The Relationships between Climate and River flow in the Alps
(1) Write a brief introduction to the study areas, data etc. (given in table 1), maybe input an annotated map of location?
The study is a presentation of statistical review of patterns in yearly flow records of the Swiss Alps for successive period of 30 years. The study sets out to determine various significant trends observed at various time series.
(2) Plot the time series of the flows of the two rivers on the same graph, and comment on the pattern of variation shown.
(3) Plot the two river flow series against each other as an xy scattergram. Comment on the distribution.
(4). Plot the histogram of annual total discharge of the Massa, using all the years. Comment on the dispersion shown. To what extent can annual discharge be considered to be normally-distributed?
(5) Calculate the Pearson’s product moment correlation coefficient between the two series. [Choose an empty cell on the spreadsheet – say G1 – and click there. In the text bar type = and then select CORREL from the drop down menu. Correlation coefficients are usually reported to two decimal places. Format cell, number, decimal places 2, enter].
How well are the two series correlated? Are there any periods when the series are not moving in tandem?
0.86909. Because the Pearson correlation value is greater than 5% at 0.869. The statistics fails to demonstrate a significant correlation between the two series.
(6) Summer discharge from glacierised basins is likely to fairly strongly influenced by energy availability for melting snow and ice. Plot the xy scattergrams and calculate the correlation coefficients between air temperature and each of the discharge series. Why are both of these values less than 1.0? Add air temperature to the time series graph and add any further comments.
(7) Winter precipitation might affect runoff in summer in two opposing tendencies. In summers following winters with higher than average amounts of snowfall, then runoff might be expected to be enhanced as the additional snow is melted and contributes to runoff. Alternatively, high levels of winter snowfall may retard the rising of the transient snowline, especially when a snowy winter is followed by a cool spring. Such accumulation of snow will reduce the amount of ice melted in summer and hence tend to limit the amount of meltwater entering the discharge. Calculate the correlation coefficients between winter precipitation and each of the discharge series. Plot the xy scattergrams. Why are both of these values fairly close to 0.0? Why are the correlation coefficients negative?
The Correlation coefficient is -0.13788: This therefore demonstrates a strong and positive correlation between the discharge series because the Pearson’s Correlation value is less than 0.005 at -0.13788. Implicitly, winter precipitation has a bearing on the discharge.
(8) Are snowy winters followed by cool summers? Yes: Summers tend to be cool owing to the cloud cover while winters are milder beyond expectations because of the latitude.
(9) By now sufficient correlation coefficients should have been calculated to produce a matrix of correlation coefficients between all the possible pairs of variables, as below:
Q5-9 Massa Q6-9 Gornera T5-9 Graechen
Q5-9 Massa – – –
Q6-9 Gornera – –
T5-9 Graechen –
P10-5 Graechen
| Column 1 | Column 2 | Column 3 | Column 4 | |
| Column 1 | 1 | |||
| Column 2 | 0.90565 | 1 | ||
| Column 3 | 0.886126 | 0.86909 | 1 | |
| Column 4 | -0.20826 | -0.11811 | -0.13788 | 1 |
Why are some cells excluded (-)? Complete the table and insert into the document.
(10) Add regression to the graphs produced.
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| Regression Statistics | ||||||||
| Multiple R | 0.989195 | |||||||
| R Square | 0.978507 | |||||||
| Adjusted R Square | 0.942793 | |||||||
| Standard Error | 296.1631 | |||||||
| Observations | 29 | |||||||
| ANOVA | ||||||||
| df | SS | MS | F | Significance F | ||||
| Regression | 1 | 1.12E+08 | 1.12E+08 | 1274.761 | 2.91E-24 | |||
| Residual | 28 | 2455952 | 87712.58 | |||||
| Total | 29 | 1.14E+08 | ||||||
| Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
| Intercept | 0 | #N/A | #N/A | #N/A | #N/A | #N/A | #N/A | #N/A |
| 379.05 | 5.222358 | 0.146269 | 35.7038 | 6.78E-25 | 4.92274 | 5.521976 | 4.92274 | 5.521976 |
| RESIDUAL OUTPUT | PROBABILITY OUTPUT | |||||||
| Observation | Predicted 1970 | Residuals | Standard Residuals | Percentile | 1970 | |||
| 1 | 2007.997 | -36.9967 | -0.12713 | 1.724138 | 1971 | |||
| 2 | 1509.993 | 462.0074 | 1.587589 | 5.172414 | 1972 | |||
| 3 | 2059.385 | -86.3847 | -0.29684 | 8.62069 | 1973 | |||
| 4 | 1650.944 | 323.0559 | 1.110112 | 12.06897 | 1974 | |||
| 5 | 1747.401 | 227.599 | 0.782095 | 15.51724 | 1975 | |||
| 6 | 1766.619 | 209.3807 | 0.719492 | 18.96552 | 1976 | |||
| 7 | 1616.947 | 360.0535 | 1.237246 | 22.41379 | 1977 | |||
| 8 | 1432.493 | 545.5072 | 1.874518 | 25.86207 | 1978 | |||
| 9 | 1830.071 | 148.9291 | 0.511763 | 29.31034 | 1979 | |||
| 10 | 1544.408 | 435.592 | 1.496818 | 32.75862 | 1980 | |||
| 11 | 1863.703 | 117.2971 | 0.403066 | 36.2069 | 1981 | |||
| 12 | 2250.262 | -268.262 | -0.92182 | 39.65517 | 1982 | |||
| 13 | 2064.45 | -81.4504 | -0.27989 | 43.10345 | 1983 | |||
| 14 | 1489.991 | 494.009 | 1.697555 | 46.55172 | 1984 | |||
| 15 | 1873.26 | 111.7402 | 0.383971 | 50 | 1985 | |||
| 16 | 2034.161 | -48.1607 | -0.16549 | 53.44828 | 1986 | |||
| 17 | 1969.56 | 17.43988 | 0.059928 | 56.89655 | 1987 | |||
| 18 | 2076.201 | -88.2007 | -0.30308 | 60.34483 | 1988 | |||
| 19 | 2064.868 | -75.8682 | -0.2607 | 63.7931 | 1989 | |||
| 20 | 2231.148 | -241.148 | -0.82865 | 67.24138 | 1990 | |||
| 21 | 2218.771 | -227.771 | -0.78269 | 70.68966 | 1991 | |||
| 22 | 2303.478 | -311.478 | -1.07033 | 74.13793 | 1992 | |||
| 23 | 2055.729 | -62.729 | -0.21555 | 77.58621 | 1993 | |||
| 24 | 2591.7 | -597.7 | -2.05387 | 81.03448 | 1994 | |||
| 25 | 1759.569 | 235.4309 | 0.809007 | 84.48276 | 1995 | |||
| 26 | 1672.095 | 323.9054 | 1.113031 | 87.93103 | 1996 | |||
| 27 | 1915.874 | 81.12572 | 0.278771 | 91.37931 | 1997 | |||
| 28 | 2210.259 | -212.259 | -0.72938 | 94.82759 | 1998 | |||
| 29 | 2498.272 | -499.272 | -1.71564 | 98.27586 | 1999 | |||
(11) What does the r2 value describe?
The Regression model explains 97.85 percent of the variation of the time series
(12) Is the regression line significantly different from one with gradient b=0 ?
Yes: The statistics shows that the slope is significantly different from zero,
The observed data indicate that the slope is significantly different from zero, thus observed statistics positively indicate that precipitation are largely dependant on various environmental factors
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