Question

xtra fees imposed automated tellet 2 The Federal an such fee is $1.14. is vielded a mean CULATIONS AND INTERPRETATIONS 10. ATM Fees. Do you hate paying the extra fe by banks when withdrawing funds from an auto machine (ATM) not owned by your bank? The Fe Reserve System reports that the mean such fee is A random sample of 36 such transactions yielder of $1.07 in extra fees. Suppose the population standa deviation of such extra fees is $0.25. a. Test using level of significance a = 0.05 wheth there has been a reduction in the population mean fee charged on such transactions. b. Which type of error is it possible that we are making • a Type I error or a Type II error? Which type of erro are we certain we are not making? 11. Alcohol-Related Fatal Car Accidents. The National Traffic Highway Safety Commission keeps statistics on the "mean years of potential life lost" in alcohol-related fatal automobile accidents. For males the mean years of life lost is 32. That is, on average, males involved in fatal drinking- and-driving accidents had their lives cut short by 32 years. A random sample of 36 alcohol-related fatal accidents had a mean years of life lost of 33.8, with a standard deviation of 6 years. a. Test whether the population mean years of life lost has changed, using a t test and level of significance a = 0.10. b. Assess the strength of the evidence against th hypothesis. sing

Answer 1

10.
Given that,
population mean(u)=1.14
standard deviation, σ =0.25
sample mean, x =1.07
number (n)=36
null, Ho: μ=1.14
alternate, H1: μ<1.14
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1.07-1.14/(0.25/sqrt(36)
zo = -1.68
| zo | = 1.68
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =1.68 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -1.68 ) = 0.05
hence value of p0.05 < 0.05, here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: μ=1.14
alternate, H1: μ<1.14
test statistic: -1.68
critical value: -1.645
decision: reject Ho
p-value: 0.05
we have enough evidence to support the claim that whether there has been reduction in the
population mean fee charged on such transactions.
b.
type 1 error is possible because it reject the null hypothesis.
type 2 error is not possible.
(if type 2 error is possible only it fails to reject the null hypothesis)

11.
Given that,
population mean(u)=32
sample mean, x =33.8
standard deviation, s =6
number (n)=36
null, Ho: μ=32
alternate, H1: μ!=32
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.69
since our test is two-tailed
reject Ho, if to < -1.69 OR if to > 1.69
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =33.8-32/(6/sqrt(36))
to =1.8
| to | =1.8
critical value
the value of |t α| with n-1 = 35 d.f is 1.69
we got |to| =1.8 & | t α | =1.69
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.8 ) = 0.0805
hence value of p0.1 > 0.0805,here we reject Ho
ANSWERS
---------------
a.
null, Ho: μ=32
alternate, H1: μ!=32
test statistic: 1.8
critical value: -1.69 , 1.69
decision: reject Ho
p-value: 0.0805
b.
we have enough evidence to support the claim that whether the population mean years of life
lost has changed