PARTS A-C B) Determine the critical value(s). The critical value(s) is(are)= (Ro

Question

PARTS A-C

B) Determine the critical value(s).

The critical value(s) is(are)=

(Round to 3 decimal places)

C) Use technology to determine the p-value for this test.

p-value=

(Round to 3 decimal places )

Score: 0.67 of 1 pt 6 of 12 (5 complete) HW Score: 18.75%, 2.25 of 12 pts %) 9.3.16-T Question Help During the winter of 2008-2009, the average utility bill for residents of a certain state was $184 per month. A random sample of 60 customers was selected during the winter of 2009-2010, and the average bill was found to be $175.35 with a sample standard deviation of $20.64. Complete parts a and b below a) Using = 0.05, does this sample provide enough evidence to conclude that the average utility bill in this state was lower in the winter of 2009-2010 than it was in the winder of 2008-2009? etermine the hypotheses 184 B. Ho: #184 and H1 : = 184 OC. HOP 184 and H1:>184 D. Ho : 184 and H1 :

Explanation / Answer

9.3.16.

Given that,
population mean(u)=184
sample mean, x =175.35
standard deviation, s =20.64
number (n)=60
null, Ho: >=184
alternate, H1: <184
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.671
since our test is left-tailed
reject Ho, if to < -1.671
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =175.35-184/(20.64/sqrt(60))
to =-3.246
| to | =3.246
critical value
the value of |t | with n-1 = 59 d.f is 1.671
we got |to| =3.246 & | t | =1.671
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -3.2463 ) = 0.00096
hence value of p0.05 > 0.00096,here we reject Ho
ANSWERS
---------------
a.
null, Ho: >=184
alternate, H1: <184
test statistic: -3.246
b.
critical value: -1.671
decision: reject Ho
c.
p-value: 0.00096
we have enough evidence to support the claim that the average bill utility in this state is lower than in the winter

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