I do not know your subject matter, and how the tests are done, but as a statistician I can say that if you are taking measurements, each one counting as one independent, random observation, and you want to estimate the mean that that measurement should be in the population, then what you need is to first estimate/guess the standard deviation (sigma) of that measurement for the population. (I'm guessing you are taking continuous - type data measurements, but there are analogous considerations for yes/no proportions.) From there you can see what sample size will be needed to obtain an estimated acceptable standard error for that mean, for which you then have a suitable confidence interval. The higher the "confidence" you want, and/or shorter the interval you want, the larger the sample size you will need.
That is your question, correct? You want to 'estimate' your sample size needed to estimate a mean?
To estimate that standard deviation, sigma, you might first need a pilot study. Sigma (the population standard deviation) is a constant to be estimated, or guessed from related information. If you do not later obtain good data, say, too much nonsampling error, then sigma will generally be effectively artificially inflated.
If you use a book such as Cochran, W.G(1977), Sampling Techniques, 3rd ed., John Wiley & Sons, or one of many others, and/or research "standard deviation," "standard error," and "confidence interval" on the internet (the NIH, and also Pennsylvania State University, often seem to be good choices when searching for explanations regarding statistics - so when searching, you might look particularly at URLs with nih or psu in them), then this might help you.
Unless you are using strictly model-based sampling and 'prediction' (which means you need a regressor), you need to use random sampling. (And even then, we start with the sigma for the estimated residuals in OLS, or the sigma for the random factors of the estimated residuals in weighted least squares (WLS) regression, to begin to estimate sample sizes.)
If your observations are not from different sources - say a number of observations from each of the fruit - you could look at "cluster sampling." (Perhaps it should be called cluster random sampling.)
Caution: On the internet you will see a number of sample size 'calculators.' They usually do not warn you that they are (usually) only for proportions (yes/no data), they usually are for the worst case (when p = q = 0.5), and usually do not use a finite population correction (fpc) factor, so they may suggest a sample size that is bigger than your population in some instances. To me, the worst part is not telling you that it is only for proportions, when there are other kinds of data. I worked with continuous data, and it sounds like you might also. (Even for qualitative data, there would be some kind of variability analogous to a sigma, I suspect, which you should consider, but that is not an area of expertise for me.)
I do not have access to the article at the link you provided, but from the title it does sound like it might be relevant. If it uses an hypothesis testing approach, however, caution should be used in a good type II error analysis. Also, note that a p-value alone is not very meaningful. A lone p-value would be inadequate.
your question is bit confusing are you asking about total sample size of each item or size in g (amount ) to carry out test. As Mps Bakshi said 1.0 g oven dried sample is needed for each sample so while collecting please collect at least 150g from market so that for actual test you can easily have enough amount to carry out test. Regarding sample size of your study its very difficult to do test for more than 50 samples if you are doing varous test in the lab. I hope it make sense to your question.
The question is the amount of sample to be taken for the estimation of dietary fiber. The 0.5 to 1.0g oven dried, finely ground sample will be sufficient for estimating fiber.
But no doubt, for the complete analysis (all other parameters), she should take 200-300g fresh fruit sample, as the fresh fruits contain hardly 10-11% DM.