QUESTION 6 4 points Save Answer (GRADED Problem) Section 9.2, Problem 2: Referri

Question

QUESTION 6 4 points Save Answer (GRADED Problem) Section 9.2, Problem 2: Referring to Problem 1, suppose that the type I error is 0.05 and the true population mean is 4.5. Determine the size of (the probability of type ll error) Probability of type ll error QUESTION 7 4 points Save Answer GRADED Problem) Section 9.3, Problem 2: A machine used for producing "quarter inch" rivets is to be checked by taking a random sample of 10 rivets and measuring their diameters. It is feared that the wear-off factor of the machine will cause it to produce rivets with diameters less than 1/4 inch. Describe the critical region in terms of X , the average of the 10 diameters, for a test at level of significance of 1%, of Ho: = 0.25 versus H1: =|

Explanation / Answer

Solution:

QUESTION 6
We want to test 0 > 4 at level = 0.05 when true mean is 1 = 4.5. true std is 0.5, n = 36
The power of right tailed test is given by:
1- = P(Z > [Z - sqrt(n)/ (1 - 0)]

= P(Z > [1.645 - sqrt(36)/0.5 (4.5 -4)])

= P(Z > -4.355)
power = 1.0000 (from z -table)
= 0.0000
  
QUESTION 7
A type II error is failing to reject the null when the alternative is true.

We need to find the probability of making a type II error.

We would fail to reject Ho if p is greater than .01.

The z score at .01 is -2.33

And the mean rivet diameter at rejection is -2.33*.0015/sqrt10 + .25 = .248895

So the null would not be rejected as long as the mean rivet diameter is greater than 0.248895

Next, we need to find the probability of getting a mean diameter greater than .248895 if the actual mean was .2490:

First convert to z:

(.248895 - .2490)/(.0015/sqrt 10) = -.22

then find the p value:

P(z > -.22) = .5871

So if the mean diameter is .249, the probability of failing to reject the null and committing a type II error is .5871.

The probability of committing a type II error is also known as beta.

Beta = .5871

Power of test = 1

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