Example of Hypothesis Testing

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Hello Researchers, In this section we will discuss with an example of Hypothesis testing. In Research Methodology Hypothesis Testing carry important roll for Researchers. It helps to take decision, which data will be taken and which not.

A sample of 225 customers was obtained and asked to indicate their perceptions of service on a five-point scale, where 1 indicates “very unfriendly” service and 5 indicates “very friendly” service. The scale is assumed to be an interval scale, and experience has shown that the previous distribution of this attitudinal measurement assessing the service dimension was approximately normal. Now, suppose Pizza-In believes the service has to be different from 3.0 before a decision about expansion can be made. In conventional statistical terminology, the null hypothesis for this test is that the mean is equal to 3.0


H0: = 3.0

The alternative hypothesis is that the mean does not equal 3.0:

H1: ≠3.0

More practically, the researcher is likely to write the substantive hypothesis (as it would be stated in a research report or proposal) something like this:

H1: Customer perceptions of friendly service are significantly greater than three.

Note that the substantive hypothesis matches the “alternative” phrasing. In practical terms, researchers do not state null and alternative hypotheses. Only the substantive hypothesis implying what is expected to be observed in the sample is formally stated.

Next, the researcher must decide on a significance level. This level corresponds to a region of rejection on a normal sampling distribution. The peak of the distribution is the theoretical expected value for the population mean. In this case it would be three. If the acceptable significance level is 0.05, then the 0.025 on either side of the mean that is furthest away from the mean forms the rejection zone (shaded orange in Exhibit 21.1). The values within the unshaded area are called acceptable at the 95 percent confidence level (or 5 percent significance level, or 0.05 alpha ‘α’ level), and if we find that our sample mean lies within this region we conclude that the means are not different from the expected value, 3 in this case. More precisely, we fail to reject the null hypothesis. In other words, the range of acceptance (1) identifies those acceptable values that reflect a difference from the hypothesized mean in the null hypothesis and (2) shows the range within which any difference is so minuscule that we would conclude that this difference was due to random sampling error rather than to a false null hypothesis. H1 would not be supported.

In our example, the Pizza-In restaurant hired research consultants who collected a sample of 225 interviews.

The mean friendliness score on a five-point scale equaled 3.78.

(If σ is known, it is used in the analysis; however, this is rarely true and was not true in this case.)

The sample standard deviation was S=1.5.

Now we have enough information to test the hypothesis. The researcher has decided that the acceptable significance level will be set at 0.05.

This means that the researcher wishes to draw conclusions that will be erroneous 5 times in 100 (0.05) or fewer.

From the table of the standardized normal distribution, the researcher finds that the Z score of 1.96 represents a probability of 0.025 that a sample mean will be above 1.96 standard errors from μ. Likewise, the table shows that 0.025 of all sample means will fall below -1.96 standard errors from μ. Adding these two “tails” α =0.05 ¸ 2(0.025+0.025) together, we get 0.05. The values that lie exactly on the boundary of the region of rejection are called critical values of μ. Theoretically, the critical values are Z =-1.96 and +1.96. Now we must transform these critical Z-values to the sampling distribution of the mean for this image study. The critical values are

n= 225

μ = 3

α =0.05 ¸ 2=(0.025+0.025)

Standard deviation (s)=1.5

Critical value-upper limit = μ + ZSx¯ or μ + Z(s÷√n)

= 3.0 + 1.96 (1.5÷√225)

= 3.0 + 1.96 (0.1)

= 3.0+0.196


Critical value-lower limit = μ – ZSx‾ or μ – Z(s÷√n)

= 3.0 – 1.96 ( 1.5 ÷√225)

= 3.0 – 1.96 (0.1)

= 3.0-0.196


Based on survey results, the sample mean (X¯ ) is 3.78. The sample mean is contained in the region of rejection (see the dark shaded areas of Exhibit 21.3). Since the sample mean is greater than the critical value of 3.196, falling in one of the tails (regions of rejection), the researcher concludes that the sample result is statistically significant beyond the 0.05 level. A region of rejection means that the thought that the observed sample mean equals the predetermined value of 3.0 will be rejected when the computed value takes a value within the range. Here is another way to express this result: If we took 100 samples from this population and the mean were actually 3.0, fewer than five will show results that deviate this much.

By the above systematic way we could test Hypothesis. There may be other way to test. This article is based on Book and internet article. 

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