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April 14

Calculating Stock Dividends

I don't want the answer, I just wanna know how to do a problem like this (I changed the numbers):

Acme Inc declared and issued a 80 percent stock dividend. The company has 884,000 shares authorized and 177,800 shares outstanding. The par value of the stock is $15 per share and the market value is 95 dollars per share. What is the general journal entry gonna look like? I know it has common stock and retained earnings as the two entries.

What do I multiply/divide? I can't get a straight answer elsewhere. Thanks. --131.156.211.13 (talk) 00:12, 14 April 2013 (UTC)[reply]

We are not licensed accountants, and cannot give accounting advice here. Sławomir Biały (talk) 00:23, 14 April 2013 (UTC)[reply]

It's for a class on accounting, nothing involving actual accounting. My book mentions nothing of this so I'm having trouble. Could you at least link me to a webpage that discusses this concept if no one can talk about it?--131.156.211.13 (talk) 00:34, 14 April 2013 (UTC)[reply]

This seems relevant, though I don't know if it answers your specific question: http://www.accountingcoach.com/online-accounting-course/17Xpg05.html What I don't understand is how a stock dividend of 80% (4 shares for every 5 held, as I see it) can apply when 80% of (884000 - 177800) exceeds the 177800 available.←86.186.142.172 (talk) 11:17, 14 April 2013 (UTC)[reply]

Mann Whitney U function

Is there a functional, non-computational description of the U-function? I can only find procedural descriptions but none using math-style functions and symbols. 93.132.160.155 (talk) 10:30, 14 April 2013 (UTC)[reply]

I'm a little confused by the use of the term "non-computational". From context, I presume you meant "algebraic, rather than algorithmic" as opposed to "conceptual". Please correct me if I'm wrong. Our article Mann–Whitney U effectively gives one, under method two:

\min \left(\sum _{i\in A}R_{i}-{N_{A}(N_{A}+1) \over 2},\sum _{i\in B}R_{i}-{N_{B}(N_{B}+1) \over 2}\right)

Where A and B are the two classes the data is split into, N_X is the number of observations in class X, and R_i is the rank of entry i. - Also note that there are several definitions running around for the Mann–Whitney/Mann–Whitney–Wilcoxon/Wilcoxon/Wilcoxon–Mann–Whitney test, each doing effectively the same thing, but with slightly different calculated statistics and critical values. -- 71.35.98.29 (talk) 18:50, 14 April 2013 (UTC)[reply]

Yes that's the way I meant it. But in the above formula, all the magic is hidden inside the

R_{i}

. Is it possible to break that open? 95.112.217.255 (talk) 08:32, 15 April 2013 (UTC)[reply]

I think perhaps you have to step back a little. You rank the observations by sticking the observations of both samples together, sorting all the values, and saying 1 for the first 2 for the second etc. The numbers you get out of that are the ranks you add up for each sample. Dmcq (talk) 08:56, 15 April 2013 (UTC)[reply]

Yes I know the algorithm. My question is how to put that prose description of what to do into some math-style formula. 95.112.217.255 (talk) 10:12, 15 April 2013 (UTC)[reply]

Well one can stick the conditions under the summation signs but I think Iverson brackets are easier here. For method 1 and denoting the values and samples as

v_{1i}\in S_{1},v_{2j}\in S_{2}

one could say:

U_{1}=\sum _{v_{1i}\in S_{1}}\sum _{v_{2j}\in S_{2}}[v_{2j}<v_{1i}]+{\frac {1}{2}}[v_{2j}=v_{1i}]

this hides how one would normally do it because it doesn't say 'sort' anywhere. Dmcq (talk) 13:03, 15 April 2013 (UTC)[reply]

Thanks, that's what I was looking for! 95.112.217.255 (talk) 14:13, 15 April 2013 (UTC)[reply]