Experience raising Labrador puppies revealed that the weight of the puppies at six months follow a normal distribution. The subsequent weights (in pounds) of a small sample of seven six-month puppies were:

$$ \begin{array}{|c|c|} \hline \mbox{Puppy #}& \mbox{Weight} \\ \hline 1 & 16 \\ 2 & 46 \\ 3 & 34 \\ 4 & 31 \\ 5 & 21 \\ 6 & 26 \\ 7 & 22 \\ \hline \mbox{Sum} & 196 \\ \hline \end{array} $$

Based on this information:

a) What is the mean weight of these seven puppies?

Which I got to be $28$.

b) Based on this small sample, what is the standard deviation?

I got this to be $9.27$.

c) Construct the 95% confidence interval for the mean weight of all six-month puppies based on this information.

I got this to be $(9.83, 46.17)$ but I was wondering could I write this as $9.83 < \mu < 46.17$?

d) Would it be reasonable to state that this litter contained two “unusual” puppies; i.e., a runt and a giant? Why or why not?

I have no clue what this is asking. What I was thinking was we could say no because all of our data falls in our CI. Is this a good way to approach it?

  • 1
    $\begingroup$ Just a quick point: You definitely can not write $9.83 < \mu < 46.17$. That implies you know for sure that $\mu$ lies in that interval. What you actually have is an interval that might span $\mu$ with some probability. Note I carefully did not say that $\mu$ lies in the interval with some probability. This is because $\mu$ is not random, so the probability of it existing within the interval is either 0 or 1. But the interval is random, so the probability of it spanning $\mu$ is a meaningful construct. $\endgroup$ Commented Feb 25, 2013 at 2:10
  • $\begingroup$ alright ill leave it as (9.83, 46.17) but its only the last question that I am stumbled on. what i wrote is. No it wouldbe reasonable to say there are unusual puppies because all our weighted puppies are in our confidence interval. Hence not unusual is that okay? $\endgroup$
    – MathGeek
    Commented Feb 25, 2013 at 2:21

2 Answers 2


Two main areas for you to look at:

First, the denominator in your estimation of the standard deviation of the population, which has to be based on the sample. As you only have a sample, you have used a degree of freedom when you estimated the mean and hence @Karim is right in this case that you want (n-1) rather than n.

Your question C has a more fundamental problem. To create a confidence interval you need the standard deviation of the estimate of the mean, often (annoyingly, to me) called the "standard error". This is not the same as the standard deviation of the population and gets smaller as the sample size gets larger. You probably have a formula for this somewhere...

Secondarily with regard to your question C - you have a small sample size, and you are estimating the variance of the mean from your population. This means that your estimate of the mean has a t distribution rather than a Normal distribution.

For the record, here is how I recreated your results with the errors hinted at above:

> labs <- c(16,46,34,31,21,26,22)
> n<- length(labs)
> labs.s.wrong <- sqrt(sum((labs-mean(labs))^2)/n) # wrong denominator
> labs.s.wrong
[1] 9.273618
> mean(labs) + qnorm(c(0.025,0.975))* labs.s.wrong # wrong distribution; wrong standard error
[1]  9.824042 46.175958

One correct series of calculations in R would instead proceed like this:

labs <- c(16,46,34,31,21,26,22)
s <- sd(labs); m <- mean(labs); n <- length(labs) # The statistics
confidence <- 0.95
z <- qt(1 - (1-confidence)/2, n - 1)              # Student t quantile
m + c(-1,1) * z * s / sqrt(n)                     # Confidence limits

The output is

[1] 18.73614 37.26386

To answer the last question (d), re-express the weights as multiples of the standard deviation away from the mean (that is, as standardized weights):

> print(sort((labs - m)/s), digits=2)
[1] -1.2 -0.7 -0.6 -0.2  0.3  0.6  1.8

None of these values is surprising: from the 68-95-99.7% rule of thumb, we expect 68% of the data (around four or five of the weights) to have standardized values between $-1$ and $1$ (and five of them do) while 95% of the data (most likely all of them) would have standardized values between $-2$ and $2$ (and all of them do). There is no evidence that any of the weights is unusually high or low relative to the set of seven weights.


For the standard deviation, you used the population standard deviation, I think the sample standard deviation would be more appropriate (divide by n-1 instead of n).

As for the last question, my understanding is that it asks, given the normal distribution with the parameters you have estimated, what are the probabilities of observing a weight : 1) 16 or less and 2) 46 or more

  • $\begingroup$ are you sure about dividing by n-1 instead? isnt the population Standard deviation what we want? and for the last question you are telling me I should find the probability of P(x<16) and P(x>46)?? $\endgroup$
    – MathGeek
    Commented Feb 25, 2013 at 0:28
  • $\begingroup$ Yes you want (to estimate) the population standard deviation but you don't have the population, you only have a small sample; dividing by n-1 (the degrees of freedom) gives you an unbiased estimator of the population standard deviation. $\endgroup$
    – Karim L
    Commented Feb 25, 2013 at 0:45
  • $\begingroup$ alright sounds goood, how about for the last question? $\endgroup$
    – MathGeek
    Commented Feb 25, 2013 at 1:01
  • 2
    $\begingroup$ Karim, are you really sure you would use an estimate of the population standard deviation in place of the standard error of the mean in the formula for a confidence interval? As far as your last remark goes, it raises a subtle but important issue: the probabilities of observing such weights are not assessed with confidence intervals but with prediction intervals. (If you haven't heard of the latter, search our site: there are explanations and formulas.) Given the elementary nature of these questions, I doubt that's the intended interpretation. $\endgroup$
    – whuber
    Commented Feb 25, 2013 at 1:49
  • 1
    $\begingroup$ "dividing by n-1 (the degrees of freedom) gives you an unbiased estimator of the population standard deviation" -- no, it doesn't. The $n-1$ denominator is unbiased for the variance. It is biased for the standard deviation. $\endgroup$
    – Glen_b
    Commented Aug 23, 2013 at 16:44

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