Question

A meteorologist has been studying the effects of Rivers and Mountains on local weather patters. She measured the amount of precipitation for the months of August - January in the Idaho cities of Bonners Ferry and Moyle Springs (approximately 10 miles east of Bonners Ferry) and Libby. Montana which is approximately 40 miles East of Bonners Ferry. Is there evidence to suggest a difference in precipitation between these three cities? Bonners Ferry 0.94 1.31 Libby 0.91 Moyie Springs 1.07 1.16 1.61 3.03 1.18 1.5 1.92 2.44 3 3.09 2.91 2.7 2.2 1.77 2.98 Yes. The data was not normal (based on the normal plots) so I used the Kruskal-Wallistest. The p-value is 0.4843 so we would reject the null hypothesis (statement of equality between all three cities) and conclude there is enough evidence to suggest a difference in percipitation between these three cities. No. The data was not normal (based on the normal plots) so I used the Kruskal-Wallis test. The p-value is 0.4843 so we would fail to reject the null hypothesis (statement of equality between all three cities) and conclude there is not enough evidence to suggest a difference in percipitation between these three cities. No. The data was not normal (based on the normal plots) so I used the ANOVA test. The p-value is 0.5133 so we would fail to reject the null hypothesis (statement of equality between all three cities) and conclude there is not enough evidence to suggest a difference in percipitation between these three cities. No. The data wasreasonably normal (based on the normal plots) so I used the ANOVA test. The p-value is 0.5133 so rejected the null hypothesis (statement of equality between all three cities) and concluded there is enough evidence to suggest a difference in percipitation between these three cities.

Answer 1

Sol:

	1	2	3
count, ni =	6	6	6
mean , x̅ i =	2.207	2.08	1.67
std. dev., si =	0.949	0.901	0.588
sample variances, si^2 =	0.900	0.813	0.346
total sum	13.24	12.48	10		35.72
grand mean , x̅̅ =	Σni*x̅i/Σni =	1.98
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	0.049	0.009	0.101
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	0.296	0.055	0.606		0.956978
SS(within ) = SSW = Σ(n-1)s² =	4.498	4.063	1.730		10.292

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   18
df within = N-k =   15
  
mean square between groups , MSB = SSB/k-1 =    0.4785
  
mean square within groups , MSW = SSW/N-k =    0.6861
  
F-stat = MSB/MSW =    0.6974

anova table
	SS	df	MS	F	p-value	F-critical
Between:	0.957	2	0.478	0.70	0.5133	3.68
Within:	10.292	15	0.686
Total:	11.249	17
					α =	0.05

Ho: µ1=µ2=µ3
H1: not all means are equal

F stat = 0.70

p value>α, do not reject Ho

there is no enough evidence to suggest that there is a significant difference

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