Question

The supervisor of a production line believes that the average time to assemble an electronic component is 18 minutes. Assume that assembly time is normally distributed with a standard deviation of 4 minutes. The supervisor times the assembly of 19 components, and finds that the average time for completion was 20.29 minutes. Is there evidence that the average amount of time required to assemble a component is something other than 18 minutes?

What are the appropriate hypotheses one should test?
H₀: X = 18,   H_a: X ≠ 18
H₀: μ = 18,   H_a: μ > 18
H₀: μ = 18,   H_a: μ < 18
H₀: μ = 18,   H_a: μ ≠ 18

The test-statistic to use here is
(p −p₀)/(√{p₀(1−p₀)/n})
(X - μ₀)/(s/√n)
(X - μ₀)/(σ/√n)

The value of the test-statistic is
-2.4955
2.4955
0.5725

We should reject at 2% level of significance if
| test-statistic | > 2.326
| test-statistic | > 2.054
test-statistic < -2.054
test-statistic > 2.214

If α= 0.02, what will be your conclusion?
Do not reject H₀
not enough information to reach a decision
Reject H₀

The p-value of the test is equal to
0.0063
0.9937
0.0126

Answer 1

Given:The supervisor of a production line believes that the average time to assemble an electronic component is 18 minutes. Assume that assembly time is normally distributed with a standard deviation of 4 minutes.  

n = 19

We have to test if there is evidence that the average amount of time required to assemble a component is something other than 18 minutes.

Since this statement is non-directional this is two tailed test.

Part i) What are the appropriate hypotheses one should test?

H₀: μ = 18,   H_a: μ ≠ 18

Part ii) The test-statistic to use here is

We use z test statistic , since population standard deviation is known and population is normally distributed.

thus correct answer is:

(X - μ₀)/(σ/√n)

Part iii) The value of the test-statistic is

Part iv) We should reject at 2% level of significance if_____.

Use following excel command:

=NORM.S.INV(0.02/2)

= -2.326

thus absolute z = 2.326

thus correct answer is:

| test-statistic | > 2.326

Part v) If α= 0.02, what will be your conclusion?

Since > 2.326 , we reject H0.

Thus correct answer is:

Reject H₀

Part vi) The p-value of the test is equal to

p-value = 2 * P( Z > z )

p-value = 2 * P( Z > 2.4955)

p-value = 2 * [ 1 - P( Z < 2.4955) ]

Use following excel command:

=2*(1-NORM.S.DIST(2.4955,TRUE))

=0.0126

Thus correct answer is:

0.0126