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

Answer:

Step-by-step 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;

$t = \dfrac{\bar X - \mu}{\dfrac{s}{\sqrt{n}}}$

$t = \dfrac{35- 38}{\dfrac{5}{\sqrt{20}}}$

$t = \dfrac{-3}{\dfrac{5}{4.472}}$

$t_o = -2.6833$

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:

$X^2= \dfrac{(n-1 )s^2}{\sigma^2}$

$X^2= \dfrac{(20-1 )25}{22}$

$X^2= \dfrac{(19)25}{22}$

$X^2= \dfrac{475}{22}$

$X^2$ = 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.