3

enter image description here

For this question shouldn't they be using a t test and the test statistic should be t and not z as the sample is small? Is this a mistake in the mark scheme?

user134785
  • 1,117

2 Answers2

4

Deciding between a z-test and a t-test has to do with whether or not the population standard deviation is known. In this case, you are told that the standard deviation is .8. When performing a t-test, the population standard deviation is unknown and thus must be estimated with the data via the sample standard deviation.

jsk
  • 543
-2

Since it is given that the distribution is normal and the standard deviation is $0.8$ mg, then the $z$-test is correct. If the location were unknown and were being inferred from the data, a $t$-test would be correct.

A $t$-test incorporates the simultaneous uncertainty in both the mean and standard deviation of a distribution. A fluctuation in the mean can be caused by a sample with the correct mean, but an unlucky set of deviations. Thus, the fluctiation in the sample conflates both the location and variance of the distribution. Since we absolutely know the mean, there is no possibility of conflation.

Edited for cross-wiring between fingers and head: "mean" -> "s.d.".

Eric Towers
  • 67,037
  • Eric, choosing between a z-test has nothing to do with whether or not the mean is known. It has to do with whether or not the population standard deviation is known! – jsk May 03 '14 at 05:01
  • @jsk: No. It has to do with whether the sample deviations are obscuring both the mean and standard deviation. In the problem, this is not the case. – Eric Towers May 03 '14 at 08:44
  • Anyone who has taken an introductory course to statistics should know the difference between a one-sample z-test and a one-sample t-test. https://www.khanacademy.org/math/probability/statistics-inferential/hypothesis-testing/v/z-statistics-vs--t-statistics – jsk May 03 '14 at 09:14
  • Yes. Gossett developed the t-test to handle data with the following properties: normally distributed, unknown mean, unknown deviation. The difference between the t-distribution and the normal distribution is that sample fluctuations, scaled by the unknown standard deviation, simultaneously distort the estimated mean and standard deviation. As a result, the t-distribution has a flatter peak than the normal distribution. Analytically, the t distribution is found by marginalizing the normal distribution versus uncertainty in the location parameter induced by individual samples. – Eric Towers May 03 '14 at 09:19
  • If you know the population standard deviation, then you don't need to use the t-test! – jsk May 03 '14 at 09:20
  • In short, computing standard error in the mean with $Z/s$ instead of $Z/\sigma$ understates the uncertainty because deviation of the sample mean from the population mean is correlated with deviation of sample variance from the population variance. – Eric Towers May 03 '14 at 09:25
  • It's becoming clear that many commentors have a "recipe" for when to use which test that is in some way divorced from the underlying theory of these test. – Eric Towers May 03 '14 at 09:26
  • Actually, the beauty of the derivation of the t-statistic is that the sample mean is independent of the sample standard deviation. – jsk May 03 '14 at 09:27
  • What's wrong with assuming the population sd $\sigma$ is known, then looking for evidence against the value of $\mu$ specified under the null hypothesis? – jsk May 03 '14 at 09:31
  • There is nothing seemingly wrong with the underlying theory of the z-test. Mathematically, it's perfectly fine if you are willing to make the bold assumption that $\sigma$ is a known constant. In reality, you are never going to know $\sigma$ anyway so it's a moot point. Personally, I don't think the one sample z-test should be taught at all, though usually it is only briefly mentioned for pedagogical reasons. – jsk May 03 '14 at 09:41
  • Your answer makes absolutely no sense "Since it is given that the distribution is normal and the mean is 0.8 mg, then the z-test is correct. If the location were unknown and were being inferred from the data, a t-test would be correct." The least of the problems with your answer is that the SD is known to be .8, not the mean. Read the problem again. And they're looking for evidence that the population mean is less than 1.64 – jsk May 03 '14 at 09:46
  • The mean may be independent, but the standard error in the mean (i.e. deviation of sample mean from populatoin mean) is not independent of the sample standard deviation. Use of the population s.d. to estimate the distribution of sample means is incorrect. I.e., use of the given s.d. to estimate likelihood of observed mean different from target mean is incorrect. – Eric Towers May 03 '14 at 21:17
  • Incorrect? Have you ever heard of a sampling distribution? Under the null hypothesis, $\bar{X} \sim N(\mu_0, \sigma^2/n)$, which means that you don't need to estimate the standard deviation of the sampling distribution of the sample mean, it's known since you know the population sd $\sigma$. Thus there is nothing wrong creating the test statistic $z = (\bar{X}-\mu_0)/(\sigma/\sqrt{n}) $ and $z\sim N(0,1)$ under the null hypothesis. There is no need to estimate the standard error. This test statistic looks for evidence again the mean defined in the null hypothesis. – jsk May 03 '14 at 21:39
  • If, miraculously the sample standard deviation matched the population standard deviation, the deviation of the sample mean from the population mean would follwot he distribution you cite. However, the distribution of deviation of sample means is not independent of the sample standard deviations. (Do the experiment: under repeated see that the deviation of sample means is correlated with sample standard deviations (with signs matching the sign of the skewness).) Since a sample has a standard deviation, marginalizing the sample distribution over this parameter is wrong. – Eric Towers May 04 '14 at 03:47
  • It is truly a shame that more members have not weighed in. – jsk May 04 '14 at 03:50
  • Do you agree that the distribution of the random variable $\bar{X}$ under the null hypothesis is $N(\mu_0, \sigma^2/n)$? Do you know what happens to a normal random variable if you standardize it by it's mean and standard deviation? – jsk May 04 '14 at 04:22
  • I have no idea if you will bother to look at this, but I hope you will because there is a nice derivation of the t-statistic on wikipedia. http://en.wikipedia.org/wiki/Student%27s_t-distribution#Derivation I sincerely hope that this will clear up some of your confusion. – jsk May 04 '14 at 04:50