In a questionnaire, I have asked some Likert scale questions. These questions were asked to the same population 6 years ago in another study. I only have the mean scores and standard deviation of the old study. Is there a way to compare the answers of my study to that of the old one statistically?

I have limited statistical knowledge and am using SPSS.

Below are the means (standard deviation) of each. Scale is from 0-4

The first study has n=170 vs the second study having n=184

Q1 2.11 (1.352) vs 1.83 (1.112)

If you could possibly help me do one and I could work out the rest. Thanks!

  • $\begingroup$ Do you have the sample size? $\endgroup$
    – Glen_b
    Commented Mar 7, 2014 at 9:51
  • $\begingroup$ question updated above @LadislavNado $\endgroup$ Commented Mar 7, 2014 at 10:58
  • $\begingroup$ Hi @dissertationhelp. This will be useful to you: stats.stackexchange.com/questions/89591/… Look on answer. $\endgroup$ Commented Mar 11, 2014 at 12:47
  • $\begingroup$ @LadislavNado but the t tests assumes the data is normally distributed. Mine is not. $\endgroup$ Commented Mar 12, 2014 at 15:04

1 Answer 1


If you only have means, sd's and n's, you have no choice ... you have to make some parametric assumption.

Fortunately, your sample sizes are large and your parent distributions are bounded. Your sd's are very similar. In this case the size of the sd's work to your advantage since it means the samples are unlikely to be jammed up against either boundary too strongly (it's hard for it to be very skew or for the kurtosis to be too odd). So you're in a good place to invoke the central limit theorem for the numerator of the t-statistic. Further, with nice large samples, you can make an argument (such as via Slutsky's theorem) that the difference in means divided by the pooled standard deviation will be approximately normal, and thus you can apply a z-test even if you know the data are non-normal.

(If you need more argument to support that, a simulation study of samples with similar characteristics to the ones you have would serve to settle arguments about the properties of the approach - both in terms of the significance level and power characteristics - but I doubt it would be necessary). You could even do a t-test (which SPSS will no doubt be able to perform from the summary statistics) under the argument that the t-test will closely approximate the z-test I just made the argument for.

In any case, you can just do the whole thing by hand.

You should start with your hypotheses, assumptions, and significance level:

$H_0: \mu_1 = \mu_2$

$H_1: \mu_1 \neq \mu_2$

Assumptions: the sample sizes are large enough (see above argument) to invoke the central limit theorem for the numerator and large enough that the estimate of the standard deviation of the difference in means is close enough that the distribution of the ratio of the two has a standard normal distribution (again, see the above argument). In addition, the usual additional assumptions of independence are required.

$\alpha=\text{??}$ ... you tell me! I'm not picking your type I error rate.

One can either base a statistic off the equal-variance form of the t-statistic or the unequal variance case (Welch-Satterthwaite form). Either way the argument for using a Z-test still goes through. Here's the second form:

$s_d=\text{sd}_{\bar X_1 - \bar X_2} = \sqrt{s_1^2/n_1+s_2^2/n_2}$

And then the test statistic is just $Z = \frac{\bar x_1 -\bar x_2}{s_d}$

With the Z-value there computed on the data you gave:

$\bar x_1=2.11, s_1=1.352, n_1=170$

$\bar x_2=1.83, s_2=1.112, n_2=184$

and with the result looked up in Z-tables to obtain a p-value.

The question Ladislav pointed to has R-code if you're prepared to use that.

For this data I got $\bar x_1 -\bar x_2=0.28$, $s_d=0.1322$, $z=2.12$ and a two-tailed p-value of $0.034$.


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