12) For the next two problems, identify the null and alternative hypotheses, cal
ID: 3324681 • Letter: 1
Question
12) For the next two problems, identify the null and alternative hypotheses, calculate the appropriate test statistic, find the p-value (or critical value(s) of the appropriate distribution), make a decision about the null hypothesis and write a final conclusion that addresses the original claim. Assume the samples come from populations that are normally distributed. In 2002, the mean expenditure for auto insurance in a certain state was $806. An insurance salesperson in this state believes that the mean expenditure for auto insurance is less today. She obtains a simple random sample of 32 auto insurance policies and determines the mean expenditure to be $781 with a standard deviation of $39.13. Is there enough evidence to support the claim that the mean expenditure for auto insurance is less than the 2002 amount at the 0.05 level of significance? 13) In 2000, 36% of adults in a certain country were morbidly obese. A health practitioner suspects that the percent is now higher. She obtains a random sample of 1042 adults and finds that 393 are morbidly obese. Is this sufficient evidence to support the practitioner's claim at the -.01 level of significance?Explanation / Answer
12.
Given that,
population mean(u)=806
sample mean, x =781
standard deviation, s =39.13
number (n)=32
null, Ho: =806
alternate, H1: <806
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.696
since our test is left-tailed
reject Ho, if to < -1.696
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =781-806/(39.13/sqrt(32))
to =-3.614
| to | =3.614
critical value
the value of |t | with n-1 = 31 d.f is 1.696
we got |to| =3.614 & | t | =1.696
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -3.6141 ) = 0.00053
hence value of p0.05 > 0.00053,here we reject Ho
ANSWERS
---------------
null, Ho: =806
alternate, H1: <806
test statistic: -3.614
critical value: -1.696
decision: reject Ho
p-value: 0.00053
we have enough evidence to support the claim that mean expenditure is less than the 2002 amount
13.
Given that,
possibile chances (x)=393
sample size(n)=1042
success rate ( p )= x/n = 0.3772
success probability,( po )=0.36
failure probability,( qo) = 0.64
null, Ho:p=0.36
alternate, H1: p>0.36
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.33
since our test is right-tailed
reject Ho, if zo > 2.33
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.37716-0.36/(sqrt(0.2304)/1042)
zo =1.154
| zo | =1.154
critical value
the value of |z | at los 0.01% is 2.33
we got |zo| =1.154 & | z | =2.33
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.15396 ) = 0.12426
hence value of p0.01 < 0.12426,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.36
alternate, H1: p>0.36
test statistic: 1.154
critical value: 2.33
decision: do not reject Ho
p-value: 0.12426
we do not have enough evidence to support the claim that practitioner suspects now that percent is higher
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