# Verifying The Action Hypothesis Statement

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### Example: Right-tailed test

An engineer measured the Brinell hardness of 25 pieces of ductile iron that were subcritically annealed. The resulting data were:

 170 167 174 179 179 156 163 156 187 156 183 179 174 179 170 156 187 179 183 174 187 167 159 170 179

The engineer hypothesized that the mean Brinell hardness of all such ductile iron pieces is greater than 170. Therefore, he was interested in testing the hypotheses:

H0 : μ = 170
HA: μ > 170

The engineer entered his data into Minitab and requested that the "one-sample t-test" be conducted for the above hypotheses. He obtained the following output:

The output tells us that the average Brinell hardness of the n = 25 pieces of ductile iron was 172.52 with a standard deviation of 10.31. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 10.31 by the square root of n = 25, is 2.06). The test statistic t* is 1.22, and the P-value is 0.117.

If the engineer set his significance level α at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t* were greater than 1.7109 (determined using statistical software or a t-table):

Since the engineer's test statistic, t* = 1.22, is not greater than 1.7109, the engineer fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the α = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.

If the engineer used the P-value approach to conduct his hypothesis test, he would determine the area under a tn - 1 = t24 curve and to the right of the test statistic t* = 1.22:

In the output above, Minitab reports that the P-value is 0.117. Since the P-value, 0.117, is greater than α = 0.05, the engineer fails to reject the null hypothesis. There is insufficient evidence, at the α = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.

Note that the engineer obtains the same scientific conclusion regardless of the approach used. This will always be the case.

### Example: Left-tailed test

A biologist was interested in determining whether sunflower seedlings treated with an extract from Vinca minor roots resulted in a lower average height of sunflower seedlings than the standard height of 15.7 cm. The biologist treated a random sample of n = 33 seedlings with the extract and subsequently obtained the following heights:

 11.5 11.8 15.7 16.1 14.1 10.5 15.2 19 12.8 12.4 19.2 13.5 16.5 13.5 14.4 16.7 10.9 13.0 15.1 17.1 13.3 12.4 8.5 14.3 12.9 11.1 15 13.3 15.8 13.5 9.3 12.2 10.3

The biologist's hypotheses are:

H0 : μ = 15.7
HA: μ < 15.7

The biologist entered her data into Minitab and requested that the "one-sample t-test" be conducted for the above hypotheses. She obtained the following output:

The output tells us that the average height of the n = 33 sunflower seedlings was 13.664 with a standard deviation of 2.544. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 13.664 by the square root of n = 33, is 0.443). The test statistic t* is -4.60, and the P-value, 0.000, is to three decimal places.

Minitab Note. Minitab will always report P-values to only 3 decimal places. If Minitab reports the P-value as 0.000, it really means that the P-value is 0.000....something. Throughout this course (and your future research!), when you see that Minitab reports the P-value as 0.000, you should report the P-value as being "< 0.001."

If the biologist set her significance level α at 0.05 and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic t* were less than -1.6939 (determined using statistical software or a t-table):

Since the biologist's test statistic, t* = -4.60, is less than -1.6939, the biologist rejects the null hypothesis. That is, the test statistic falls in the "critical region." There is sufficient evidence, at the α = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.

If the biologist used the P-value approach to conduct her hypothesis test, she would determine the area under a tn - 1 = t32 curve and to the left of the test statistic t* = -4.60:

In the output above, Minitab reports that the P-value is 0.000, which we take to mean < 0.001. Since the P-value is less than 0.001, it is clearly less than α = 0.05, and the biologist rejects the null hypothesis. There is sufficient evidence, at the α = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.

Note again that the biologist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

### Example: Two-tailed test

A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. On one production run, he took a random sample of n = 10 pieces of gum and measured their thickness. He obtained:

 7.65 7.6 7.65 7.7 7.55 7.55 7.4 7.4 7.5 7.5

The quality control specialist's hypotheses are:

H0 : μ = 7.5
HA: μ ≠ 7.5

The quality control specialist entered his data into Minitab and requested that the "one-sample t-test" be conducted for the above hypotheses. He obtained the following output:

The output tells us that the average thickness of the n = 10 pieces of gums was 7.55 one-hundredths of an inch with a standard deviation of 0.1027. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 0.1027 by the square root of n = 10, is 0.0325). The test statistic t* is 1.54, and the P-value is 0.158.

If the quality control specialist sets his significance level α at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t* were less than -2.2622 or greater than 2.2622 (determined using statistical software or a t-table):

Since the quality control specialist's test statistic, t* = 1.54, is not less than -2.2622 nor greater than 2.2622, the qualtiy control specialist fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the α = 0.05 level, to conclude that the mean thickness of all of the manufacturer's spearmint gum differs from 7.5 one-hundredths of an inch.

If the quality control specialist used the P-value approach to conduct his hypothesis test, he would determine the area under a tn - 1 = t9 curve, to the right of 1.54 and to the left of -1.54:

In the output above, Minitab reports that the P-value is 0.158. Since the P-value, 0.158, is greater than α = 0.05, the quality control specialist fails to reject the null hypothesis. There is insufficient evidence, at the α = 0.05 level, to conclude that the mean thickness of all pieces of spearmint gum differs from 7.5 one-hundredths of an inch.

Note that the quality control specialist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

### In closing

In our review of hypothesis tests, we have focused on just one particular hypothesis test, namely that concerning the population mean $$\mu$$. The important thing to recognize is that the topics discussed here — the general idea of hypothesis tests, errors in hypothesis testing, the critical value approach, and the P-value approach — generally extend to all of the hypothesis tests you will encounter.

## FORMULATION OF ACTION HYPOTHESIS

This section helps you in understanding how to formulate an action hypothesis.

Objectives

 After reading this material and performing the activities listed, you will be able to acquire knowledge about hypothesis and Action Hypothesis. write action hypothesis in three forms. select an appropriate form of writing an action hypothesis.

## Overview

 In the last chapter we discussed how a problem could be selected for Action Research. We also saw how we can go about identifying the specific problem, At this stage we need to think and generate a list of alternative causes for the pin-pointed problem. From this list we can choose a cause which we think is the most likely one and start working on it. What we undertake is an intellectual exercise of considering all the causes for the problem and deciding which cause needs to be tackled in solving the problem. What we initially have is a launch. Later we form a hypothesis. Formulating a hypothesis gives precision to our work and helps us to be objective. In this chapter we will know more about the techniques of formulating an Action Hypothesis.

## WHAT IS A HYPOTHESIS?

You might be wondering what an action hypothesis is?

The processes, an investigator may use to examine a problem in the field of education are similar to the ones we use to attack our day to day problems.

Look at the following example.

A teacher notices that one of her Students in the IV grade does not show progress in learning “addition of two digit numbers”. Careful observation of this child in the classroom may suggest several possible causes for this problem. This in turn will help the teacher think of suitable remedies.

Based on these possible causes the teacher states HYPOTHESES which are the guessed strategies for solving the problem. Then the teacher designs and carries out a programme aimed at testing each hypothesis and checking the child’s progress.

Without ‘guessing’ the possible causes the teacher can not plan any remedy for the problem.

Definition

 A Hypothesis is a hunch or a shrewd guess or a tentative solution or an inference or sub-position to be tested by empirical evidences.

Once the investigator diagnoses the causes of the pinpointed/specific problems, he/she starts thinking about what concrete action, if taken, would bring about the desired change/solution.

Then he/she formulates hypothesis specifying the immediate ‘actions’ that could be taken to solve the problems.

The hypotheses formulated in action research are called ACTION HYPOTHESES

## CHARACTERISTICS OF A GOOD ACTION HYPOTHESIS

A good action hypothesis should be

1. Logically related to the problem
2. Testable in classrooms situations
3. Clearly stated without ambiguity
4. Directly stated in terms of the expected outcome (should not be a generalized statement)
5. Testable within a considerably short time (maximum of three months)

## DIFFERENT FORMS FOR STATING ACTION HYPOTHESIS

a) Declarative form: An action hypothesis may be formulated as a statement with a positive relationship between the two factors identified, one being the cause and the other being the effect. This is also called a directional hypothesis.

b) Predictive form: An action hypothesis clearly predicting the expected out come which would emerge after the action plan is implemented. This can be stated using ‘if and then’ statement.

c) Question form: Questions can be raised as action hypotheses as what would be the result of the intended action plan.

d) Null form: A null hypothesis states that no relationship exists between the factors considered in the problems. This form is mostly used when rigorous statistical techniques are to be used.(A thoroughly worked out example for all these forms is given in the next unit.) Thus, an action hypothesis provides clarity and direction to solve a problem. Hence it is considered an important stage in action research.

## FORMULATION OF AN ACTION HYPOTHESIS

To form a hypothesis the investigator should

1. Have a thorough knowledge about the problem
2. Be clear about the desired goal (solution)
3. Make a real effort to look at the problem in new ways other than the regular practices (come out form conventional thinking)
4. Give importance for imagination and speculation
5. Think of many alternative solutions.
6. Thoroughly examine the conditions/contexts in which the problem exists and then
7. State the hypothesis

## Illustration of an action hypothesis in four different forms

Activity Sheet on Formulation of action hypothesis

## Reflection

 In our day to day activities we are often faced with problems. We undertake a number of activities to solve them. First we try to identity possible reasons for the problem. Then we think of possible intervention strategies that would solve the problem. We try to find a solution to the problem through logical reasoning. These intelligent and logical “guesses” about possible differences, relationships, causes and solutions are called HYPOTHESES.

Activity

 Give an example for each type of hypothesisDeclarative formPredictive formQuestion formNull form

Activity

 Eight students in class IV are not able to identify directions on a map. You have realized that inadequate exposure to map reading is the cause for the problem. Now write a hypothesis for finding a solution to this problem in all the four forms. Your course of remedial action should be reflected in the hypothesis.

## Summary

 Hypotheses are intelligent and logical guesses about possible differences, relationships, causes and solutions. Hypotheses could be tested by empirical evidence. An action hypothesis is one that is formed in an action research. A hypothesis could be in declarative forms, predictive form, question form or null form. An action hypothesis is formed after listing all possible causes and choosing the most likely cause among them.

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