A letter arrives at the office of a law enforcement agency stating: \"This is to

Question

A letter arrives at the office of a law enforcement agency stating: "This is to inform you that I cracked the security code of one of your city's largest bank. I will continue to transfer deposits. So far I retrieved $78.20, 118.87, 1.25, 83.00, 123.5, 34.90, 10.00, 55.40, 100.80, 62.85 If you can tell me which bank I broke in, I shall return the money. Otherwise they are mine." (a) A police officer checks whether the money could have come from the Capital Bank where the average deposit is $86. Should the money belong to the bank he wants to be 95% sure not to declare it cam from another bank. (b) What was the power of the police officer's test if the bank from which the deposits were actually stolen has an average deposit differing from that of the Capital Bank by $23? (c) Another possible bank is the Citizens Bank where deposits have a standard deviation of $108. The officer decides to test H0 : 108 versus H1 : 108 at the 5% level (d) Using the sample standard deviation as the best guess of , estimate the power of the test in (c (Notc: In casc it is not covercd yct in class, for parts (c) and (d), you nced to rcad Scction 8.2.3 from the class notes.

Explanation / Answer

a.
Given that,
population mean(u)=86
sample mean, x =66.877
standard deviation, s =42.375
number (n)=10
null, Ho: =86
alternate, H1: !=86
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.2622
since our test is two-tailed
reject Ho, if to < -2.2622 OR if to > 2.2622
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =66.877-86/(42.375/sqrt(10))
to =-1.427
| to | =1.427
critical value
the value of |t | with n-1 = 9 d.f is 2.2622
we got |to| =1.427 & | t | =2.2622
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.4271 ) = 0.1873
hence value of p0.05 < 0.1873,here we do not reject Ho
ANSWERS
---------------
null, Ho: =86
alternate, H1: !=86
test statistic: -1.427
critical value: -2.2622 , 2.2622
decision: do not reject Ho
p-value: 0.1873
b.
Given that,
Standard deviation, =42.375
Sample Mean, X =66.877
Null, H0: =23
Alternate, H1: !=23
Level of significance, = 0.05
From Standard normal table, Z /2 =1.96
Since our test is two-tailed
Reject Ho, if Zo < -1.96 OR if Zo > 1.96
Reject Ho if (x-23)/42.375/(n) < -1.96 OR if (x-23)/42.375/(n) > 1.96
Reject Ho if x < 23-83.055/(n) OR if x > 23-83.055/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 10 then the critical region
becomes,
Reject Ho if x < 23-83.055/(10) OR if x > 23+83.055/(10)
Reject Ho if x < -3.264 OR if x > 49.264
Implies, don't reject Ho if -3.264 x 49.264
Suppose the true mean is 66.877
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(-3.264 x 49.264 | 1 = 66.877)
= P(-3.264-66.877/42.375/(10) x - / /n 49.264-66.877/42.375/(10)
= P(-5.234 Z -1.314 )
= P( Z -1.314) - P( Z -5.234)
= 0.0944 - 0 [ Using Z Table ]
= 0.094
For n =10 the probability of Type II error is 0.094
power of the test = 1- beta = 1-0.094 = 0.906
c.
Given that,
population standard deviation ()=108
sample standard deviation (s) =42.375
sample size (n) = 10
we calculate,
population variance (^2) =11664
sample variance (s^2)=1795.640625
null, Ho: =108
alternate, H1 : <108
level of significance, = 0.05
from standard normal table,left tailed ^2 /2 =16.919
since our test is left-tailed
reject Ho, if ^2 o < -16.919
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(10 - 1 ) * 1795.640625 / 11664 = 9*1795.640625/11664 = 1.386
| ^2 cal | =1.386
critical value
the value of |^2 | at los 0.05 with d.f (n-1)=9 is 16.919
we got | ^2| =1.386 & | ^2 | =16.919
make decision
hence value of | ^2 cal | < | ^2 | and here we do not reject Ho
^2 p_value =0.9979
ANSWERS
---------------
null, Ho: =108
alternate, H1 : <108
test statistic: 1.386
critical value: -16.919
p-value:0.9979
decision: do not reject Ho
d.
Given that,
Standard deviation, =108
Sample Mean, X =66.877
Null, H0: =86
Alternate, H1: !=86
Level of significance, = 0.05
From Standard normal table, Z /2 =1.96
Since our test is two-tailed
Reject Ho, if Zo < -1.96 OR if Zo > 1.96
Reject Ho if (x-86)/108/(n) < -1.96 OR if (x-86)/108/(n) > 1.96
Reject Ho if x < 86-211.68/(n) OR if x > 86-211.68/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 10 then the critical region
becomes,
Reject Ho if x < 86-211.68/(10) OR if x > 86+211.68/(10)
Reject Ho if x < 19.061 OR if x > 152.939
Implies, don't reject Ho if 19.061 x 152.939
Suppose the true mean is 66.877
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(19.061 x 152.939 | 1 = 66.877)
= P(19.061-66.877/108/(10) x - / /n 152.939-66.877/108/(10)
= P(-1.4 Z 2.52 )
= P( Z 2.52) - P( Z -1.4)
= 0.9941 - 0.0808 [ Using Z Table ]
= 0.913
For n =10 the probability of Type II error is 0.913
power of the test = 1- beta = 1-0.913 = 0.087

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