Please solve (a) and (e) with the solutions. A manufacturer of semiconductor dev
ID: 2935583 • Letter: P
Question
Please solve (a) and (e) with the solutions.
A manufacturer of semiconductor devices takes a random sample of size n of chips and tests them, classifying each chip as defective or nondefective. Let x, = 0 if the chip is nondefective and X-1 if the chip is defective. The sample fraction defective is X1 X2+ + Xn After collecting a sample, we are interested in computing the error in estimating the true value p. For each of the sample sizes and estimates of p, compute the error at the 90% confidence level. Round your error answers to 3 decimal places. (a)n 60 and = 0.10 (b)n = 70 and = 0.10 (c)n = 120 and = 0.10 (d) Compare your results from parts (a)-(c) and comment on the effect of sample size on the error in estimating the true value of p and the 90% confidence level (e) Repeat parts (a)-(c), this time using a 95% confidence level. (f) Examine your results when the 90% confidence level and then the 95% confidence level are used to compute the error and explain what happens to the magnitude of the error as the percentage confidence increases.Explanation / Answer
a.
TRADITIONAL METHOD
given that,
possibile chances (x)=6
sample size(n)=60
success rate ( p )= x/n = 0.1
I.
sample proportion = 0.1
standard error = Sqrt ( (0.1*0.9) /60) )
= 0.039
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.039
= 0.064
III.
CI = [ p ± margin of error ]
confidence interval = [0.1 ± 0.064]
= [ 0.036 , 0.164]
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b.
TRADITIONAL METHOD
given that,
possibile chances (x)=7
sample size(n)=70
success rate ( p )= x/n = 0.1
I.
sample proportion = 0.1
standard error = Sqrt ( (0.1*0.9) /70) )
= 0.036
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.036
= 0.059
III.
CI = [ p ± margin of error ]
confidence interval = [0.1 ± 0.059]
= [ 0.041 , 0.159]
c.
TRADITIONAL METHOD
given that,
possibile chances (x)=12
sample size(n)=120
success rate ( p )= x/n = 0.1
I.
sample proportion = 0.1
standard error = Sqrt ( (0.1*0.9) /120) )
= 0.027
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.027
= 0.045
III.
CI = [ p ± margin of error ]
confidence interval = [0.1 ± 0.045]
= [ 0.055 , 0.145]
d.
from a to c sample size increases then it should be effect on stanadard error and margin error both are decreased at 90% confidence interval
e.
TRADITIONAL METHOD
given that,
possibile chances (x)=6
sample size(n)=60
success rate ( p )= x/n = 0.1
I.
sample proportion = 0.1
standard error = Sqrt ( (0.1*0.9) /60) )
= 0.039
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.039
= 0.076
III.
CI = [ p ± margin of error ]
confidence interval = [0.1 ± 0.076]
= [ 0.024 , 0.176]
TRADITIONAL METHOD
given that,
possibile chances (x)=7
sample size(n)=70
success rate ( p )= x/n = 0.1
I.
sample proportion = 0.1
standard error = Sqrt ( (0.1*0.9) /70) )
= 0.036
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.036
= 0.07
III.
CI = [ p ± margin of error ]
confidence interval = [0.1 ± 0.07]
= [ 0.03 , 0.17]
TRADITIONAL METHOD
given that,
possibile chances (x)=12
sample size(n)=120
success rate ( p )= x/n = 0.1
I.
sample proportion = 0.1
standard error = Sqrt ( (0.1*0.9) /120) )
= 0.027
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.027
= 0.054
III.
CI = [ p ± margin of error ]
confidence interval = [0.1 ± 0.054]
= [ 0.046 , 0.154]
f.
from a to c repeats procedure but changes the confidence interval 95% increases then it will compare the 90% confidence interval.then estimating errors are inmproved