3. A rando m sample of 60 measurements of the helium porosity (in percentage)tak

Question

3. A rando m sample of 60 measurements of the helium porosity (in percentage)taken from coal samples in a particular seam. Th e sample mean was 4.4. The owner of the land wants to say that the true mean andard dev = 1.3. percentage of helium porosity for that seam is no larger than 5. Assume the population a) (5 marks) Set up the appropriate Hypothesis to be tested. Remember what is to be specified in H b) (5 marks) Find the Test of size =0.01 and indicate whether your test rejects Ho or not. c) (5 marks) Compute the probability of a type II error for your test in (b) when -4.5 and d) (5 marks) Find the minimum sample size that would give =0.01 and a -0.05 ( beta is calculated using -4.5). st

Explanation / Answer

Question 3

(a) sample mean x = 4.4

population standard deviation = 1.3

(a H0 : > 5

Ha : =< 5

standard error of the sample se0 = /sqrt(n) = 1.3/ sqrt(60) = 0.168

Z = (x - H)/ se0 = (4.4 - 5.0)/ 0.168 = -3.571

Here p - value = Pr(Z < -3.571) = 0.0002

os p - value < 0.01 so we shall reject the null hypothesis and can conclude that true mean percentage of helium porosity is not larger than 5.

(c) True mean = 4.5

Here we shall not be able reject the null hypothesis when sample mean will be greater than 5 - Z0.01 (s/sqrt(n) = 5 - 2.33 * (1.3/sqrt(60 ) = 5 - 2.33 * 0.1678 = 4.609

so Pr(x > 4.609 ) = Pr(Type II error) = Pr(x > 4.609 ; 4.5 ; 0.1678)

Z= (4.609- 4.5)/ 0.1678 = 0.65

Pr(Type II error) = Pr( Z > 0.65) = 0.258

(d) Here let say sample size = n

Here we shall not be able reject the null hypothesis when sample mean will be greater than 5 - 2.33 * (1.3/ sqrt(n) = 5 - 3.03/ n

so here the type II error = 0.05

Pr(Type II error ) = 0.05

Pr(x > 5 - 3.03/ n ; 4.5 ; 1.3/ n) = 0.05

Z- value = 1.96

so,

[(5 - 3.03/ n) - 4.5] / (1.3/ n) = 1.96

0.5 - 3.03/   n = 2.548/ n

0.5 = 5.578/ n

n = 124.45 or say 125

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