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I need to analyze this hypothesis h0:Men use cell phone for calls more than woman do. Made a quesstionaire and asked on the likert scale

How often do you use your cell phone for calls

1=hourly 2=at least once per day 3=at least once every few days 4=once a week 5=never used

Can i analyse the data using one way ANOVA by making phone call dependent variable and gender:male/female as independent variable. What other method can i use.Thanks.

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  • $\begingroup$ 1. Are you sure that hypothesis isn't the alternative (H1)? It doesn't contain an equality. 2. Either way the questionnaire does not deal with the stated hypothesis. One hypothesis that would correspond to the questionnaire would (in words) be "Men report using cell phones for calls more than ... " (etc). To test the hypothesis you mentioned would require measuring whether men actually do that, not that they say they do. $\endgroup$
    – Glen_b
    Commented Aug 10, 2016 at 11:56
  • $\begingroup$ Thanks for the reply.I want to know who uses cell phones for call more men or women.So ill make it a research question rather than hypothesis.Can you ans my question regarding the analysis that can i use anova for it even though i have likert scale type data $\endgroup$
    – addikted
    Commented Aug 10, 2016 at 12:03
  • $\begingroup$ Anova is not a good choice. How many participants in total, and what is the range of counts per response for each group ? $\endgroup$
    – Robert Long
    Commented Aug 10, 2016 at 12:21
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    $\begingroup$ I don't think you can call this a Likert scale, as this is just one item. In fact it is simply a dependent categorical variable with 5 values, with an independent variable that is dichotomous. You would need a categorical-with-categorical association measure, strictly speaking. In social sciences, it would be a widespread (though incorrect, statistically speaking) practice to treat your dependent as continuous. Then you could use ANOVA, but I'd recommend against that, as more appropriate methods exist. Odds ratios could be used as well. A Chi-square test of independence too, I imagine. $\endgroup$
    – Maxim.K
    Commented Aug 10, 2016 at 12:27
  • $\begingroup$ 400 participants 380 male 80 female $\endgroup$
    – addikted
    Commented Aug 10, 2016 at 18:16

1 Answer 1

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First off, as others pointed out, $H_0$ should be that there is no difference.

Next you don't need ANOVA as you have only two groups, which would be a t test. However you cannot do a t-test because the scale is categorical.

One possibility is to convert your answers from 1,2,3,4,5 to "times per week" so then 1 = 70 (or something) 2 = 7 3 = 2 4 = 1 5 = 0

then you could do a t-test, but I don't recommend this here as your categories are pretty poorly chosen.

I'd say you want some ordinal test that does not assume equal spacing. One such is the Jonckheere-Terpstra test. In SAS, it is available in PROC FREQ with the JT option. In R it is in the clinfun package.

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