# How safe is that to ignore homogeneity of variances test and continue with Post hoc test?

I want to know if average time people spending on their favourite social media are statistically different. I consider here five groups, and the total number of participant is 700 persons.

I have already performed the Shapiro-Wilk normality test, on each group (Facebook, Twitter ...) to see if the time people spend on social media follows a normal distribution, according to the test, the null hypothesis fails so the time isn't normally distributed. Though, I decided to consider the time as a normal distributed parameter, because the average time is approximately normally distributed as $$n=700$$ is large enough.

Next, I carry out the 'Test of homogeneity of Variances', according to the the results the equality of variances is ruled out.

Meanwhile, I looked at the 'Robust test of equality of means' or 'Welch and Brown-Forsythe', both tests results in p-value $$0,000$$.

Now, I want to know if I can trust the results of Post hoc? Or I frequently violated the rules? If so how I can eventually compare means?

I have a very few experience in data analyses, so I apologise here for carelessness of my argument.

• Do you have 700 subjects altogether? Or 700 in each group (3500 altogether)? If the former, are the numbers in the various groups all about equal $(\approx 140$ each), or do group sizes vary greatly? OR is this a 'block design' in which all 700 spend time on all 5 media? May 1, 2019 at 18:12
• @BruceET I should have added those info, anyway edited my question, the assumption is every individual spends time only in one of the platforms. May 1, 2019 at 19:11
• So it seems my Answer is applicable. The smallest group still has 64 subjects, so Shapiro-Wilk should alert you to aggressive non-normality. Variances are somewhat, but not greatly, different. All should be OK. May 1, 2019 at 19:17
• So it seems my Answer is applicable. The smallest group still has 64 subjects, so Shapiro-Wilk should alert you to aggressive non-normality. (But I'd look at boxplots of all 5 groups to see if there are any far outliers.) // Variances are somewhat, but not greatly, different. All should be OK. May 1, 2019 at 19:27
• Jul 5, 2021 at 19:14

As you can see from the question in my Comment, I'm just guessing that you have 700 people altogether, with approximately $$n_i = 140$$ in each (social media) group.

If that is true, Shapiro-Wilk normality tests on individual groups of over 100 each should pick up the degree of non-normality that would invalidate a Welch (separate-variances) ANOVA.

You say you have found significant differences among the five groups with overall a tiny overall P-value.

In that case it should be fruitful to look at ad hoc paired comparisons using two-sample Welch t tests. You could mitigate 'false discovery' by using Bonferroni criteria.

For example, if you do all $${5 \choose 2} = 10$$ comparisons, you could look for comparisons significant at the $$0.5\%$$ level. (However, you may find that you don't need to do all possible comparisons in order to get a good idea what the pattern of significant differences might be.)

Note: If you are using R, perhaps see this link for the functions to use, and some ideas on methods (including, but not limited to, 'Bonferroni') avoiding false discovery in making ad hoc comparisons.

• Thank you, i use SPSS, i'm trying to avoid the ten comparison test to save time and effort, but if i have to then i don't know which test should i carry out in case of not equal variances. May 1, 2019 at 19:14
• One strategy: Rank the sample means from largest to smallest. Start by looking for gaps where differences seem large enough to be of practical importance and check those pairs to see if statistically signif. // Don't know about SPSS, but some software will just do them all at once and give you the results in a tidy form. (Look at link.) May 1, 2019 at 19:24
• For future readers, the link inside the answer contains great stuff(SPSS , R ...), again thank you. May 1, 2019 at 19:29

Look at

           mean st.dev        cv
Snapchat   93.6   94.6 1.0106838
Instagram  78.1   78.0 0.9987196