Part A –JohananOttensooser10873305, HiralPatel 10769127, Austin Formosa10747032

Question A1

1) In this question we are trying to identify whether positive or negative growth in January stock figures will correspond with a like outcome for the entire year. We hope to obtain an objective outcome by using probability tables that represent; observed frequency, joint probability, expected frequency and bringing them together to perform a Chi Square test of independence (Black et al., 2007, p285). We have chosen to use these binomial tables in order for us to determine the Chi Square test of independence result. This result will then be contrasted against the critical value we have calculated in order for us to determine whether or not the January Barometer holds to be true.

We have used the chi square test of independence because it can be used to analyze our two variables the positive or negative growth using the categories of January and year to determine whether the two variables are independent.

Observed Frequency Table
January + / January - / Total
Year + / 29 / 11 / 40
Year - / 4 / 10 / 14
Total / 33 / 21 / 54
Expected Frequency Table / =Column*Row/Total
(J+) / (J-) / Total
(Y+) / 24.44444 / 15.555556 / 40
(Y-) / 8.555556 / 5.4444444 / 14
Total / 33 / 21 / 54
Chi-Square Test of Independence =[(obs-exp)^2/exp]
(J+) / (J-) / Total
(Y+) / 0.84899 / 1.334126984 / 2.183116883
(Y-) / 2.425685 / 3.811791383 / 6.237476809
Total / 3.274675 / 5.145918367 / 8.420593692
Chi-p-value = 0.00371
(95% probability, 2 degrees of freedom)
Chi Critical Value =CHIINV(0.05,1) / 3.841459149

3) In conclusion as drawn from the above tables, since the Chi-Square Value is greater than the Chi-Square critical value the data sets are dependent thus rejecting the null hypothesis. This means that the growth that occurs for the year is dependent on the January growth thus holding the January Barometer to be true. We would recommend to the international investment firm based on our outcome, that is should invest for the entire year as the January 2010 growth figure was positive 0.9%.

Question A2

1)Hypothesis that the magnitude of January growth “Jgrowth” predicts annual growth “Agrowth”. This hypothesis is intentionally broad to allow for different methods of statistical testing: i.e. through qualitative and quantitative methods. Qualitative is defined as relating to or involving comparisons based on qualities (Princeton, 2010). In this case, we will use a pie chart to represent data within a specific hypothesis nonrejection region. This will show, loosely, whether or not there may be a correlation. Quantitative is defined as expressible as a quantity or relating to or susceptible of measurement (Princeton, 2010). Here, we will use the Pearson product-moment correletion coefficient to extract statistical correlation in the magnitute of the two growth figures (Black et al., 2007, p96). This will distinclty prove the strength of statistical correlation in the figures.

2) In order to roughly test correlation, first, we roughly graphed (in a scatter plot) Jgrowth as a function of Agrowth (fig. 1). This did seem to suggest that in most cases . This is quite a wide test: if January growth is 5%, this allows values between -5% and 27.5%. However, it is clear from figure 1 that a tighter test would exclude too many values. Thus, we plotted the above as the nonrejection region of a null hypothesis, as per Black(Black et al., 2007, p309) (fig. 2). Since this hypothesis was accepted only slightly more than half the time (29 out of 54 times), it seems statistically irrelevant.

In order to ascertain magnitudal, not merely sign-based correlation, a chi2 test, as in question 1 would not suffice. Thus, we used excel’s Pearson product-moment correlation coefficient (“r value”) to test the relationship of the two data sets as plotted in figure 3 (Black et al., 2007, p95-97). This revealed an r-value of 0.348757, which is classed as a weak-moderate positive correlation.

3) There is a weak positive correlation between January and annual growth, satisfying the hypothesis above. However, the weakness of the correlation, as expressed by the low r value and the wide qualitative bracket without a large majority result, stresses that it is limited as a statistical tool. Any correlation, however, is better than chance. As such, considering a January growth rate of +0.9, I would recomend an international businessman to invest in Australia, since there is a greater than 50% chance (as proved by a positive correlation) that annual growth will, thus, be positive(tradingeconomics.com, 2010).

Question A3 p285

January Growth / Annual Growth
Mean / 1.136111111 / Mean / 6.16537037
Standard Error / 0.669833068 / Standard Error / 1.966145397
Median / 1.58 / Median / 8
Mode / 3.3 / Mode / 6.5
Standard Deviation / 4.922247691 / Standard Deviation / 14.44815895
Sample Variance / 24.22852233 / Sample Variance / 208.749297
Kurtosis / -0.118470482 / Kurtosis / 0.46992798
Skewness / 0.292990088 / Skewness / -0.688494117
Range / 20.8 / Range / 66.88
Minimum / -7.6 / Minimum / -34.48
Maximum / 13.2 / Maximum / 32.4
Sum / 61.35 / Sum / 332.93
Count / 54 / Count / 54

Bullet 3) To identify any anomalies in the data we must meet the following criterion: the anomalies must be 1.5 times the inter-quartile range and added or subtracted to the upper and lower quartiles in order to identify the lower and upper limits (Black et al., 2007, p91).

Bullet 4)

Upper and Lower Limits
January Growth / Annual Growth / Position / Proportion
MIN / -7.6 / -34.48
Quartile 1 / -2.635 / -4.96 / 13.5 / 0.25
Quartile 2 / 1.58 / 8
Quartile 3 / 4 / 15.3 / 40.5 / 0.75
MAX / 13.2 / 32.4
Inter-quartile Range / 6.635 / 20.26
Lower Limit / -17.5525 / -35.35 / Quartile 1 - 1.5* Inter-quartile Range
Upper Limit / 23.1525 / 45.69 / Quartile 3 + 1.5* Inter-quartile Range

Using the figures that have been calculated in table 3.1, no outliers have been identified as the minimum and maximum values of both January and Annual growth figures all fall within the upper and lower limits of the outlier criterion.

Bullet 5) Using the histogram and informally identifying outliers we can deduce that by removing the January data for years 2006-2008 , the graph will become more normally distributed. Similarly using the yearly growth histogram, we can clearly see that if we remove the 1955-1957 data the histogram will reflect a more evenly distributed bell curve.

Bullet 6)Based on the listed criterion there are no outliers and anomalies. This means we would not change anything in the descriptive summary, thus it would remain the same.

Bullet 7) Using z-scores is less than appropriate opposed to the one and a half times the inter-quartile range values, seeing that the data is not normal. However, this is most probably due to the small quantity of data used. If more data was used, the distribution would probably start to form, that of a bell-curve distribution. This would then make the z score test more applicable, However, neither tests identifies any outliers. Only when we start to use a range of ±2stddev opposed to 3 stddevwe then yield 3 outliers for both the January and annual growths. These outliers occur in 2006, 2007, 2008 and in 1955, 1956, 1957, respectively (Black et al., 2007, p77).

Z Score Outlier test / January / Annual / Z Score Outlier test / January / Annual
Average / 1.136111 / 6.16537 / Average / 1.136111 / 6.16537
Std Dev. / 4.922248 / 14.44816 / Std Dev. / 4.922248 / 14.44816
lower limit (avg -3StdDev) / -13.6306 / -37.1791 / lower limit (avg -2StdDev) / -8.70838 / -22.7309
upper limit (avg +3StdDev) / 15.90285 / 49.50985 / upper limit (avg +2StdDev) / 10.98061 / 35.06169
Outliers / FALSE / FALSE / Outliers / 3 / 3

Fig 1 Fig 2

Since z-scores and one and a half times the inter-quartile range have similar results, both can be said to be valid. Although, given that normality has not been established, we are of the opinion that one and half times the inter-quartile range is a more valid test.

Question A4

The January Barometer is said to predict that once the first month of the year’s trade is over the rest of the year will follow the growth pattern accordingly. In this task we went about trying to examine the data provided to confirm or deny this theory. In the first test we examined and tested the data based on the growth being positive or negative. In the second test we qualitatively and quantitatively analyzed the magnitude of growth. In the third test we went about trying to identify any anomalies in the data provided.

The Chi-squared test proved a strong correlation (to greater than 0.95 certainty) that January growth and December growth are correlated regarding sign.

The Pearson test proved a moderate to weak positive correlation between the magnitude of January growth to the magnitude of annual growth

The anomalies test proved that there were no real outliers according to the criteria we applied, this suggests that the other tests are valid in terms that the data does not contain any extreme values. When using the z-scores and applying three standard deviations from the mean no outliers where identified either.

In conclusion the “January Barometer” is a valid signal to determine the yearly growth to a certain extent. We would recommend to the international investment firm that is looking to invest in the Australian share market in 2010 to invest for the year seeing that the growth is positive, the magnitude has a week positive correlation and the data has no outliers.

Bibliography

BLACK, ASAFU-ADJAYE, KHAN, PERERA, EDWARDS & HARRIS 2007. Australian Business Statistics, Wiley.

PRINCETON 2010. Wordnet. Princeton.

TRADINGECONOMICS.COM. 2010. Australia GDP growth rate [Online]. Available: [Accessed 12/04/2010].

1 | Page`