Perform a hypothesis testing using Kendall’s Tau method to check if the distribution of peak flows does not change as a function of time. Data is below please show step by step. I will rate afterwards.
Year | Peak Flow |
1947 | 15500 |
1948 | 27400 |
1949 | 27500 |
1950 | 43000 |
1951 | 19100 |
1952 | 17600 |
1953 | 15800 |
1954 | 8600 |
1955 | 9950 |
1956 | 35400 |
1957 | 34700 |
1958 | 37000 |
1959 | 26000 |
1960 | 27000 |
1961 | 42300 |
1962 | 26300 |
1963 | 41600 |
1964 | 50500 |
1965 | 15500 |
1966 | 6710 |
1967 | 31000 |
1968 | 31700 |
1969 | 36100 |
1970 | 16900 |
1971 | 26700 |
1972 | 23700 |
1973 | 17500 |
1974 | 22900 |
1975 | 27500 |
1976 | 20300 |
1977 | 13200 |
1978 | 26600 |
1979 | 28700 |
1980 | 21600 |
1981 | 23800 |
1982 | 19500 |
1983 | 15200 |
1984 | 18400 |
1985 | 29800 |
1986 | 29200 |
1987 | 21800 |
1988 | 18500 |
1989 | 27000 |
1990 | 31300 |
1991 | 48600 |
1992 | 20400 |
1993 | 26600 |
1994 | 49500 |
1995 | 20900 |
1996 | 27600 |
1997 | 22900 |
1998 | 39600 |
1999 | 34600 |
2000 | 10600 |
2001 | 16800 |
2002 | 38300 |
2003 | 65700 |
2004 | 30800 |
We firstly look at the distribution of the peak flow:
Now we use the Kendall’s Tau method to check if the distribution of peak flows does not change as a function of time.
Here, in the image we can observe that (Prob>|z| = 0.351) which is simply the p-value of the test.
Now, the null hypothesis states that Peak flow and year are independent. Here, since p-value is greater than 0.05, thus, we fail to reject the null hypotheisis. Hence, peak flow and time are independent.
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