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. 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 and a sample standard deviation s= 0.3.
(a) Do these statistics provide sufficient evidence to halt the process? Use a=0.05.
(b) 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?
(c) Construct a 90% confidence interval for the true mean PCB amount discharged in the water.

Answer 1

Answer:

a. Z = 2.6667

b.

c. Upper limit = 3.39

Step-by-step explanation:

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

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 = 90%

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