Question

In: Statistics and Probability

Suppose the Annual rainfall (in inches) in 5 different locations in Hawaii are: 19, 27, 18,...

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?

Solutions

Expert Solution

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


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