Example: Benzene is a toxic chemical used in the manufacturing of medicinal chem

Question

Example: Benzene is a toxic chemical used in the manufacturing of medicinal chemicals, dyes, artificial leather, and linoleum. To assess the benzene content of the exit water, 10 independent water samples were collected and found to have an average of 7906 ppm of benzene. Assume a known standard deviation of 80 ppm and use a significance level of 0.01.

a) A manufacturer claims that its exit water meets the federal regulation with a mean of less than 7980 ppm of benzene. Test the manufacturer’s claim.

b)What is the –value if the true mean is 7920?

c)What sample size would be necessary to detect a true mean of 7920 with a probability of at least 0.90?

d)Find a 99% one-sided upper confidence bound.

Explanation / Answer

Here H0 : = 7980 ppm

Ha : < 7980 ppm

(a) Here known standard deviation = 80 ppm

sample mean x = 7906 ppm

sample size n = 10

standard error of the sample mean se0 = / sqrt(n) = 80/ sqrt(10) = 25.298 ppm

here test statistic

Z = (x - H)/ se0 = (7906 - 7980)/ 25.298 = -2.925

Here critical value t for dF = 9 and alpha = 0.01 is

Zcritical = 2.326 (one tailed test)

so here Z > Zcritical so we will accept the manufactures's claim.

(b) Here we have to sample mean above which we cannot reject the null hypothesis.

Let say it is xo

xo = H - Zcritical * se0 = 7980 - 2.326 * 25.298 = 7921.1568 ppm

so if true mean if 7920 ppm then we shall not reject the null hypothesis if the sample mean x > xo

So Pr(Type II error) = Pr(x > 7921.1568) = NORMAL (x > 7921.1568; 7920 ; 25.298)

Z = (7921.1568 - 7920)/ 25.298 = 0.0457

so Pr(Type II error) = 1 - Pr(Z < 0.0457) = 1 - 0.5182 = 0.4818

(c) Here let say sample size = n

standard error of sample mean se0 = 80/ n

so the critical sample mean above which we cannot reject the null hypothesis is.

xo = H - Zcritical * se0 = 7980 - 2.326 * 80/ n = 7980- 186.08 /n

Power of the test here = 0.90

so here the type II error = 1 - power = 1 - 0.90 = 0.10

so Now,

Pr(TYpe II error) = Pr(x > 7980- 186.08 /n) = NORMAL (x > 7980- 186.08 /n; 7920 ; 80/ n) = 0.10

so Z - value for p- vlaue = 0.90 is

Z = 1.282

1.282 = (7980- 186.08 /n - 7920)/80/ n

60 = 186.08/n + 102.56/n

60 = 288.56/n

n = 288.56/60 = 4.81066

n = 23.1425

n = 24

(d) 99% one sided upper onfidence bound = x + Z99% se0 = 7906 + 2.326 * 25.298 = 7964.84 ppm

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