Question

(1 point) Samples were collected from two ponds in the Bahamas to compare salinity values (in parts per thousand). Several samples were drawn at each site.

Pond 1: 37.02, 36.72, 37.03, 38.85, 36.75, 37.54, 37.32

Pond 2: 38.71, 38.53, 39.21, 39.05, 38.89
Use a 0.050.05 significance level to test the claim that the two ponds have the same mean salinity value.

(a) The test statistic is  .

(b) The conclusion is

A. There is not sufficient evidence to indicate that the two ponds have different salinity values.
B. There is sufficient evidence to indicate that the two ponds have different salinity values.

(c) We should

A. not take the results too seriously since neither sample is big enough to be meaningful.
B. remove the largest and smallest values from the larger data set and only test equal size samples.
C. check to see if the data appear close to Normal since the sum of the sample sizes is less than 15.
D. All of the above.

Answer 1

For Pond 1 :

∑x = 261.23

∑x² = 9751.979

n1 = 7

Mean , x̅1 = Ʃx/n = 261.23/7 = 37.3186

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(9751.9787-(261.23)²/7)/(7-1)] = 0.7358

For Pond 2 :

∑x = 194.39

∑x² = 7557.784

n2 = 5

Mean , x̅2 = Ʃx/n = 194.39/5 = 38.8780

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(7557.7837-(194.39)²/5)/(5-1)] = 0.2689

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Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((7-1)*0.7358² + (5-1)*0.2689²) / (7+5-2) = 0.3538

a) Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (37.3186 - 38.878) / √(0.3538*(1/7 + 1/5)) = -4.4777

df = n1+n2-2 = 10

p-value :

Two tailed p-value = T.DIST.2T(ABS(-4.4777), 10) = 0.0012

Conclusion:

Answer B. There is sufficient evidence to indicate that the two ponds have different salinity values.

c) Answer: C. check to see if the data appear close to Normal since the sum of the sample sizes is less than 15.