Question

In: Statistics and Probability

Park Acreage A state executive claims that the average number of acres in western Pennsylvania state...

Park Acreage A state executive claims that the average number of acres in western Pennsylvania state parks is less than 2000 acres. A random sample of 12 parks is selected, and the number of acres is shown. 2075 2065 2174 1934 2265 1936 2277 1968 2183 2029 1976 2151

a. At = 0.01, is there enough evidence to support the claim? Use the P-value method

b. Find the 99% confidence interval of the true mean.

Solutions

Expert Solution

Mean X̅ = Σ Xi / n
X̅ = 25033 / 12 = 2086.0833


Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 165338.9162 / 12 -1 ) = 122.6002

To Test :-

H0 :- µ >= 2000
H1 :- µ < 2000

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 2086.0833 - 2000 ) / ( 122.6002 / √(12) )
t = 2.4323

Decision based on P value
P - value = P ( t > 2.4323 ) = 0.9834
Reject null hypothesis if P value < α = 0.01 level of significance
P - value = 0.9834 > 0.01 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

There is insufficient evidence to support the claim that the average number of acres in western Pennsylvania state parks is less than 2000 acres.

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 12- 1 ) = 3.106
2086.0833 ± t(0.01/2, 12 -1) * 122.6002/√(12)
Lower Limit = 2086.0833 - t(0.01/2, 12 -1) 122.6002/√(12)
Lower Limit = 1976.1569
Upper Limit = 2086.0833 + t(0.01/2, 12 -1) 122.6002/√(12)
Upper Limit = 2196.0097
99% Confidence interval is ( 1976.1569 , 2196.0097 )


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