Question

2. [20 marks] Fresh water from rivers and lakes varies in acidity/alkalinity depending on the rocks and soils it runs over. Granite country yields water that is slightly acidic, while water from limestone country is slightly alkaline. Acidity and alkalinity are measured on a pH scale, on which 7 is neutral, values below 7 are acidic and values above 7 are alkaline. pH can change due to pollution or due to waste from biomass such as algae. A certain river, fed by various tributaries, usually has a pH of about 6.3 (slightly acidic). The file Water_pH.csv contains pH values from randomly-selected sites on this river and its tributaries. (a) Provide a graph that displays the distribution of the data. Bear in mind that distribution shape and outliers may be important. [5] (b) How much evidence is there that the average pH of water in this river system has changed from the usual value of 6.3? Evaluate the evidence using a hypothesis test with a significance level of 5%. Take care to state and evaluate all assumptions, and provide an appropriate conclusion. [12] (c) Provide a 99% confidence interval for the mean pH in this river system.

Answer 1

Answer: a) Date: ---24/5/2019 Boxplot ph 0 4 there are no outliers in the data. the distribution is approximately normally distributed. b) Ho: the average pH of water in this river system has NOT changed from the usual value of 6.3. u-6.3 hl: the average pH of water in this river system has changed from the usual value of 6.3. u6.3 6.00 [Excel function 0.631 [Excel function used-> STDEV] sd= u 6.30 n= 29.00 [Excel function used-> COUNT]

alpha 5% critical value, t(a/2,n-1) t(0.05/2,29-1) 2.048 test statistic, t - (mean-u)/(sd/sqrt(n) (5.9966-6.3)/0.6305/sqrt(29)) 2.5914 p-value 2(1-P(T-t) 2*(1-P(T<abs(-2.5914)) T.DIST.2T(abs(-2.5914),29-1) 0.0150 With t=-2.59, p<5%, i reject ho and conclude that the average pH of water in this river system has changed from the usual value of 6.3 critical value, t(a/2,n-1) (0.01/2,29-1) 2.763 Cl mean + t(a/2,n-1) (sd/sqrt(n)) ower 5.9966 - 2.763*(0.6305/sqrt(29)-5.67 upper5.9966 2.763 (0.6305/sqrt(29)6.32