In: Statistics and Probability
Suppose the Annual rainfall (in inches) in 5 different locations in Hawaii are: 19, 27, 18, 16, 13. A statistician wants to use this data to test the claim that the average rainfall in Hawaii is more than 16 inches. Assume significance level α = 0.05 .
a) What is the alternative hypothesis H 1? What tailed test is this?
b) Determine the value of the Test Statistic.
c) What is the P-value?
d) Should the statistician Reject or Fail to Reject the Null Hypothesis?
e) What is the conclusion in plain English? (use the summary table provided in class)
f) Calculate the 90% Confidence Interval for the average rainfall in Hawaii
g) Are the confidence limits (in part (f)) consistent with the conclusion in part (e)? Why or why not?
Solution:
a)
H1 : > 16
Right tailed test(One tailed right sided)
b)
n = 5
19, 27, 18, 16, 13
Using calculator ,
Sample mean = = 18.6
Sample SD = s = 5.22494019
The test statistic t is
t = = [18.6 - 16]/[5.22494019/5] = 1.113
The value of the test statistic t = 1.113
c)
d.f. = n - 1 = 5 - 1 = 4
One tailed right sided test
t = 1.113
p value = 0.1640
d)
Fail to reject H0
(Because p value 0.1640 is greater than given significance level α = 0.05 )
e)
Conclusion: There is not sufficient evidence to support the claim that the average rainfall in Hawaii is more than 16 inches.
f)
c = 90% = 0.90
= 1- c = 1- 0.90 = 0.10
/2 = 0.10 2 = 0.05
Also, d.f = n - 1 = 5 - 1 = 4
= = 0.05,4 = 2.132
( use t table or t calculator to find this value..)
The margin of error is given by
E = /2,d.f. * ( / n)
= 2.132 * (5.22494019/ 5)
= 4.9818
Now , confidence interval for mean() is given by:
( - E ) < < ( + E)
(18.6 - 4.9818) < < (18.6 + 4.9818)
13.6182 < < 23.5818
Required 90% confidence interval is (13.6182 , 19.5124)
g)
Results are consistent