Question

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: 37.01, 37.71, 37.36, 36.72, 37.02, 37.45, 37.32

Pond 2: 38.24, 38.53, 38.66, 38.71, 39.21
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

Answer 1

as the population is different we use test statitistic in this case as

where X_bar1 and X_bar2 are the sample mean, S1^2nd S2^2re the sample variances and n_1 and n_2 are the number of sample

X_bar1 = 37.22714

X_bar2 = 38.67

S1^2 = 0.1093905

S2^2 = 0.12445

n1 = 7 n2 = 5

H0: mu1 = mu2 vs H1: mu1 not equals to mu2

after inserting all the values, we get t0 = -7.168092

and t_critical at alpha = 0.05 and n1 + n2 - 2 = 10 degrees of freedom is

t_critical = 2.228139

as t_critical is less than absolute value of t0 we reject H0, that implies they don't have same mean salinity at 5% level of significance