do u have to u use t distubition? and how do i solve thanks Problem 6 (8 points)

Question

do u have to u use t distubition? and how do i solve thanks

Problem 6 (8 points) A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally distributed, with standard deviation 0.25 Volts, and the manufacturer wishes to test using n-8 units. a) The acceptance region is 4.85 5.15 . Find the value of. b) Find the power of the test for detecting a true mean output voltage of 5.1 Volts. c) Find the boundary of the critical region if the type 1 error probability is -001. d) Find the probability of a type II error if the true mean output is 5.05 volts and -001 and n-16

Explanation / Answer

6.

All i used Z test only
a.
given that,
standard deviation, =0.25
mean U =5
population size (n)=8
= [ 4.85 < X <5.15 ] region then we get level of significance
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
Assume
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
margin of error = 1.645 * 0.088
= 0.145
[ 4.85 < X <5.15 ] region we get when level of significance, = 0.1

b.
Given that,
Standard deviation, =0.25
Sample Mean, X =5.1
Null, H0: =5
Alternate, H1: !=5
Level of significance, = 0.1
From Standard normal table, Z /2 =1.6449
Since our test is two-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-5)/0.25/(n) < -1.6449 OR if (x-5)/0.25/(n) > 1.6449
Reject Ho if x < 5-0.411/(n) OR if x > 5-0.411/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 8 then the critical region
becomes,
Reject Ho if x < 5-0.411/(8) OR if x > 5+0.411/(8)
Reject Ho if x < 4.855 OR if x > 5.145
Implies, don't reject Ho if 4.855 x 5.145
Suppose the true mean is 5.1
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(4.855 x 5.145 | 1 = 5.1)
= P(4.855-5.1/0.25/(8) x - / /n 5.145-5.1/0.25/(8)
= P(-2.772 Z 0.509 )
= P( Z 0.509) - P( Z -2.772)
= 0.6946 - 0.0028 [ Using Z Table ]
= 0.692
For n =8 the probability of Type II error is 0.692
power of the test = 1- beta =1-0.692
power of the test = 0.308

c.
Given that,
Standard deviation, =0.25
Sample Mean, X =5.1
Null, H0: =5
Alternate, H1: !=5
Level of significance, = 0.01
From Standard normal table, Z /2 =2.576
Since our test is two-tailed
Reject Ho, if Zo < -2.576 OR if Zo > 2.576
Reject Ho if (x-5)/0.25/(n) < -2.576 OR if (x-5)/0.25/(n) > 2.576
Reject Ho if x < 5-0.644/(n) OR if x > 5-0.644/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 8 then the critical region
becomes,
Reject Ho if x < 5-0.644/(8) OR if x > 5+0.644/(8)
Reject Ho if x < 4.772 OR if x > 5.228
Suppose the true mean is 5.1
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(4.772 < x OR x >5.228 | 1 = 5.1)
= P(4.772-5.1/0.25/(8) < x - / /n OR x - / /n >5.228-5.1/0.25/(8)
= P(-3.711 < Z OR Z >1.448 )
= P( Z <-3.711) + P( Z > 1.448)
= 0.0001 + 0.0738 [ Using Z Table ]
= 0.074

d.
Given that,
Standard deviation, =0.25
Sample Mean, X =5.05
Null, H0: =5
Alternate, H1: !=5
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-5)/0.25/(n) < -2.5758 OR if (x-5)/0.25/(n) > 2.5758
Reject Ho if x < 5-0.644/(n) OR if x > 5-0.644/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 16 then the critical region
becomes,
Reject Ho if x < 5-0.644/(16) OR if x > 5+0.644/(16)
Reject Ho if x < 4.839 OR if x > 5.161
Implies, don't reject Ho if 4.839 x 5.161
Suppose the true mean is 5.05
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(4.839 x 5.161 | 1 = 5.05)
= P(4.839-5.05/0.25/(16) x - / /n 5.161-5.05/0.25/(16)
= P(-3.376 Z 1.776 )
= P( Z 1.776) - P( Z -3.376)
= 0.9621 - 0.0004 [ Using Z Table ]
= 0.962
For n =16 the probability of Type II error is 0.962

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