Question

In: Statistics and Probability

A manufacturer is interested in the output voltage of a power supply used in a PC....

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 volt. The manufacturer wishes to test H0: μ=5 volts against H1: μ≠5 volts using α=0.10. Suppose a sample size of 8 power supply units is used and you are interested in computing the probability the test will fail to detect a true mean output voltage of 5.1 volts. What is the appropriate notation for the referred-to probability?

Selected Answer:

power

Answers:

P-value

power

Solutions

Expert Solution

a.
Given that,
population mean(u)=5
standard deviation, σ =0.25
sample mean, x =5.1
number (n)=8
null, Ho: μ=5
alternate, H1: μ!=5
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 5.1-5/(0.25/sqrt(8)
zo = 1.131
| zo | = 1.131
critical value
the value of |z α| at los 10% is 1.645
we got |zo| =1.131 & | z α | = 1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.131 ) = 0.258
hence value of p0.1 < 0.258, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=5
alternate, H1: μ!=5
test statistic: 1.131
critical value: -1.645 , 1.645
decision: do not reject Ho
p-value: 0.258
we do not have enough evidence to support the claim that population mean is equal to 5.
b.
Given that,
Standard deviation, σ =0.25
Sample Mean, X =5.1
Null, H0: μ=5
Alternate, H1: μ!=5
Level of significance, α = 0.1
From Standard normal table, Z α/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 < 5-0.4112/√(n) OR if x > 5-0.4112/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 8 then the critical region
becomes,
Reject Ho if x < 5-0.4112/√(8) OR if x > 5+0.4112/√(8)
Reject Ho if x < 4.8546 OR if x > 5.1454
Implies, don't reject Ho if 4.8546≤ x ≤ 5.1454
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(4.8546 ≤ x ≤ 5.1454 | μ1 = 5.1)
= P(4.8546-5.1/0.25/√(8) ≤ x - μ / σ/√n ≤ 5.1454-5.1/0.25/√(8)
= P(-2.7764 ≤ Z ≤0.5136 )
= P( Z ≤0.5136) - P( Z ≤-2.7764)
= 0.6962 - 0.0027 [ Using Z Table ]
= 0.6935
For n =8 the probability of Type II error is 0.6935
power = 1-type 2 error
power = 1-0.6935 =0.3065


Related Solutions

A manufacturer is interested in the output voltage of a power supply used in a PC....
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 V, and the manufacturer wishes to test H0: μ = 9 V against H1: μ ≠ 9 V with a sample of n = 10 units. (a) The critical region is defined as , find the type I error probability α. (b) Find the power of the test for detecting a true...
Problem 4. A high-voltage power supply should have a nominal output voltage of 350 V. A...
Problem 4. A high-voltage power supply should have a nominal output voltage of 350 V. A sample of four units is selected each day and tested for process-control purposes. The data (see sheet Problem4 in Homework 4.xlsx) give the difference between the observed reading on each unit and the nominal voltage times ten; that is, ?? = (observed voltage on unit ? - 350)*10 (a) Set up ?̅ and ? charts on this process. Is the process in statistical control?...
A high-level voltage power supply should have a nominal output voltage of 350 V. A sample...
A high-level voltage power supply should have a nominal output voltage of 350 V. A sample of four units is selected each day and tested for process-control purposes. The data (20 days, 1 sample/day) shown in the table below give the difference between the observed reading on each unit and the nominal voltage times ten; that is, xi = (observed voltage on unit 350)*10. Voltage Data Voltage Data Sample # xi Sample # xi 1 6 11 8 1 9...
A quality control engineer wishes to estimate the mean output voltage of a DC power supply...
A quality control engineer wishes to estimate the mean output voltage of a DC power supply for different loads. The engineer measures output voltage when the supply is connected to 11 different loads and computes a mean 98.4 V and standard deviation 7 V. Assume that the output voltages follow a normal distribution. (a) Construct and interpret the 99.9% confidence interval. (b) Construct and interpret the 90% prediction interval.
The output voltage of a power supply unit has an unknown distribution. Using a sample size...
The output voltage of a power supply unit has an unknown distribution. Using a sample size of 36, sixteen samples are taken with the following sample-mean values: 10.35 V, 9.30 V, 10.00 V, 9.96 V, 11.65 V, 12.00 V, 11.25 V, 9.58 V, 11.54 V, 9.95 V, 10.28 V, 8.37 V, 10.44 V, 9.25 V 9.38 V and 10.85 V. Let µ and σ2 denote the mean and the variance of the output voltage of the power supply unit. (a)...
An AC power supply produces a maximum voltage ΔVmax = 88 V. This power supply is...
An AC power supply produces a maximum voltage ΔVmax = 88 V. This power supply is connected to a resistor R = 27.0 Ω, and the current and resistor voltage are measured with an ideal AC ammeter and voltmeter as shown in the figure below. An ideal ammeter has zero resistance, and an ideal voltmeter has infinite resistance. (a) What is the reading on the ammeter? A (b) What is the reading on the voltmeter? V
For a power supply with full wave bridge rectifier, input voltage is 230 V and required...
For a power supply with full wave bridge rectifier, input voltage is 230 V and required output voltage is 12 V DC Load resistance is of 100 . Determine output rms voltage, dc current, rms current, power, ac power, efficiency of the rectifier, form factor, ripple factor, transformer utilization factor, and size of the transformer Draw complete diagram of the supply circuit
A power supply produces voltage described as V= 100sin(60t). what is the average value of this...
A power supply produces voltage described as V= 100sin(60t). what is the average value of this voltage and what is the average voltage squared? The average voltage is supposed to be 0V and then the avg voltage squared is supposed to be 5000 V^2 which makes zero sense to me. Thanks!
A collector-modulated power has a supply voltage of 48V and an average collector current of 600...
A collector-modulated power has a supply voltage of 48V and an average collector current of 600 mA. Determine: a) the input power to the transmitter? b) the modulating signal power needed to produce 100% modulation? 2) …. receivers convert all incoming signals to a lower, fixed frequency. 3) List all the frequencies that emerge from a mixer that is fed with an incoming frequency of 107.1 MHz and another frequency of 96.4 MHz What is the Image Frequency that results...
Design a power supply to deliver 4 to 9 Vdc adjustable voltage to control DC motor...
Design a power supply to deliver 4 to 9 Vdc adjustable voltage to control DC motor speed. Motor power at maximum load is 20W at maximum speed (maximum voltage). Supply is 12V car battery with 70Ah capacity. How long the battery will last at maximum load and maximum speed? Add a 13.8Vdc communications power supply to continuously charge the battery at 500mA.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT