graphics ...

Tests; Hypothesis, or Laboratory etc., address the Proposition from certain Benchmarks, and try establish the desired, New Benchmark Median, within Variables.

In the context of the case studies produced above ... I, a long time Consumer of Tetra Packaged Milks, was dismayed to observe my favorite Brand, being Gradually eased out from the Market, by another Multinational Brand Competitor.

The Corporate wrangling has now reached a point, where perishable products are being deliberately retained by 'greased' distributors, or being variously delayed, so that by the time, these reach the shelf ... these are either nearing expiry, or have gone into C or D grade of Quality.

Or ...

These have been Tampered with, by the "Competing" Multinational's Agent at Source, by such criminal means as 'doctored' emulsifiers, and Pathogens introduced in the Packaging system !

These criminal Acts have full support of the balls of these Multinationals ... Buttered [Yellow] Print media. The latest example is of the criminal Wrangling, to hit the Bottled Water Sector ... where Except the Multinational Products, all other Tested samples were found to be "Tainted" with pathogens ... I am not putting the Exclamation mark, because, I am not surprised.

Not that I am a fan of "Packaged" foods, but am forced by the Multinational "created" circumstance, to shift my Brand preference, and buy their products, to avoid risking my health.

I reckon, in Developing Countries ... it is easy to "Manipulate" a few, into doing these doctorings on behalf of the Multinationals, now eying the World's Food market. [after having conquered Banking, Insurance, Housing, Health, Cloth and Apparel, Water, Trade and Industry Sectors, etc.].

Is it not time you spared your time and exposed these Wranglings, for the good of Pakistanis, and the developing countries at large ?

• jessica

9 years ago

sir

i dont know how to calculate s when the question not given.

s i mean is confidence level for population mean.thanks

• kamran

9 years ago

Dear Sir, impressive work. Ive used this method in my office for the newly arrived trainees i.e to see what is their deviation and z-score i.e how much deviated. i found it excellent. moreover, it is easy to look into bell curve (Ive given this method a name in my office) rather to see data sheet for each candidate. thnx

• ubedullah khan mahsoood

9 years ago

Dear Hafeez ur Rehman malik, i read the article very deeply and found the importance of statistic in research, before studying the article i was in trouble about the application of statistic in research,know i learn the application of Z-score, T -table and F-test,

specially your given example are impressive, hope u will share ur precious knowledge with us in future,

• AUTHOR

hafeezrm

9 years ago from Pakistan

Dear Rufi,

F-Test compares variances between two samples. (Z and t measures differences between means).

Variance is an indicator of risk. The higher the variance, the higher the risk.

9 years ago from Karachi

Dear Sir,

STATISTICAL methods and Cricketers example you have used is very much impressive and interesting. Thanks for adding such research knowledge.

I need to know the real application of F-Test, as you quote the example of Shares, but I need to know whether we will consider only the Standard Deviation in such case?

How F-Test could be viable tool for such kind of decision making?

Thanks a lot!