(1 point) Below are samples that were collected from two ponds in the Bahamas to
ID: 3062550 • Letter: #
Question
(1 point) Below are samples that were collected from two ponds in the Bahamas to compare salinity values (in parts per thousand) Pond 1: 37.54, 36.72, 38.85, 37.02, 37.45, 37.36, 37.71 Pond 2: 38.53, 39.04, 39.21, 38.51, 40.08 use a 0.05 sgnaicance evettea o a same mean sainity aue The test statistic is 4.010278161 The conclusion is A. There is sufficient evidence to indicate that the two ponds have different salinity values. o B. There is not sufficient evidence to indicate that the two ponds have different salinity values. We shoulod O 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. O 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.Explanation / Answer
The statistical software output for this problem is:
Two sample T hypothesis test:
1 : Mean of Pond 1
2 : Mean of Pond 2
1 - 2 : Difference between two means
H0 : 1 - 2 = 0
HA : 1 - 2 0
(without pooled variances)
Hypothesis test results:
Hence,
Test statistic = -4.0468334
Conclusion: Option A is correct.
We should: Option D is correct.
Difference Sample Diff. Std. Err. DF T-Stat P-value 1 - 2 -1.5525714 0.38365094 9.0464295 -4.0468334 0.0029Related Questions
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