Statistics homework question. The maintenance costs of fixing all 3 of HP motors

Question

Statistics homework question. The maintenance costs of fixing all 3 of HP motors is normally distrawned with a standard deviation and their mean maintenance from the maintenance center located in Cleveland. cost was found to be S20. Alela n.OS motors the data in this sample support the elaim of of fixing all fixing all NHP is per hour? Perform a Hypothesis test, list all the required 5 steps to answer of the question including the standard normal d B. Repeat only the three required steps to test the same Hypothesis as in part A but uss ONIN ths r: JALUH as a reiectionnale here and include the standard normal distribution Points c actual mean ofthe population of hourly maintenance costs offixing all 3 HP motors is equal t If the information given in partA Points the lype ll error probability Bfor the

Explanation / Answer

Given that,
population mean(u)=21
sample mean, x =20
standard deviation, s =3
number (n)=25
null, Ho: =21
alternate, H1: !=21
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.064
since our test is two-tailed
reject Ho, if to < -2.064 OR if to > 2.064
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =20-21/(3/sqrt(25))
to =-1.667
| to | =1.667

PART A
critical value
the value of |t | with n-1 = 24 d.f is 2.064
we got |to| =1.667 & | t | =2.064
make decision
hence value of |to | < | t | and here we do not reject Ho

PART B
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.6667 ) = 0.1086
hence value of p0.05 < 0.1086,here we do not reject Ho

ANSWERS
---------------
null, Ho: =21
alternate, H1: !=21
test statistic: -1.667
critical value: -2.064 , 2.064
decision: do not reject Ho
p-value: 0.1086

PART C
Given that,
Standard deviation, =3
Sample Mean, X =20
Null, H0: =21
Alternate, H1: !=21
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-21)/3/(n) < -1.96 OR if (x-21)/3/(n) > 1.96
Reject Ho if x < 21-5.88/(n) OR if x > 21-5.88/(n)
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Suppose the size of the sample is n = 25 then the critical region
becomes,
Reject Ho if x < 21-5.88/(25) OR if x > 21+5.88/(25)
Reject Ho if x < 19.824 OR if x > 22.176
Implies, don't reject Ho if 19.824 x 22.176
Suppose the true mean is 23
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(19.824 x 22.176 | 1 = 23)
= P(19.824-23/3/(25) x - / /n 22.176-23/3/(25)
= P(-5.2933 Z -1.3733 )
= P( Z -1.3733) - P( Z -5.2933)
= 0.0848 - 0 [ Using Z Table ]
= 0.0848
For n =25 the probability of Type II error is 0.0848

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