1
$\begingroup$

The following is an exercise from a friend who asked if I can help. Since it is has passed a long time from when I studied some statistics I was wondering if some could help me a little.

Exercise: We measure the weight of 45 lambs after their birth and we find that the average is $4 \textrm{kg}$ and standard deviation 1kg. For this race of lambs we know that the average of the birth is 3,5kg and the variance $44 \textrm{kg}^2$. Estimate the level of significance of the difference between the two mean values, in three different ways.

Here are some thought's/ questions I have:

Attempt 1: My first thought was to apply the Student $t$-test. I consider the hypothesis of the form

$$ \mathrm{H}_0: \mu= \mu_0 \quad \mathrm{against} \quad \mathrm{H}_a: \quad \mu \neq \mu_0, $$ with $ \mu_0= 3,5 \textrm{kg}.$ While working like this I realized that I didn't use at any stage the variance/standard deviation of the race of the lambs, namely the number $ 44 \textrm{kg}^2.$

Attempt 2: After the above, I looked more carefully on an old textbook of statistics and I found that we had one more statistical test for the difference of the means. The problem here, is that the idea of that test is that we have two groups, one of which is the "testing group", while the other is the "control group". This is again not the case my problem.

Question 1: Could anyone give any hind/idea on how to work this problem ? My main difficulty is to recognize which type of statistical test I have to apply.

In a more general context, my question is the following:

Question 2: Assume we are given a sample $A,$ with mean $ \mu_1$ and standard deviation $ \sigma_1$, Suppose that this sample cames from a polulation with mean $ \mu_2$ and standard deviation $ \sigma_2.$ Which statistical test do we use if we want to examine the level of significance of the difference of $ \mu_1 - \mu_2 $ ?

I suppose there would be something to treat this kind of problem. I would really appreciate if someone could help me, giving some advices also in which sources I could find something useful.

Thank you in advance!

$\endgroup$
1
$\begingroup$

Without knowing the context of the problem in the book from which it came, it is difficult to know what three tests the author of the question may have in mind. I will give details about a couple of possibilities that seem reasonable, and I hope that will be helpful.

Background and data. It is important to note that we have a sample of size $n = 45,$ mean $\bar X = 4.0,$ and standard deviation $S = 1.$ And we have a population (unknown size, presumably very large) with mean $\mu = 3.5$ and standard deviation $\sigma = \sqrt{44} = 6.633.$ We are asked to 'compare the two mean values'.

I think you are right that we are supposed to compare the mean $\mu_1$ of the actual population from which the 45 lambs were chosen (perhaps one cooperative of farms) with the population mean $\mu_2$ for the entire 'race'. That would be a one sample test of $H_0: \mu_1 = \mu_2$ against the alternative $H_a: \mu_1 \ne \mu_2.$

But an unusual question arises: Do we use the SD $S = 1$ from the relatively small sample, or the 'race' SD $\sigma = 6.633$. Without knowing a lot more that I do about lambs and the factors that may affect variability in their weights, I don't know which to choose.

T-test. If I use $\bar X$ and $S$ from the sample, I have a one-sample t test. The test statistic is $T = (\bar X - \mu_2)/(S/\sqrt{n}) = (4 - 3.5)/(1/\sqrt{45}) = 3.354.$ Here $T$ has Student's t distribution with $n - 1 =44$ degrees of freedom, so the 'critical value' for a test at the 5% level of significance is 2.014. This means we would reject $H_0$ because $|T| = 3.3364 > 2.014.$

In Minitab 17 statistical software, the test looks like this:

One-Sample T 
Test of μ = 3.5 vs ≠ 3.5

 N   Mean  StDev  SE Mean      95% CI         T      P
45  4.000  1.000    0.149  (3.700, 4.300)  3.35  0.002

From this information, we would reject $H_0$ because the P-value is less than 0.5 = 5%.

Z-Test. If I use $\bar X$ from the sample and $\sigma = 6.633$ from the sample, I have a one-sample t test. The test statistic is $Z = (\bar X - \mu_2)/(\sigma/\sqrt{n}) = (4 - 3.5)/(6.633/\sqrt{45}).$ Here $Z$ has a standard normal distribution, so the 'critical value' for a test at the 5% level of significance is 1.96. One would reject $H_0$ if $|Z| > 1.96, but not here.$ Minitab output is as follows:

One-Sample Z 
Test of μ = 3.5 vs ≠ 3.5
The assumed standard deviation = 6.633

 N   Mean  SE Mean      95% CI         Z      P
45  4.000    0.989  (2.062, 5.938)  0.51  0.613

Here, we would not reject $H_0$ at the 5% level because the P-value exceeds 0.05.

I'm not quite sure about the meaning of the phrase 'estimate the level of significance'. It is possible that this means to compare the two P-values. Some authors refer to P-values as 'achieved' significance levels.

Interpretation. We got different results from the two tests. How might that make sense?

  • From the point of view of the co-op where the 45 lambs were born, they might say their lambs weigh higher than average for 'race'. Moreover, because of the very small SD, they might believe this is not a one-time 'fluke', and that they are doing something that gets heavier lambs.

  • From the point of view of raisers of lambs nationwide, they might say nothing special is going on here. The larger SD for the 'race' reflects differences in local methods of breeding and caring for lambs.

  • Other tests? I don't immediately see what third test the textbook author may have in mind. Perhaps s/he is thinking of using a one sided-alternative. Perhaps the normality of the data are in doubt and some kind of nonparametric test is intended.

  • Disparity in variability. Finally, I should mention that another major difference between the sample of 45 and the 'race' is the SD. There are formal statistical tests to check whether $S = 1$ is significantly smaller than $\sigma = 6.633.$ But I won't go into that because the question is clearly focused on comparing means.

Addendum (pursuant to Comments): Here is a possible third test. Theoretically, it is a bit of a reach, so consider it with care.

Take the sample of $n_1 = 45$ sheep as Sample 1. Then consider that the information about this 'race' of sheep must have originated with a very large sample, say of size $n_2 = 2000$ (purely fictional, possibly reasonable) with $\bar X_2 = 3.5$ and $S_2 = 6.633.$ Also, consider that $n_1$ and $n_2$ are both large enough to use the sample SDs as excellent estimates of population SDs.

With these (somewhat shaky and potentially controversial) assumptions, we could consider the test of a significant difference between $\bar X_1 = 4.0$ and $\bar X_2 = 3.5$ to be a two sample z-test. Then the test statistic is $Z = (\bar X_1 - \bar X_2)/SE,$ where $SE^2 = \sigma_1^2/n_1 + \sigma_2^2/n_2$ and one would reject $H_0$ if $|Z| > 1.96.$ The Minitab output is as follows:

As far as I know, there is no Minitab procedure for this. Arithmetic (and probability computation) in R statistical software follows: I get $Z = 2.378,$ (which you might check on a calculator). So we reject at the 0.05 level. The P-value is $0.017 < 0.05.$

(Notice that the value of $S_2$ doesn't matter much because $S_2^2$ is divided by a huge $n_2.$ But if I take $n_2 = 1000,$ then $Z = 1.94,$ and we are just a bit short of rejection.)

a1 = 4;  s1 = 1;  n1 = 45
a2 = 3.5; s2 = 6.633;  n2 = 2000
SE = sqrt(s1^2/n1 + s2^2/n2)
z = (a1 - a2)/SE;  z
## 2.377704
2*pnorm(-2.377704)
## 0.0174208
$\endgroup$
  • $\begingroup$ Dear BruceET, first of all, let me thank you for your time to reply my question. Your answer is very nice and clear. I followed your hints, to solve the problem, but I still have a question. In general if we have to study the difference of the mean values of a sample(known size) and the population(unknown size) what is the best method ? For example, could I consider a second sample of the population of the same size, and do a $Z$ test ? Any other way to tackle this kind of problem ? Please refer to any books/sources that you think could help me. Thank you in advance. $\endgroup$ – passenger Jan 22 '17 at 17:21
  • $\begingroup$ Glad this makes sense to you. Usually the info provided is simpler. In general, when data are nearly normal, it would be either a one-sample t test (using an estimate $S$ of $\sigma$) or a one-sample z test (using a known $\sigma$). Also, for this particular situation, I have thought of a possible third test, which I will append later, when I have more time. $\endgroup$ – BruceET Jan 22 '17 at 17:40
  • $\begingroup$ Thank's for your quick comment! I think I start to understand it better.... $\endgroup$ – passenger Jan 22 '17 at 17:47
  • $\begingroup$ Please see addendum. $\endgroup$ – BruceET Jan 23 '17 at 1:44
  • $\begingroup$ Thank you very much for your time! That's also a good way to test the difference of the means. I have only two questions: 1) Why did you take $n_2=2000 ,$ and not $45$ again ? 2) This is about the first part of your answer on the $Z$-Test : Before the code of Minitab, you write "we reject $H_0$ because $|Z| > 1.96 .$ " and after you write "we don't reject $H_0$ since $P$-value $ > 0.05 .$ " So, eventually what are we doing: we keep $H_0$ or we reject it ? Thank you in advance. $\endgroup$ – passenger Jan 25 '17 at 11:24

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.