Question

1. Cloud seeding has been studied for many decades as a weather modification procedure. The rainfall in acre-feet from 20 clouds that were selected at random and seeded with Silver Nitrate follows:

18.0        30.7        19.8        27.1        22.3        18.8        31.8        23.4        21.2        27.9       

31.9        27.1        25.0        24.7        26.9        21.8        29.2        34.8        26.7        31.6       

We need to test if the true mean rainfall exceeds 25 acre-feet.

a. Write down the null and alternative hypotheses to be tested. Clearly define the terms used.

b. What is the type of statistical test procedure that should be used to test the hypotheses? Explain.

c. Construct a 95% confidence interval. Test the hypotheses using the confidence interval. Interpret the test result clearly

d. Test the hypotheses using the Test Statistic / Critical Value method (you must clearly indicate the test statistic, and the critical value(s) use Take α=5%). Do you get the same answer as part (c)? If your answer is the same as (c), you do not have to interpret again. Otherwise, interpret the test findings.

e. What is the P value of the test? Test the hypotheses using the P value. Take α=5%. Do you get the same answer as parts (c) and (d)? If your answer is the same as (c) and (d), you do not have to interpret again. Otherwise, interpret the test findings.

Answer 1

a.
Given that,
population mean(u)=25
sample mean, x =26.035
standard deviation, s =4.663
number (n)=20
null, Ho: μ=25
alternate, H1: μ>25
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.729
since our test is right-tailed
reject Ho, if to > 1.729
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =26.035-25/(4.663/sqrt(20))
to =0.993
| to | =0.993
critical value
the value of |t α| with n-1 = 19 d.f is 1.729
we got |to| =0.993 & | t α | =1.729
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 0.9926 ) = 0.16668
hence value of p0.05 < 0.16668,here we do not reject Ho
ANSWERS
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a.
null, Ho: μ=25
alternate, H1: μ>25
b.
T test for single mean with unknown population standard deviation
d.
test statistic: 0.993
critical value: 1.729
decision: do not reject Ho
e.
p-value: 0.16668
we do not have enough evidence to support the claim that if the true mean rainfall exceeds 25 acre-feet.
c.
TRADITIONAL METHOD
given that,
sample mean, x =26.035
standard deviation, s =4.663
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.663/ sqrt ( 20) )
= 1.043
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.093
margin of error = 2.093 * 1.043
= 2.182
III.
CI = x ± margin of error
confidence interval = [ 26.035 ± 2.182 ]
= [ 23.853 , 28.217 ]
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DIRECT METHOD
given that,
sample mean, x =26.035
standard deviation, s =4.663
sample size, n =20
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.093
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 26.035 ± t a/2 ( 4.663/ Sqrt ( 20) ]
= [ 26.035-(2.093 * 1.043) , 26.035+(2.093 * 1.043) ]
= [ 23.853 , 28.217 ]
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interpretations:
1) we are 95% sure that the interval [ 23.853 , 28.217 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean