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Hypothesis Testing

Updated: Dec 24, 2022




The process of research typically involves six steps in the particular order.

The research starts with a problem, observed or experienced by the investigator. The initiation of researcher may also be due to his/her interest in studying a particular scenario or problem. For example, an investigator may observe a problem in declining sales of a particular product.

After identification of the problem researcher may choose to find the existing literature on the particular problem, may talk to experts for formulating the hypothesis. For example, the researcher finds sufficient literature where causes of declining sales of similar products are iterated by the researchers in the past or some sales experts have provided some insights on declining sales.

After conducting sufficient research, the researcher may formulate one or many hypotheses. This is to note that hypothesis is a pre-conceived idea on the basis of research, observation or experience. For example, the researcher may formulate a hypothesis that consumers are getting a better product of competitor at a lesser cost and, hence driving more value.

After formulating the hypotheses, the researcher uses certain statistical tools to collect the observations. In the example above the researcher may conduct a primary survey using a sample of consumers for testing the hypothesis.

After formulating the hypotheses, the researcher uses certain statistical tools to collect the observations. In the example above the researcher may conduct a primary survey using a sample of consumers for testing the hypothesis.

After the observations are recorded, a suitable statistical process is used for analyzing the data and a conclusion is drawn. The researcher in the above example may reject or accept the null hypothesis on the basis of observed data and statistical analysis.

As mentioned above the hypothesis can be defined as a pre-conceived idea about the population parameter. An investigator may be interested in knowing the difference between male and female IQ at a university. Depending on research, observation or expert opinion the researcher may formulate some hypothesis.

Fundamentally the hypothesis is divided into null and alternate hypothesis. The null hypothesis maintains the status quo and alternate hypothesis challenges the status quo. In the example mentioned above the researcher may formulate null and alternate hypothesis such as

Null Hypothesis H0: there is no significant difference between the IQ scores of males and females.

Here the population parameter under study is average IQ of males and females in the university. The null hypothesis may also be written as

Type I and Type II error

One tailed and two tailed tests



The six-step process of Hypothesis testing

z – test for testing population parameters '








Since the z-calculated is less than + 1.64 we do not reject the null hypothesis. Diagrammatically

Illustration: Daily sales figures of 40 shopkeepers showed that their average sales and standard deviation were Rs 528 and Rs 60 respectively. Is the assertion that daily sales on the average is Rs 400, contradicted at 5% level of significance by the sample. As the incentives and penalties are be decided on the basis of less of more sales hence test the null hypothesis at both the tails.



t – test for testing the hypothesis


Illustration: A machine is designed to produce insulating washers for electrical devices of average thickness of 0.025 cm. A random sample of 10 washers was found to have an average thickness of 0.024 cm with a S.D. of 0.002 cm. Test the deviation of mean from the population at 5% level of significance.

Solution:



the t-critical values in above illustration can be calculated as described in the following figure. At 9 degrees of freedom (checked in the row) at 5% level of significance checked in two tail entries (as the null hypothesis indicates that it is a two-tailed test) in the column we find that the critical value 2.262 i.e., - 2.262 on left tail and +2.262 on the right tail.



illustration: The mean weekly sales of the chocolate bar in a candy store was 146.3 bars per store. After an advertisement campaign the mean weekly sales in 22 stores for a typical week increased to 153.7 and showed a standard deviation of 17.2. Was the advertisement campaign successful.

Solution:

This time as suggested by the hypothesis the t-critical value will be observed from single tail values of level of significance.



t - test for difference of means



illustration: Two samples of 6 and 5 items respectively gave the following data:


Mean of first sample 40

SD of first sample 8

Mean of second sample 50

SD of second sample 10


Is the difference between means is significant. The value of t for 9 degrees of freedom at 5% level of significance is 2.26.

Illustration: A group of 5 patients treated with medicine A weight 42, 39, 48, 60 and 41 Kgs. Second group of patients from the same hospital treated with medicine B weight 38, 42, 56, 64, 68, 69 and 62 Kgs. Do you agree with the claim that, medicine B increases the weight significantly.

Solution



Significance Test for Dependent Samples or Paired Observations



Illustration: An IQ test is administered to 5 persons before and after they are trained. The results are given below.


Does training program ensured a change in IQ of the candidates


Solution


calculating the deviations



Chi-square test


Illustration: The number of automobile accidents per week in a certain area were as follows;

Are these frequencies in agreement with the belief that accidents conditions were the same during las 10-week period?

Solution:



Illustration: Suppose that, in a public opinion survey answers the questions

a. Do you drink?

b. Are you in favour of local option on sale of liquor?

Were as tabulated below:


Analysis of Variance

When there are more than two variables under study and a significance of comparison is to be checked analysis of variance comes really handy. for example, if the performance of sales from three stores for one week are to be compared against the null hypothesis that there is no significant difference between the sales of the stores during past week analysis of variance is used for testing the hypothesis.

Illustration: The following data relate to the yield of four varieties of wheat each sown on three plots. Find whether there is a significant difference between the mean yield of these varieties.

Here the null and alternate hypothesis is given by.

following entries table can be used to test the hypothesis using analysis of variance.


for calculation of SSA and SSW we can use the following formula


for the problem stated above following calculation is carried out for SSA and SSW




further the calculations can be complied by following table. the F-value can be observed from F-table for 5% level of significance, where we check the degrees of freedom between the groups in row and degrees of freedom in column and find the intersecting value (as given in the diagram below calculations).



F-critical value



Two-way Analysis of Variance

As in calculations in one-way analysis of variance. following table can be used for the calculation


The calculation for SSC, SSR and SST can be made using the following formula.



Illustration: To study the performance of three detergents and three different water temperatures, the following whiteness readings were obtained with specially designed equipment.



YouTube Tutorial



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