75% bullish in December

[kr1zh]

Yes,

Believe or not All Ordinaries market has bullish probability of 75% in December (was also claimed by ChartTV.com). April is sitting the second highest after December with bulish probability of 71 pc

Analysis was done based on methodology used by ChartTV.com and analysis was done using data from 1986 to 2006 provided by Yahoo Finance.


Market Guru,

I am a newbie in share trading, could you share your thoughts in this regards.

Thanks


Cheers

kr1zh

 

[nizar]

Hmm.... Im suprised May and October were both green (bullish) in that chart?

What happened to October being historically red and "sell in may and go away" ?

 

[kr1zh]

Nizar,

Refer to attached table for the "Red" zone historically. As you said, you're right about Octobers. The table has shown that the cummulative score in octobers is -790.3 pts

However, is we look at all points for October since 1986, the probability to be bulish is 60 pc compared the 40 pc to be bearish.

This year in 2006, the point score in May is -208.1 However, the bulish probability might still seen as bigger than the bearish.

It's been tough to predict accurate buy & sell timing. however, thanks for your input nizar, hope the table give us a bit of idea.

 

[kr1zh]

It's also interesting to see the gain in pc for a particular month.

It seems October is the worst month ever !

So watch your step guys..

 

[chemist]

It's also interesting to see the gain in pc for a particular month.

It seems October is the worst month ever !

So watch your step guys..

The statistics in this thread are worthless. No significance levels, no confidence intervals.

A more rigorous and convincing approach would be to separately compare each month's returns data with the returns data for the remaining months. Using standard methods test the the null hypothesis that the population mean return of the month in question is the same as that of the other months. The analyst should also test for equality of the monthly variances.

If you post your data I will provide a worked analysis.

cheers,
Chemist

 

[kr1zh]

Chemist,

That's a very kind of you.. My data is based on the one provided by finance yahoo. you could download in following link.

http://au.rd.yahoo.com/finance/quotes/internal/historical/download/*http://ichart.finance.yahoo.com/table.csv?s=%5EAORD&a=747&b=3&c=1984&d=04&e=27&f=2006&g=d&ignore=.csv


Please share us your statistical expertise to help us analysing the correct and accurate data.


Many Thanks

The statistics in this thread are worthless. No significance levels, no confidence intervals.

A more rigorous and convincing approach would be to separately compare each month's returns data with the returns data for the remaining months. Using standard methods test the the null hypothesis that the population mean return of the month in question is the same as that of the other months. The analyst should also test for equality of the monthly variances.

If you post your data I will provide a worked analysis.

cheers,
Chemist

 

[tech/a]

kr1zh
I hope you supply the data,would be interested in the analysis.

 

[kr1zh]

ok now the data i'm using is not much different with the one in post #3 in this thread.

the monthly score is taking the sum of scores for particular month. the definition of score is the difference between the close and the open points.

 

[kr1zh]

So where is our statistical expertise ? The data I am using is the one I posted earliar. So please chemist, share your statistical expertise if you really are and intend to share knowledge to share traders in this forum...

regards

Krishna

 

[chemist]

Your data is not suitable. Raw (points) index changes now cannot be compared with those 20 years ago. You must look at relative changes (returns). I sourced my data from Yahoo! I used only the closing prices.

I constructed monthly relative returns by calculating the differences between the logs of monthly closing prices. A cursory examination of the data indicated that Oct 87 was an outlier, with a value of -0.55. After excluding this point the returns data were adequately normal. See attached aoxm.csv.

aoxm.csv

I then constructed return vectors for each month and compared them to the whole sample using the standard R function t.test. The (rather tedious) results are shown below.

In summary, the only month significantly different from the whole sample was December, which was more profitable than average at the 5% confidence level. However, we've mined through 12 different months to find this apparent anomaly, so the real significance level is more like 41%, ie completely insignificant.

Cheers,
Chemist

> t.test(ret.dec,Return)

Welch Two Sample t-test

data: ret.dec and Return
t = 2.133, df = 24.952, p-value = 0.04294
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.0005114338 0.0292618736
sample estimates:
mean of x mean of y
0.023929489 0.009042835

> t.test(ret.nov,Return)

Welch Two Sample t-test

data: ret.nov and Return
t = -0.8923, df = 22.159, p-value = 0.3818
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.02718518 0.01082427
sample estimates:
mean of x mean of y
0.000862381 0.009042835

> t.test(ret.oct,Return)

Welch Two Sample t-test

data: ret.oct and Return
t = -0.202, df = 19.569, p-value = 0.842
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.02826812 0.02328408
sample estimates:
mean of x mean of y
0.006550815 0.009042835

> t.test(ret.sep,Return)

Welch Two Sample t-test

data: ret.sep and Return
t = -1.2074, df = 21.934, p-value = 0.2401
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.031100232 0.008215084
sample estimates:
mean of x mean of y
-0.002399739 0.009042835

> t.test(ret.aug,Return)

Welch Two Sample t-test

data: ret.aug and Return
t = -0.4159, df = 21.573, p-value = 0.6816
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.02506644 0.01669987
sample estimates:
mean of x mean of y
0.004859554 0.009042835

> t.test(ret.jul,Return)

Welch Two Sample t-test

data: ret.jul and Return
t = 0.9792, df = 21.336, p-value = 0.3385
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.01154739 0.03213362
sample estimates:
mean of x mean of y
0.019335951 0.009042835

> t.test(ret.jun,Return)

Welch Two Sample t-test

data: ret.jun and Return
t = -1.1463, df = 24.237, p-value = 0.2629
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.023620609 0.006745979
sample estimates:
mean of x mean of y
0.0006055197 0.0090428351

> t.test(ret.may,Return)

Welch Two Sample t-test

data: ret.may and Return
t = 0.0518, df = 24.565, p-value = 0.9591
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.01442015 0.01516286
sample estimates:
mean of x mean of y
0.009414191 0.009042835

> t.test(ret.apr,Return)

Welch Two Sample t-test

data: ret.apr and Return
t = 1.6191, df = 24.83, p-value = 0.1181
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.003433481 0.028638043
sample estimates:
mean of x mean of y
0.021645116 0.009042835

> t.test(ret.mar,Return)

Welch Two Sample t-test

data: ret.mar and Return
t = 0.0877, df = 22.852, p-value = 0.9309
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.01946730 0.02119027
sample estimates:
mean of x mean of y
0.009904317 0.009042835

> t.test(ret.feb,Return)

Welch Two Sample t-test

data: ret.feb and Return
t = -0.8648, df = 23.714, p-value = 0.3958
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.02557027 0.01047567
sample estimates:
mean of x mean of y
0.001495534 0.009042835

> t.test(ret.jan,Return)

Welch Two Sample t-test

data: ret.jan and Return
t = 0.3597, df = 25.203, p-value = 0.7221
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.01281183 0.01823589
sample estimates:
mean of x mean of y
0.011754867 0.009042835

 

[chemist]

Is any month more likely to out-perform? The average monthly return in the 20 year sample is about +0.9%. I counted how often each month did better than that and applied the standard R function prop.test (see below).

The resultant p-value 0.2088 is only weakly suggestive of an effect.

Overall, I don't think month of the year effects are worth much for trading the all ords.

cheers,
Chemist

> plus
 [1] 13  9 10 13 11  6 12 10  8 10 10 16
> plus.n
 [1] 21 21 21 21 20 20 20 20 20 19 20 20
> prop.test(plus,plus.n)

        12-sample test for equality of proportions without continuity
        correction

data:  plus out of plus.n
X-squared = 14.4557, df = 11, p-value = 0.2088
alternative hypothesis: two.sided
sample estimates:
   prop 1    prop 2    prop 3    prop 4    prop 5    prop 6    prop 7    prop 8
0.6190476 0.4285714 0.4761905 0.6190476 0.5500000 0.3000000 0.6000000 0.5000000
   prop 9   prop 10   prop 11   prop 12
0.4000000 0.5263158 0.5000000 0.8000000

 

[tech/a]

Chemist.

Can you break that down into a little clearer/plainer English?

Thanks.

 

[Ageo]

Chemist.

Can you break that down into a little clearer/plainer English?

Thanks.

I thought i was back in school for a second there

If you could break it down chemist that would be fantastic.


Adrian