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a certain type of oak tree has a mean growth of 14.3 inches in 4 years. a forest biologist believes a new variety has a greater mean growth during the same length of time. a 36 tree sample of the new variety showed a mean growth of 15.9 inches with a sample standard deviation of 1.6 inches. can it be concluded the biologist is correct at alpha 5% or 1% 
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The method I use requires a graphing calculator with the t-test function. A TI-84 or TI-89 should work.
The p-value is essentially the probability that the new variety does NOT have a greater mean growth than the old variety. The answer depends on whether p<0.01; if it is, then the biologist is correct at alpha=1%. If 0.05>p>0.01, then the biologist is correct at alpha=5%.
Because population standard deviation is unknown, use a one-sample t-test. The given information is:
x-bar=15.9, s=1.6, n=36, u0=14.3
From this the following can be found:
SE = s/sqrt(n) = 1.6/sqrt(36)
df = n-1 = 35
Use a t-test with x-bar, SE, u0and df as given or found. The appropriate test is >u0. Find the p-value to find your answer. I get p=3.86*10^-7. This is much lower than 0.01, so the biologist is correct even at alpha=1%.