Homework: Week 7 Homework 40 pts Sa Score: 125 of 5 pts 10 of 14 (14 complete) H

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Homework: Week 7 Homework 40 pts Sa Score: 125 of 5 pts 10 of 14 (14 complete) HW Score: 40 54%,16 22 of 40 7.2.35-T tio is 63 years Ato-001, is there enough ovide ce n fthe da fa The len hs o ine nyears at took arandom sa deo 32fo mers kers toqit smoking pe maren ya eli ted the mean time it takes smokers to quit smoking permanenly is 14 years? Complete parts (a) through (e) methe population tandard d 173 153 158 162 199 199 179 225 213 117 211 116 179 211 138 88 142 198 73 183 202 112 188 13.8 209 71 98 81 224 92 97 91 (a) Identily the claim and state the null hypothvesis and altenative hypothesh OB, Ho ) 14(claim) H, ps 14 (daim) H. p> 14 b) ldenily the standardized test staisic Use technokogy z(Round to two decimal places as needed ) Emer your answer in the answer box and then click Check Ar Clea All

Explanation / Answer

Given that,
population mean(u)=14
standard deviation, =6.3
sample mean, x =15.36875
number (n)=32
null, Ho: =14
alternate, H1: <14
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.33
since our test is left-tailed
reject Ho, if zo < -2.33
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 15.36875-14/(6.3/sqrt(32)
zo = 1.23
| zo | = 1.23
critical value
the value of |z | at los 1% is 2.33
we got |zo| =1.23 & | z | = 2.33
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : left tail - ha : ( p < 1.23 ) = 0.89
hence value of p0.01 < 0.89, here we do not reject Ho
ANSWERS
---------------
null, Ho: =14
alternate, H1: <14
test statistic: 1.23
critical value: -2.33
decision: do not reject Ho
p-value: 0.89
we do not have enough evidence to support the claim

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