In: Statistics and Probability
A second question arises from the data: “Is secchi depth related to P concentrations in the study lakes?” Using excel, choose the appropriate test to address this research question and interpret the results. Produce an appropriate figure for these data showing all relevant information
Catchement Type | Lake | P (ug/L) | Secchi Depth (m) |
Forested | Vaudray | 1.0 | 7.9 |
1.1 | 7.4 | ||
1.2 | 8.9 | ||
Hannah | 1.3 | 7.3 | |
1.3 | 7.2 | ||
1.4 | 7.6 | ||
Rice | 3.7 | 7.0 | |
3.0 | 7.3 | ||
3.5 | 7.2 | ||
Agriculture | Chemong | 23.0 | 2.3 |
22.0 | 2.2 | ||
23.5 | 2.3 | ||
Buckhorn | 19.7 | 3.1 | |
20.1 | 3.5 | ||
18.5 | 3.5 | ||
Long | 26.3 | 2.2 | |
28.1 | 2.3 | ||
27.0 | 2.2 | ||
Kushog | 26.1 | 2.0 | |
25.9 | 1.9 | ||
27.2 | 1.7 | ||
Residential | Osisko | 15.8 | 3.5 |
13.5 | 3.3 | ||
15.1 | 3.7 | ||
Whitney | 13.5 | 3.6 | |
14.1 | 4.2 | ||
14.5 | 3.0 | ||
Eel | 9.3 | 3.5 | |
10.7 | 4.0 | ||
8.9 | 4.3 |
The scatterplot is:
There is a negative relationship between the variables.
The hypothesis being tested is:
H0: β1 = 0
H1: β1 ≠ 0
The p-value from the output is 0.0000.
Since the p-value (0.0000) is less than the significance level (0.05), we can reject the null hypothesis.
Therefore, we can conclude that the slope is significant.
r² | 0.884 | |||||
r | -0.940 | |||||
Std. Error | 0.782 | |||||
n | 30 | |||||
k | 1 | |||||
Dep. Var. | P | |||||
ANOVA table | ||||||
Source | SS | df | MS | F | p-value | |
Regression | 130.2468 | 1 | 130.2468 | 212.99 | 1.29E-14 | |
Residual | 17.1229 | 28 | 0.6115 | |||
Total | 147.3697 | 29 | ||||
Regression output | confidence interval | |||||
variables | coefficients | std. error | t (df=28) | p-value | 95% lower | 95% upper |
Intercept | 7.4202 | |||||
Depth | -0.2201 | 0.0151 | -14.594 | 1.29E-14 | -0.2510 | -0.1892 |