The word “hypothesis” is rarely found in everyday conversation. However, we often hear the words “idea” or “(educated) guess”. An idea is not necessarily correct. To find out whether the idea is right, we conduct an investigation. The same is true of hypotheses. What is hypothesized is not necessarily correct. In statistics, the formulation of a hypothesis needs to be followed up with sampling. If the sampling results prove the hypothesis is incorrect, then the hypothesis is rejected.
The following are case examples that illustrate how hypothesis testing is applied in communication science.
Case 1
A communication researcher believes that increased time spent on social media leads to a decrease in face-to-face communication skills among young adults. They hypothesize that the mean/average score of face-to-face communication skills (measured on a standardized scale) for young adults who spend more than 3 hours daily on social media is lower than the national average score of 75. Here is a sample, taken to test the hypothesis, in .sav format: (click here) At the 0.05 level of significance, does the sample support the hypothesis that the mean score is lower than 75?
The following describes the steps required to answer the questions in Case 1.
Step 1: Formulate the hypotheses
H0: μ ≥ 75
H1: μ < 75
Explanation of Step 1
In statistical hypothesis testing, hypotheses are formulated in pairs, namely the null hypothesis (H0) and the alternative hypothesis (H1). The statement in H1 is the statement that will be used as a conclusion if H0 is then rejected. The sign of relation used in the alternative hypothesis must be one of <, >, or ≠. In this case, what will be proven is “the average score of face-to-face communication skills of young adults who spend more than 3 hours per day on social media is lower than the national average score (which is 75)”. Therefore, the statement in H1 is μ < 75. The relation used in H0 is the negation of the relation used in H1. Thus, in this case, the statement in H0 is μ ≥ 75.
Step 2: Determine the significance level (α)
In this case, the significance level has been determined, which is 0.05. In real research practices, the significance level is determined by the researcher. The significance level is the probability of an error occurring if the null hypothesis is rejected. Here, if the null hypothesis is rejected, the conclusion will be: “the average face-to-face communication skills score of young adults who spend more than 3 hours per day on social media is lower than the national average score,” and our probability of incorrectly concluding this is only 5%.
Step 3: Determine the test statistic and rejection region
The test statistic used here is 
with degrees of freedom 
. This statistic is used when the population is normally distributed. In theory, there are two other statistics, namely 
and 
. However, the statistic is rarely used in practice because in practice, 
(population standard deviation) is rarely known. (The data in this case has been generated in such a way that it can be assumed that the population is normally distributed, so the t statistic can be used.)
To determine the rejection region, first find the critical t value. In any cell of the Excel sheet, type: =T.INV(0.05;99). See Figure 1.
Figure 1
After the Enter key is pressed, Excel will show the value -1.660 (if rounded to 3 decimal places). Furthermore, the rejection area can be illustrated as follows.
Figure 2
In Figure 2, the rejection area is indicated by the light blue shaded area. This region is located on the left side. That is why this test is called the left-tailed test. In this test, the left tail is used because, in the alternative hypothesis (H1), the relation sign “<” is used. The objective of determining this rejection region is to decide whether there is sufficient evidence to reject H0. If the computed t value (obtained from the sample) has a value of less than -1.660 (falls in the rejection region), then H0 is rejected.
Step 4: Calculating the t-value of the sample
The computed t (i.e., the t value from the samples) is obtained using the formula 
. The following table shows the SPSS output related to the required calculation. (To find out how to display the following SPSS output, click here.)
SPSS OUTPUT TABLE
From the table above, we get the computed t value (tcomputed), i.e., -4.208, and this value is smaller than the critical value, i.e., -1.660. See Figure 3.
Step 5: Drawing conclusions
As Figure 3 shows, the computed t value in this case, -4.208, falls in the rejection area.
Figure 3
Therefore, we reject the null hypothesis and then accept its alternative hypothesis.
CONCLUSION:
The mean score of face-to-face communication skills for young adults who spend more than 3 hours per day on social media is significantly lower than the national average. The researcher has successfully supported their opinion that increasing time spent on social media leads to a decline in face-to-face communication skills among young adults.
Note
The results of the hypothesis test may be significant or insignificant. If we succeed in rejecting H0, then the test is said to be significant, and the probability of incorrectly rejecting it is α. Conversely, if we fail to reject H0, then the test is said to be insignificant. In the case where H0 is not rejected, the probability of incorrectly accepting it cannot be determined. In Case 1, we successfully reject the null hypothesis, so in the concluding sentence, there is the word “significantly”. The probability that we are incorrect in making this conclusion is the same value as the significance level. Therefore, the probability that the researcher is incorrect in concluding it is 5%.
ANOTHER WAY TO DRAW CONCLUSIONS
In addition to using critical values, there is another way to draw conclusions, namely by using p-values. This method is sometimes preferred because it does not require us to determine the rejection/critical region in Step 3. With this method, the criterion of rejecting H0 is as follows: Reject H0 if p < α. Otherwise, do not reject H0.
How do we use the p-value in Case 1? Look at the SPSS OUTPUT TABLE section entitled One-Sample Test above. In the column entitled Sig. (2-tailed), there is a number 5.662E-005, which is equal to 0.00005662. It is the p-value for a 2-tailed test. Because in Case 1 we perform a one-tailed test, the value must be divided by two beforehand, thus, in this case, the p-value is 0.00002831. Since this value (0.00002831) is less than α, we reject H0.
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