Question

The time to assemble a certain type of a computer board from acertain assembly line, has a normal distribution. The assembly times for a random sample of 20 boards are measured. The sample mean and sample standard deviation of observed times are: X-35 minutes and s-5 minutes.

a. The manager of the assembly line claims the true average time, μ, for assembling a board is less than 38 minutes. Test the manager's claim at 1% level of significance and write your conclusion.
b. Test at 5% level of significance if the true variance of the assembly time, σ, is more than 22 and write your conclusion.

Answer 1

Explanation:

(a)

From the given information; we can compute the null and the alternative hypothesis as follows:

`H_o : \mu = 38`

`H_1 : \mu < 38`

Level of significance ∝ = 1% = 0.01

The critical values of t distribution since the sample size n = 20 is:

n - 1

= 20 - 1

= 19 degree of freedom

Assuming the population is normally distributed:

The t test can be computed by using the EXCEL FUNCTION

= TINV(0.01, 19 )

= 2.539483

`t_(0.01,19) = 2.539483`

However;

we were also given the sample mean X to be = 35 minutes

the standard deviation SD = 5 minutes

Thus; the test statistics can be computed as;

`t = (\bar X - \mu)/((s)/(√(n)))`

`t = (35- 38)/((5)/(√(20)))`

`t = (-3)/((5)/(4.472))`

`t_o = -2.6833`

The P-value P ( t <
`t_o`) = P( t < - 2.633)

= 0.007355

P-value
`\approx` 0.0074

Decision Rule: If P - value is less than the level of significance; we are to reject the null hypothesis.

Conclusion: P-value < level of significance ; i.e 0.0074 < 0.01; so we reject the null hypothesis and accept the alternative hypothesis.

Thus; we conclude that the average time for assembling the computer board is less than 38 minutes at 0.01 level of significance.

b).

Given that:

Sample size n = 20

level of significance = 0.05

The population variance σ² is more than 22

Thus null hypothesis and the alternative hypothesis can be computed as follows:

`H_0 : \sigma^2 = 22`

`H_1 : \sigma^2 < 22`

From above;

degree of freedom df = 19

The critical value of
`X^2` at df = 19 and ∝ = 0.05 is = 30.14353 at the right tailed region.

`X^2_(0.05,19) =` 30.14353

The test statistics
`X^2` for the sample variance is computed as:

`X^2= ((n-1 )s^2)/(\sigma^2)`

`X^2= ((20-1 )25)/(22)`

`X^2= ((19)25)/(22)`

`X^2= (475)/(22)`

`X^2` = 21.5909

The P-value for the test statistics is :

= 1 - P(
`X^2` < 21.5909)

= 1 - 0.694914

= 0.305086

The P-value = 0.305086

Decision Rule: If P - value is less than the level of significance; we are to reject the null hypothesis.

Conclusion: SInce the P-value is greater than the level of significance ; i.e

0.305086 > 0.05 ; Therefore; we do not reject the null hypothesis.

Therefore the data does not have sufficient information to conclude that the population variance is more than 22 at 5% level of significance.

I hope that helps a lot.