Question

NSA electronics is experimenting with the manufacture of a new type of transistor that is very difficult to mass produce at an acceptable quality level. Every hour a supervisor takes a random sample of 8 transistors produced on the assembly line. From the past records, 15 per cent of transistors fail in quality inspection.

4a) NSA wants to know the probability of 0 to 8 defectives if the percentage of defective is 15%

4b) At 95% confidence level how many samples will be defective?

4c) At 99% confidence level how many samples will be defective?

Answer 1

Proportion of failure = 15% = 0.15

sample size, n = 8

This is binomial distribution since we have two events failure or success and are independent.

Let, X denotes the number of defectives

P[ defective ] = 0.15

P[ not defective ] = 1 - P[ defective ] = 1- 0.15 = 0.85

P( X = 0 ) = (8C0​)*(0.15^0)*(0.85)^(8 - 0) =  (8C0​)*(0.15^0)*(0.85^8) = 0.2725

P( X = 1 ) = (8C1​)*(0.15^1)*(0.85)^(8 - 1) =  (8C1​)*(0.15^1)*(0.85^7) = 0.3847

P( X = 2 ) = (8C2)*(0.15^2)*(0.85)^(8 - 2) =  (8C2​)*(0.15^2)*(0.85^6) = 0.2376

P( X = 3 ) = (8C​3)*(0.15^3)*(0.85)^(8 - 3) =  (8C3​)*(0.15^3)*(0.85^5) = 0.0839

P( X = 4 ) = (8C4)*(0.15^4)*(0.85)^(8 - 4) =  (8C4​)*(0.15^4)*(0.85^4) = 0.0185

P( X = 5 ) = (8C5)*(0.15^5)*(0.85)^(8 - 5) =  (8C5​)*(0.15^5)*(0.85^3) = 0.0026

P( X = 6 ) = (8C6)*(0.15^6)*(0.85)^(8 - 6) =  (8C6​)*(0.15^6)*(0.85^2) = 0.0002

P( X = 7 ) = (8C7)*(0.15^7)*(0.85)^(8 - 7) =  (8C7​)*(0.15^7)*(0.85^1) = 0

P( X = 8 ) = (8C8)*(0.15^8)*(0.85)^(8 - 8) =  (8C8​)*(0.15^8)*(0.85^0) = 0

95% confidence interval

We need to construct the 95% confidence interval for the population proportion. We have been provided with the following information about the sample proportion:

Sample Proportion p =	0.15
Sample Size, n	8

The critical value for α = 0.05 is z_c = 1.96. The corresponding confidence interval is computed as shown below:

CI(Proportion) ​= ​( p​ − z_c​*sqrt(p*(1−p​)​​/n) , p​ + z_c​*sqrt(p*(1−p​)​​/n) )

CI(Proportion) ​= ( 0.15 - 1.96*sqrt(0.15*0.85/8) , 0.15 + 1.96*sqrt(0.15*0.85/8) )

CI(Proportion) ​= (−0.097,0.397)

Therefore, based on the data provided, the 95% confidence interval for the population proportion is −0.097 < p < 0.397, which indicates that we are 95% confident that the true population proportion p is contained by the interval (−0.097,0.397).

99% confidence interval

We need to construct the 99% confidence interval for the population proportion. We have been provided with the following information about the sample proportion:

Sample Proportion p =	0.15
Sample Size, n	8

The critical value for α = 0.01 is z_c = 2.576. The corresponding confidence interval is computed as shown below:

CI(Proportion) ​= ​( p​ − z_c​*sqrt(p*(1−p​)​​/n) , p​ + z_c​*sqrt(p*(1−p​)​​/n) )

CI(Proportion) ​= ( 0.15 - 2.576*sqrt(0.15*0.85/8) , 0.15 + 2.576*sqrt(0.15*0.85/8) )

CI(Proportion) ​= (−0.175,0.475)

Therefore, based on the data provided, the 99% confidence interval for the population proportion is −0.175 < p < 0.475, which indicates that we are 99% confident that the true population proportion p is contained by the interval (-0.175, 0.475).