A major manufacturing firm producing PCB for electrical insulation discharges sm

Question

A major manufacturing firm producing PCB for electrical insulation discharges small amounts from the plant. We assume that the amount of PCB discharge per water specimen is normally distributed with known standard deviations =0-3. Production will be halted if there is evidence that the mean PCB amount discharged in the water exceeds 3 ppm (parts per million). A random sample of 16 water specimens produced x=3.2 ppm. Do these statistics provide sufficient evidence to halt the process? Use Briefly discuss the consequences of type I and type II errors. From your point of view, which of the two errors is the most serious? Construct a 99% confidence interval for the true mean PCB amount discharged in the water.

Explanation / Answer

Part a

We are given

Population mean = 3

Population standard deviation = 0.3

Sample mean = 3.2

Sample size = n = 16

Level of significance = 0.05

Null hypothesis: H0: µ = 3

Alternative hypothesis: Ha: µ > 3

The test statistic formula is given as below:

Z = (sample mean – population mean) / [SD / sqrt(n)]

Z = (3.2 – 3)/[0.3/sqrt(16)]

Z = 2.6667

Z Test of Hypothesis for the Mean

Data

Null Hypothesis                       m=

3

Level of Significance

0.05

Population Standard Deviation

0.3

Sample Size

16

Sample Mean

3.2

Intermediate Calculations

Standard Error of the Mean

0.0750

Z Test Statistic

2.6667

Upper-Tail Test

Upper Critical Value

1.6449

p-Value

0.0038

Reject the null hypothesis

Part b

Type I error is the probability of rejecting the null hypothesis that the population mean is 3 ppm when actually it is 3 ppm. Type II error is the probability of do not rejecting the null hypothesis that the population mean is 3 ppm when actually it is exceeding 3 ppm. Type II error is the serious in this scenario.

Part c

We are given

Population mean = 3

Population standard deviation = 0.3

Sample mean = 3.2

Sample size = n = 16

Confidence level = 99%

Critical z value = 2.3263

Formula is given as below:

Lower limit =sample mean – z*SD/sqrt(n)

Upper limit = sample mean + z*SD/sqrt(n)

Lower limit = 3.2 – 2.5758*0.3/sqrt(16)

Lower limit = 3.01

Upper limit = 3.2 + 2.5758*0.3/sqrt(16)

Upper limit = 3.39

Z Test of Hypothesis for the Mean

Data

Null Hypothesis                       m=

3

Level of Significance

0.05

Population Standard Deviation

0.3

Sample Size

16

Sample Mean

3.2

Intermediate Calculations

Standard Error of the Mean

0.0750

Z Test Statistic

2.6667

Upper-Tail Test

Upper Critical Value

1.6449

p-Value

0.0038

Reject the null hypothesis

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