Question

In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.

At five weather stations on Trail Ridge Road in Rocky Mountain National Park, the peak wind gusts (in miles per hour) for January and April are recorded below.

Weather Station	1	2	3	4	5
January	139	120	126	64	78
April	108	115	100	88	61

Does this information indicate that the peak wind gusts are higher in January than in April? Use α = 0.01. (Let

d = January − April.)(a) What is the level of significance?


State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?

H₀: μ_d = 0; H₁: μ_d ≠ 0; two-tailed H₀: μ_d = 0; H₁: μ_d > 0; right-tailed     H₀: μ_d = 0; H₁: μ_d < 0; left-tailed H₀: μ_d > 0; H₁: μ_d = 0; right-tailed

(b) What sampling distribution will you use? What assumptions are you making?

The standard normal. We assume that d has an approximately uniform distribution. The Student's t. We assume that d has an approximately uniform distribution.     The standard normal. We assume that d has an approximately normal distribution. The Student's t. We assume that d has an approximately normal distribution.

What is the value of the sample test statistic? (Round your answer to three decimal places.)

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud > 0
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =3.747
since our test is right-tailed
reject Ho, if to > 3.747
we use Test Statistic  
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 15
We have d = 15
pooled standard deviation = calculate value of Sd= √S^2 = sqrt [ 1967-(75^2/5 ] / 4 = 14.509
to = d/ (S/√n) = 2.312
critical Value
the value of |t α| with n-1 = 4 d.f is 3.747
we got |t o| = 2.312 & |t α| =3.747
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :right tail - Ha : ( p > 2.3118 ) = 0.04094
hence value of p0.01 < 0.04094,here we do not reject Ho
ANSWERS
---------------
a.
level of significance =0.01
null, H0: Ud = 0
alternate, H1: Ud > 0
b.
We assume that d has an approximately normal distribution.
test statistic: 2.312
critical value: reject Ho, if to > 3.747
decision: Do not Reject Ho
p-value: 0.04094
we do not have enough evidence to support the claim that information indicate that the peak wind gusts are higher in January than in April