An oceanographer wants to test, on the basis of the mean of a random sample of size ? = 35 and at the 0.05 level of significance, whether the average depth ocean in a certain area is 72.4 fathoms, as has been recorded. What will she decide if she gets ?̅ = 73.2 fathoms, and she can assume from information gathered in similar studies that σ = 2.1 fathoms?
Solution:
Z = (X – µ)/(σ/√n)
Z = (73.2–72.4)/ (2.1/√35)
Z = 0.8/0.354964787
Z = 2.253744679
Z = 2.2537 (rounded)
=> For significance level of 0.05, Z is greater than or equal to 1.64, hence the null hypothesis should be rejected. The value is 2.3 fathoms, which is greater than the population mean of 2.1 fathoms.
An oceanographer wants to test, on the basis of the mean of a random sample of size ? = 35 and at the 0.05 level of sign...
An oceanographer wants to test, on the basis of the mean of a random sample of size ? = 35 and at the 0.05 level of significance, whether the average depth ocean in a certain area is 72.4 fathoms, as has been recorded. What will she decide if she gets ?̅ = 73.2 fathoms, and she can assume from information gathered in similar studies that σ = 2.1 fathoms?
An oceanographer has an old bathymetric map of an area and thinks that the depth it shows may be incorrect. She has modern equipment with her, and decides to conduct a test to see if the map’s value of 72.4 fathoms is correct. She conducts a random sample of depth at 35 locations in the area. Her sample mean was found to be 73.2 fathoms. Based on previous studies with this equipment, we can assume that σ = 2.1 fathoms....
please tell me if its a z test ke t test and why it is :) thanks! CHECK ANSWERS! You should use a Z test because of the large sample size 30 < and we know the mean and the standard deviation. 6. The Graded Naming Test (GNT) requires respondents to name objects in a set of 30 black-and-white drawings. The test is often used to detect brain damage in the UK. The test begins with easy words like kangaroo,...
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