I have a case study where a company made new golf balls with stronger coatings than current golf balls. One of the technicians is concerned that the new coating may decrease the driving distance. An experiment was conducted in which 40 types of each ball was hit by a machine to determine mean driving distance of the two balls. The distances for each ball were given in a spreadsheet. Here are the samples:

Current: 264.00 261.00 267.00 272.00 258.00 283.00 258.00 266.00 259.00 270.00 263.00 
         264.00 284.00 263.00 260.00 283.00 255.00 272.00 266.00 268.00 270.00 287.00 
         289.00 280.00 272.00 275.00 265.00 260.00 278.00 275.00 281.00 274.00 273.00 
         263.00 275.00 267.00 279.00 274.00 276.00 262.00
New;     277.00 269.00 263.00 266.00 262.00 251.00 262.00 289.00 286.00 264.00 274.00 
         266.00 262.00 271.00 260.00 281.00 250.00 263.00 278.00 264.00 272.00 259.00 
         264.00 280.00 274.00 281.00 276.00 269.00 268.00 262.00 283.00 250.00 253.00 
         260.00 270.00 263.00 261.00 255.00 263.00 279.00

These are the questions:

  1. Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls
  2. The CEO tells you (in contrast to what Bill said) that he would like to sell the new ball unless there is overwhelming evidence that the cut‐resistant ball is slower than the old ball so you should frame your hypothesis accordingly. Hint: you are more willing to make a Type II error. Analyze the data to provide the hypothesis testing conclusion and statements (at .05). Also include narrative versions of what ever of the above forms you use such as: Ho: The mean is ... What is the p‐value for your test? What is your recommendation for Par, Inc.?
  3. Provide the following descriptive summaries of the data for each model: ONLY the means and variances and the z values.
  4. What is the 95% confidence interval for the difference between the means of the two populations?
  5. Do you see a need for larger sample sizes and more testing with the golf balls? Discuss.

I've looked for help elsewhere and everyone seems to use the t-test to conduct the analysis. Shouldn't the z-test be used since the sample sizes are both over 30? I would also like some direction in understanding how to answer the other questions but my main question is the most important.

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    $\begingroup$ Is this a homework problem? If so, please add the homework tag. This is discussed in the faq, right at the top, which you may want to read. $\endgroup$ May 2, 2012 at 23:24
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    $\begingroup$ Please do not deface your question upon receiving a suitable answer. This site is intended to serve as a persistent store of knowledge regarding topics within its stated scope. $\endgroup$
    – cardinal
    May 3, 2012 at 20:07

2 Answers 2


If we can assume the driving distances are normally distributed with possibly different means but a common variance then the t test is the exact best test. A sample size of 30 just means that the t distribution is getting close to the normal so the z test approximates the t test. But that does not mean it should be preferred. Sometimes it is used on the grounds that it is simpler. If the two variances differ you have the famous Behrens-Fisher problem and the normalized distribution of the mean difference is neither normal nor t. However there is an approximate t distribution with non-integer degrees of freedom that approximates the null distribution of the test statistic. If the two population distributions are very non-normal and the sample size is not large a test of the null hypothesis that the two distributions are the same versus that they come from different distributions with different medians then a nonparametric test is appropriate. The Wilcoxon rank sum test would be preferred.

If the data are paired there is a paired t test in the first situation and a signed rank test in the last case.

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    $\begingroup$ This is a common pitfall among data analysis for lab data: normality is not a requirement for the T-Test to be justified. In fact, choosing a test based on the distribution of the error terms leads to a p value which doesn't mean what you think it means, so the test as a whole is very hard to interpret. The t-test is in fact robust to departures from normality and is uniformly most powerful among unbiased tests such as a rank sum test. $\endgroup$
    – AdamO
    May 2, 2012 at 23:04
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    $\begingroup$ @AdamO The t-test is UMP assuming normality. That claim no longer is valid (in general) if there is any departure from normality in the underlying distribution. One frequently cited reason for preferring the Wilcoxon test is that it cedes only a little power to the t-test for normal distributions yet, for many other distributions, is more powerful than the t-test. $\endgroup$
    – whuber
    May 3, 2012 at 15:11

With respect to your main question, there is no 'bright line' between small $N$ and large $N$, not at 30, 50 or 100 (where most tables stop before jumping to $\infty$). If you are estimating the SD from your data, the t-test is appropriate. After a certain point, the results you get will be indistinguishable from a z-test, but from a theoretical perspective, the t-test remains appropriate.


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