i JJ (JJ complete) HW Score: 91.67%, 30 25 of 33 pts 8.4.15-T Question Help * As
ID: 3358314 • Letter: I
Question
i JJ (JJ complete) HW Score: 91.67%, 30 25 of 33 pts 8.4.15-T Question Help * Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and altermative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. simple random sample of 25 filtered 100 mm cigarettes is obtained, and the tar content of each cigareteis measured. The sample has a mean of 196 mg and a ndard deviation of 3.25 mg. Use a 0.05 significance level to test the claim that the mean tar content of filtered 1 00 mm cigarettes is less than 21.1 mg, which is the mean for unfiltered king size cigarettes. What do the results suggest, if anything, about the effectiveness of the filters? What are the hypotheses? A. Ho H21.1 mg H1Explanation / Answer
8.4.15.T.
Given that,
population mean(u)=21.1
sample mean, x =19.6
standard deviation, s =3.25
number (n)=25
null, Ho: =21.1
alternate, H1: <21.1
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.711
since our test is left-tailed
reject Ho, if to < -1.711
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =19.6-21.1/(3.25/sqrt(25))
to =-2.3077
| to | =2.3077
critical value
the value of |t | with n-1 = 24 d.f is 1.711
we got |to| =2.3077 & | t | =1.711
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -2.3077 ) = 0.01497
hence value of p0.05 > 0.01497,here we reject Ho
ANSWERS
---------------
i)
null, Ho: =21.1
alternate, H1: <21.1
ii)
test statistic: -2.3077 = -2.308
critical value: -1.711
decision: reject Ho
iii).
p-value: 0.01497 = 0.0145
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