# Buying the best hard drive, or: Confidence interval for difference between means, variance of ordinal data, etc.

I think my hard drive is on the fritz, and so I've been sorting through my options on Amazon. However, the ratings on Amazon don't really make distinctions as nicely as I'd like; the only readily available piece of information is a mean rounded to one decimal place. This has led me to try to glean more from the numbers available, namely variance.

My first question is the last one that occurred to me: Is it even appropriate to calculate the variance of ordinal data? After all, it's not like the difference between a four-star rating and five-star rating can be measured in any meaningful way. Then again, without running the kind of extensive study on some large enough $n$ of hard drives — the kind of study that Samsung and Crucial and Kingston probably have done, examining a specific variable, but would never release the results of — perhaps the best we can do is a collection of consumers' self-reported measurements of their own perceptions.

Less, uh, philosophically — and assuming that the answer to the previous question is yes — is this the kind of data set for which an expression like

$$\left(\bar{x_1}-\bar{x_2} \right) \pm z_{\alpha / 2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$$

even makes sense? My textbook says that $X_1$ and $X_2$ are normally distributed. Even if the actual quality of each hard drive is normally distributed, the ratings sure aren't. I think I'm making an assumption here, that even though I'm using all of the ratings, they really constitute only a sample of the opinions on a given piece of hardware. Therefore, I could invoke the central limit theorem and say that it doesn't matter how the ratings are distributed: the distribution of its sample means is normal. Would this reasoning be correct?

(By the way, for the comparison of most interest to me, $n_1 = 1151$ and $n_2 = 547$; hence the $z$-test.)

## 1 Answer

You are surely making an assumption which is the core of deriving the confidence interval. Namely, that the ratings are normally distributed. I do not think it is erroneous to assume so. If, say, ratings are values between 0 and 10 and the "service" is bad, it is quite plausible to assume that the ratings will be normally distributed, but probably with a low mean and if in addition the service is really bad, probably with a little variance too. So I do not see anything too wrong by assuming normality, in any case, it should not be difficult to carry out a normality test, specially with so big data set. You also need Independence which you can easily assume, and also that the means and variances are Equal and unknown, which you then estimate and since the sample is quite big one can still use the normal distribution to carry out inference.

In any case, as you very well mentioned, the data is big enough to assume normality by virtue of the Central Limit Theorem.