1. ## C.L.T hypothesis testing

People who diet can expect to lose an average of 3kg in a month. In a book, the authors claim that people who follow a new diet will lose an average of more than 3kg in a month. The weight losses of the 180 people in a random sample who had followed the new diet for a month were noted. The mean was 3.3 kg and the standard deviation was 2.8kg.

(i)Test the authors claim at the 5% significance level, stating your null and alternative hypothesis.

I didnt think I had problems but the example answer given divides the standard deviation by 179 when standardizing to the normal distribution. Why not 180. Is there some other technique requiring a loss of 1 degree of freedom???
It mentions that if 180 is used then the answer will be sligtly less.

Thanks.

2. With a normal approximation to an integer based random variable, quite often one will add or subtract .5. That's splitting the difference between the two sets. BUT I don't see why you would do that here. This is not an integer based random variable like the binomial.
Nor why use 179 and not 179.5.

3. Originally Posted by woollybull
People who diet can expect to lose an average of 3kg in a month. In a book, the authors claim that people who follow a new diet will lose an average of more than 3kg in a month. The weight losses of the 180 people in a random sample who had followed the new diet for a month were noted. The mean was 3.3 kg and the standard deviation was 2.8kg.

(i)Test the authors claim at the 5% significance level, stating your null and alternative hypothesis.

I didnt think I had problems but the example answer given divides the standard deviation by 179 when standardizing to the normal distribution. Why not 180. Is there some other technique requiring a loss of 1 degree of freedom???
It mentions that if 180 is used then the answer will be sligtly less.

Thanks.
I suspect that the standard deviation quoted is:

$s=\sqrt{\frac{1}{180}\sum_i (x_i-\overline{x})^2}$

but you need the square root of the unbiased estimator for the population variance:

$s^*=\sqrt{\frac{1}{179}\sum_i(x_i-\overline{x})^2}$

So while you need to use $s^*/\sqrt{180}$ you get this by taking $s/\sqrt{179}$

Not that any of this makes much difference given the accuracy of the approximations and assumptions involved (or rather if it did make any difference the test is questionable and you will need more data)

CB

4. Thanks. That's what I was missing.