Newly manufactured automobile tires of a certain type are supposed to be filled
ID: 3311802 • Letter: N
Question
Newly manufactured automobile tires of a certain type are supposed to be filled to a pressure of 34 psi. Toavoid over-filling, the manufacturer wants to perform a test on the average pressure filled. To do so, theytake a random sample of 36 tires and average pressure is calculated to be 34.66 psi. Assume that the tirepressures are normally distributed, with standard deviation 2.53 psi.
1. What is the population of interest? What is the parameter of interest for the population? Denote itby.
2. State the null and alternative hypotheses in terms of the parameterabove.
3. Which method would you use? (a) 1-sample T(b) 1-sample Z(c) 2-sample T(d) 2-sample Z
4. Calculate thepvalueof the above testing procedure and make your decision based on= 0.01 in thecontext of the problem.
5. Interpret the abovepvaluein the context.
6. What is the rejection region in terms of the sample average X?
7. Suppose that the true average filled pressure is actually 35. For the test above, what is the power tosuccessfully detect the alternative.
Need help on 4, 5, 6, 7
Explanation / Answer
4.
Given that,
population mean(u)=34
standard deviation, =2.53
sample mean, x =34.66
number (n)=36
null, Ho: =34
alternate, H1: !=34
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 34.66-34/(2.53/sqrt(36)
zo = 1.57
| zo | = 1.57
critical value
the value of |z | at los 1% is 2.576
we got |zo| =1.57 & | z | = 2.576
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.57 ) = 0.12
hence value of p0.01 < 0.12, here we do not reject Ho
ANSWERS
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null, Ho: =34
alternate, H1: !=34
test statistic: 1.57
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.12
5.
we conclude that newly manfcatured tires are filled to a pressure of 34 psi
6.
reject Ho, if zo < -2.576 OR if zo > 2.576
7.
Given that,
Standard deviation, =2.53
Sample Mean, X =34.66
Null, H0: =34
Alternate, H1: !=34
Level of significance, = 0.01
From Standard normal table, Z /2 =2.5758
Since our test is two-tailed
Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758
Reject Ho if (x-34)/2.53/(n) < -2.5758 OR if (x-34)/2.53/(n) > 2.5758
Reject Ho if x < 34-6.516774/(n) OR if x > 34-6.516774/(n)
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Suppose the size of the sample is n = 36 then the critical region
becomes,
Reject Ho if x < 34-6.516774/(36) OR if x > 34+6.516774/(36)
Reject Ho if x < 32.913871 OR if x > 35.086129
Implies, don't reject Ho if 32.913871 x 35.086129
Suppose the true mean is 35
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(32.913871 x 35.086129 | 1 = 35)
= P(32.913871-35/2.53/(36) x - / /n 35.086129-35/2.53/(36)
= P(-4.9473415 Z 0.2042585 )
= P( Z 0.2042585) - P( Z -4.9473415)
= 0.5809 - 0 [ Using Z Table ]
= 0.5809
For n =36 the probability of Type II error is 0.5809
Power of test = 1- Type II error = 1 - 0.5809 = .4191