I was attempting a self-study question on Margin of Error and managed to get the answer right. The answer key states 2.77, but I managed to get at 2.76. Close enough I guess.

But I thought that it seems pretty weird that I did not have to use many of the provided variables such as mean or even the confidence interval.

Although I managed to get the right answer, I am not sure if I did really answer the question correctly. Appreciate some guidance please.


A portable DVD manufacturer based in Shanghai provides battery perfor- mance information in order to inform customers when making a purchasing decision. One of the important measures of battery performance is the length time average batteries last compared with a standard battery with a base of 100. This mean that a set of batteries with a grade of 200 should last consumers twice as long, on average, as a set of batteries graded with a base of 100. A Chinese manufacturer’s organization wants to estimate the actual battery life of brand-name batteries that claim to be graded 200. A random sample of n = 400 indicates a sample mean battery life of 195.3 and a standard deviation of 21.4. A 99% confidence interval for the population mean battery life for batteries produced by this manufacturer under this brand name is constructed. Find the margin of error in this estimation.

My Attempt

enter image description here

  • $\begingroup$ According to Wikipedia, the margin of error for a particular statistic of interest is usually defined as the radius (or half the width) of the confidence interval for that statistic. Thus, the above calculation is correct. Probably, the mean is also reported for completeness. $\endgroup$ – Epaminondas May 4 '14 at 14:08

Actually the confidence interval is computed using a critical value from the t distribution, not the z distribution. In this example with so many degrees of freedom, the two are almost the same, but the difference is enough to explain the discrepancy between your result and the posted correct result.

The critical value of t for alpha=0.01 and 399 degrees of freedom (400-1) is 2.588207164 as computed by the Excel formula "=T.INV.2T(0.01,399)".

Use that value instead of the value from the z distribution (2.5758) and the result is 2.76938, which can be rounded to 2.77.

In this example it doesn't really matter if you use t or z. But with smaller sample sizes it would matter substantially.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.