Question

I struggle with Rstudio and would like to know how I can complete these equations by using Rstudio and a .csv file rather than manually. Thankyou

1. A certain river, fed by various tributaries, usually has a pH of about 6.3 (slightly acidic).
Water_pH.csv contains pH values from randomly-selected sites on this river and its tributaries.

a) Is there evidence the average pH of this river has changed from the usual value of 6.3? Evaluate the evidence using a hypothesis test with a significance level of 5%. State and evaluate all assumptions, and provide an appropriate conclusion.

b) Provide a 99% confidence interval for the mean pH in this river system.

pH
6.2
6.3
5
5.8
4.6
4.7
4.7
5.4
6.2
6
5.4
5.9
6.2
6.1
6
6.3
6.2
5.8
6.2
6.3
6.3
6.3
6.4
6.5
6.6
6.1
6.3
7.4
6.7

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	6.2	0.0414
	6.3	0.0921
	5	0.9932
	5.8	0.0387
	4.6	1.9505
	4.7	1.6812
	4.7	1.6812
	5.4	0.3559
	6.2	0.0414
	6	0
	5.4	0.3559
	5.9	0.0093
	6.2	0.0414
	6.1	0.0107
	6	0
	6.3	0.0921
	6.2	0.0414
	5.8	0.0387
	6.2	0.0414
	6.3	0.0921
	6.3	0.0921
	6.3	0.0921
	6.4	0.1627
	6.5	0.2534
	6.6	0.3641
	6.1	0.0107
	6.3	0.0921
	7.4	1.9695
	6.7	0.4948
Total	173.9	11.1301

Mean X̅ = Σ Xi / n
X̅ = 173.9 / 29 = 5.9966
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 11.1301 / 29 -1 ) = 0.6305

To Test :-

H0 :-  $\mu = 6.3$

H1 :-  

μメ6.3

Test Statistic :-
$t = ( \bar{X} - \mu ) / (S /\sqrt{n})$

t = -2.5914

t- (5.9966- 6.3)/(0.6305/V29)

Test Criteria :-
Reject null hypothesis if $| t |\; > \;t_{\alpha /2, n-1}$
$t_{\alpha /2, n-1} = t_{0.05 /2, 29-1} = 2.048$

Result :- Reject null hypothesis

리 > ta/2.n-1=2.5914 > 2.048

Decision based on P value
P - value = P ( t > 2.5914 ) = 0.015
Reject null hypothesis if P value < level of significance
P - value = 0.015 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

a-0.05

There is sufficient evidence to support the claim that the average pH of this river has changed from the usual value of 6.3.

Confidence Interval
$\bar{X} \pm t_{\alpha /2, n-1} S/\sqrt{n}$

o/2,n-1 -to.o1/2,29-1 2.763

$5.9966 \pm t_{\ 0.01/2, 29 -1} * 0.6305/\sqrt{ 29}$
Lower Limit = $5.9966 - t_{\ 0.01/2, 29 -1}0.6305/\sqrt{ 29}$
Lower Limit = 5.6731
Upper Limit = $5.9966 + t_{\ 0.01/2, 29 -1}0.6305/\sqrt{ 29}$
Upper Limit = 6.3201
99% Confidence interval is ( 5.6731 , 6.3201 )