Question

A quality control engineer at a particular lcd screen manufacturer is studying the mean number of defects per screen. Based on historical evidence, the mean number of defects per screen was thought to be 1.83. There have recently been changes to the manufacturing process, and the engineer now feels that the mean number of defects per screen may be significantly larger than 1.83. Using the number of defects on each of 50 sample screens shown below, conduct the appropriate hypothesis test using a 0.05 level of significance. Assignment 7q1 data a) What are the appropriate null and alternative hypotheses? Ho: 1.83 versus Ha: 1.83 Ho: 1.83 versus Ha: > 1.83 Ho:A1-1.83 versus Hai #1.83 Ho: X-1.83 versus Ha: x >1.83 b) What is the test statistic? Give your answer to four decimal places. c) What is the P-value for the test? Give your answer to four decimal places. d) What is the appropriate conclusion? Reject the claim that the mean number of defects per screen is 1.83 because the P-value is larger than 0.05 Fail to reject the claim that the mean number of defects per screen is ї.83 because the p-value is larger than 0 05 Fail to reject the claim that the mean number of defects per screen is 1.83 because the P-value is smaller than 0.05 Reject the claim that the mean number of defects per screen is 1.83 because the P-value is smaller than 0.05

defects
1
4
1
0
3
4
2
1
3
1
0
5
1
3
3
2
1
1
4
1
1
2
3
3
1
2
2
2
2
2
4
3
1
1
3
0
6
1
1
3
2
1
3
3
0
1
0
1
3
5

Answer 1

	Defects ( X )	$\Sigma (X_{i} - \bar{X})^{2}$
	1	1.1236
	4	3.7636
	1	1.1236
	0	4.2436
	3	0.8836
	4	3.7636
	2	0.0036
	1	1.1236
	3	0.8836
	1	1.1236
	0	4.2436
	5	8.6436
	1.0	1.1236
	3	0.8836
	3	0.8836
	2	0.0036
	1	1.1236
	1	1.1236
	4	3.7636
	1	1.1236
	1	1.1236
	2	0.0036
	3	0.8836
	3	0.8836
	1	1.1236
	2	0.0036
	2	0.0036
	2	0.0036
	2	0.0036
	2	0.0036
	4	3.7636
	3	0.8836
	1	1.1236
	1	1.1236
	3	0.8836
	0	4.2436
	6	15.5236
	1	1.1236
	1	1.1236
	3	0.8836
	2	0.0036
	1	1.1236
	3	0.8836
	3	0.8836
	0	4.2436
	1	1.1236
	0	4.2436
	1	1.1236
	3	0.8836
	5	8.6436
Total	103	98.82

Mean $\bar{X} = \Sigma X_{i} / n$
$\bar{X} = 103 / 50 = 2.06$
Standard deviation $S_{X} = \sqrt{\Sigma (X_{i} - \bar{X})^{2}/n-1}$
$S_{X} = \sqrt{ 98.82 / 50 -1} = 1.4201$

To Test :-

H0 :- $\mu = 1.83$

H1 :-  $\mu > 1.83$

Test Statistic :-
$t = ( \bar{X} - \mu ) / (S /\sqrt{n})$
$t = ( 2.06 - 1.83 ) / ( 1.4201 /\sqrt{ 50 })$
t = 1.1452

Test Criteria :-
Reject null hypothesis if $t \; > \;t_{\alpha, n-1}$
$t_{\alpha, n-1} = t_{0.05 , 50-1} = 1.677$
$t > t_{\alpha, n-1} = 1.1452 < 1.677$
Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 1.1452 ) = 0.1288
Reject null hypothesis if P value < $\alpha = 0.05$   level of significance
P - value = 0.1288 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

Fail to reject the claim that the mean number of defects per screen is 1.83, because P value is larger than 0.05.