The question is in the image above.

Part b is what I'm having trouble understanding the answer to.

My working: (a) unbiased estimate for mean $\mu$ is $\bar X = 4.06$

Unbiased estimate for variance $S^2 = 0.1542^2 =0.02379.$

(b) $X_i \sim N(4.02,\sigma^2).$ $\bar X \sim N(4.02,\sigma^2/12),$ where $\sigma^2/12$ is standard error.

I thought that because $n$ is small, we cannot use $S$ as a good estimate of $\sigma.$ (We could use $S \approx \sigma,$ if $n$ is large). Therefore we must use the t-distribution:

Test statistic, $T=(\bar X - \mu)/(S/\sqrt{n}) = (4.06 - 4.02)/(0.1542/\sqrt{12}) = 0.899.$

For $t_{crit},$ 2-tailed test @1% significance, $p = 99.5,$ $\nu=n-1=11,$ $t_{crit}= \pm 3.106.$

As $T < t_{crit},$ accept null hypothesis.

However, the answers say to use $Z=(\bar X-\mu)/(\sigma/\sqrt{n}) = (4.06-4.02)/(0.1542/\sqrt{12}),$ but they are saying that $\sigma = 0.1542,$ when this is not true. $S = 0.1542$ and as $n$ is small, $\sigma \ne 0.1542.$

Can someone clarify whether I should use the t or normal distribution? Thanks!

  • $\begingroup$ Perhaps the point of the question is for you to naively use a z test and then assert that you should actually be using a t test. It seems like a bit of a weird question. $\endgroup$ – Ian Nov 26 '16 at 18:25
  • $\begingroup$ They have said that the answer to (c) is "The large variation in the original readings means that an inaccurate claim could still be accepted". $\endgroup$ – John Nov 26 '16 at 18:38
  • $\begingroup$ Formatted your Question using Jax (TeX) codes. Please check to see if I misinterpreted anything. $\endgroup$ – BruceET Nov 26 '16 at 21:52

Your analysis seems to be correct, insofar as I can check it without re-computing $\bar X$ and $S$.

Whenever $\sigma$ is unknown and estimated by $S,$ you should use the t distribution. If you were testing at the 5% level with more than $n = 30$ observations, then critical value from standard normal and $T(n-1)$ would be similar (both near 2.0). But at the 1% level, $n$ has to be more like 60 or 70 for the two critical values to be approximately the same (both near 2.6). [In my view, the "rule of 30" (incorrect except near the 5% level) is confusing and out-of-date in our age of statistical software.]

I take your values $n=12,$ $\bar X = 4.06,$ and $S = 0.1542$ as correct. Then the test statistic is $$ T = \frac{\bar X - \mu_0}{S/\sqrt{n}} = \frac{4.06 - 4.02}{0.1542/\sqrt{12}} = 0.899,$$ as you say.

The critical value for testing $H_0: \mu = 4.02$ against $H_a: \mu \ne 4.02$ at the 1% level is $t^* = 3.1058,$ which cuts probability $.005$ from the upper tail of $T(11).$ Because $|T| = 0.899 < 3.1058,$ you do not reject the null hypothesis.

Below is a printout of this one-sample t test from Minitab. Note that $\sigma$ is unknown and that the software prompted me to enter its estimate $S.$

 Test of μ = 4.02 vs ≠ 4.02

  N    Mean   StDev  SE Mean       99% CI          T      P
 12  4.0600  0.1542   0.0445  (3.9217, 4.1983)  0.90  0.388

Here you fail to reject because the P-value is not smaller than 5%. You cannot get an exact P-value from printed tables of the t distribution. The P-value is $P(|T| > 0.899),$ assuming that $T \sim T(df=11).$

Notice that the 99% confidence interval for $\mu$ is $(3.92, 4.20)$ which includes the hypothetical value $\mu_0 = 4.02.$ One can view this CI as an interval of "believable" values of $\mu$, values which would not be rejected.

Note: If you have further questions about the distinction between z tests and t tests, please leave a Comment. I will check back in several hours.

Addendum on Power: The power of a test for a particular alternative value $\mu_a$ of the population mean is the probability of rejecting $H_0$ given that $\mu_a$ is the correct value. The computation requires you to specify the significance level (here 1%) and to guess $\sigma$ (I used $S$ as my guess). Here is output from Minitab's 'Power and Sample Size' procedure; other statistical software packages have similar procedures--all of them based on the "non-central" t distribution.)

Power and Sample Size 

1-Sample t Test

Testing mean = null (versus ≠ null)
Calculating power for mean = null + difference
α = 0.01  Assumed standard deviation = 0.154

Difference    Size     Power
      0.05      12  0.053957
      0.10      12  0.256326
      0.15      12  0.613519
      0.20      12  0.888939

So, unless the difference between $\mu_0$ and $\mu_a$ is as large as 2 units, you have less than a 90% chance of detecting that difference. The larger the difference, the surer the rejection. Below is a 'power curve' from Minitab. (To be fussy, one point on the 'power curve' is not a power value; it is the height of the curve at 0, which is the significance level 1% = 0.01. Perhaps a better lable for the vertical axis would be "Probability of Rejection.")

Just to show the effect of a larger sample size, I added a second power curve (not mentioned in the printout above) for $n = 36;$ three times as much information, higher power.

enter image description here

Hote: I realize that this whole technical discussion of 'power' may be a step beyond where you are prepared to go right now. But you asked, so I tried to give a reasonably complete answer. Try to get what you can from it.

  • $\begingroup$ So it is correct to use the t distribution for this situation then? Also, I thought if sigma is unknown but we have a large sample size, then we can use s as a good estimate of sigma, and hence use a z test? $\endgroup$ – John Nov 26 '16 at 18:35
  • 1
    $\begingroup$ Yes. Sometimes approximately correct to use z instead of t for large n. But 30 is a 'large' n for testing at the 5% level, and not a 'large' n for testing at 1%. Best just to use t wheneven $\sigma$ is unknown. (Both z and t tests depend on having normal data.) $\endgroup$ – BruceET Nov 26 '16 at 18:48
  • $\begingroup$ Ah ok then. Thanks :) sorry one last question. For (c) I thought it would be something like sigma is unknown hence a z test isn't suitable, but they have said "The large variation in the original readings means that an inaccurate claim could still be accepted". $\endgroup$ – John Nov 26 '16 at 18:52
  • $\begingroup$ I think that's it. In other words, your test, based on only 12 observations, may not have sufficient power to detect a meaningful difference between the true $\mu$ and the hypothetical $\mu_0 = 4.02.$ The CI is a rough guide as to what other values might be right. I have made an addendum to my Answer, which may or may not be helpful to you. $\endgroup$ – BruceET Nov 26 '16 at 21:09
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    $\begingroup$ Promise this is the last question: Does a higher significance level increase the chance of the difference being detected, as the TS is more likely to lie in the critical region. So therefore it is more likely that Ho is rejected whether Ho is true OR false. So, a higher SL does not just increase the probability of rejecting Ho when the hypothetical mu is true, (the Power), but also the probability of rejecting Ho wrongly (hypothetical mu is not true) i.e., a Type 1 error. In contrast, a greater difference or greater sample size only increases the chance of rejecting Ho if hyp mu is correct. $\endgroup$ – John Nov 26 '16 at 22:42

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