A quality control engineer at a particular lcd screen manufa

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.

a) What are the appropriate null and alternative hypotheses?

a)H₀: = 1.83 versus H_a: > 1.83

b)H₀: = 1.83 versus H_a: < 1.83    

c)H₀: x = 1.83 versus H_a: x > 1.83

d)H₀: = 1.83 versus H_a: 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?

a)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.

b)Reject the claim that the mean number of defects per screen is 1.83 because the P-value is smaller than 0.05.   

c)Reject the claim that the mean number of defects per screen is 1.83 because the P-value is larger than 0.05.

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

Solution

Set Up Hypothesis
Null, H0: U<1.83
Alternate,mean number of defects per screen may be significantly larger than 1.83 H1: U>1.83
Test Statistic
Population Mean(U)=1.83
Sample X(Mean)=2.06
Standard Deviation(S.D)=1.4201
Number (n)=50
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =2.06-1.83/(1.4201/Sqrt(49))
to =1.145
| to | =1.145
Critical Value
The Value of |t | with n-1 = 49 d.f is 1.677
We got |to| =1.145 & | t | =1.677
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Right Tail - Ha : ( P > 1.1452 ) = 0.12884
Hence Value of P0.05 < 0.12884,Here We Do not Reject Ho

ANS)
1.c)H0: x = 1.83 versus Ha: x > 1.83
2. to =1.1452
3. ( P > 1.1452 ) = 0.12884
4.
d)Fail to reject the claim that the mean number of defects per screen is 1.83 because the P-value is larger than 0.05