A paper company needs to ship paper to a large printing business. The paper will be shipped in small boxes and large boxes. Each
1 answer:
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Answer:
Step-by-step explanation:
The mean SAT score is , we are going to call it \mu since it's the "true" mean
The standard deviation (we are going to call it ) is
Next they draw a random sample of n=70 students, and they got a mean score (denoted by ) of
The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.
- So the Null Hypothesis
- The alternative would be then the opposite
The test statistic for this type of test takes the form
and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.
With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.
Answer:
Given the data in the question;
Samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.
a)
For a p chart ( control chart for fraction nonconforming), the center line and upper and lower control limits are;
UCL = p" + 3√[ (p"(1-P")) / n ]
CL = p"
LCL = p" - 3√[ (p"(1-P")) / n ]
here, p" is the average fraction defective
Now, with the 30 samples of size 100
p" = [∑(6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.)] / [ 30 × 100 ]
p" = 234 / 3000
p" = 0.078
so the trial control limits for the fraction-defective control chart are;
UCL = p" + 3√[ (p"(1-P")) / n ]
UCL = 0.078 + 3√[ (0.078(1-0.078)) / 100 ]
UCL = 0.078 + ( 3 × 0.026817 )
UCL = 0.078 + 0.080451
UCL = 0.1585
LCL = p" - 3√[ (p"(1-P")) / n ]
LCL = 0.078 - 3√[ (0.078(1-0.078)) / 100 ]
LCL = 0.078 - ( 3 × 0.026817 )
LCL = 0.078 - 0.080451
LCL = 0 ( SET TO ZERO )
Diagram of the Chart uploaded below
b)
from the p chart for a) below, sample 28 violated the first western electric rule,
summary report from Minitab;
TEST 1. One point more than 3.00 standard deviations from the center line.
Test failed at points: 28
Hence, we conclude that the process is out of statistical control
Lets Assume that assignable causes can be found to eliminate out of control points.
Since 28 is out of control, we should eliminate this sample and recalculate the trial control limits for the P chart.
so
p" = 0.0745
UCL = p" + 3√[ (p"(1-P")) / n ]
UCL = 0.0745 + 3√[ (0.0745(1-0.0745)) / 100 ]
UCL = 0.0745 + ( 3 × 0.026258 )
UCL = 0.0745 + 0.078774
UCL = 0.1532
LCL = p" - 3√[ (p"(1-P")) / n ]
LCL = 0.0745 - 3√[ (0.0745(1-0.0745)) / 100 ]
LCL = 0.0745 - ( 3 × 0.026258 )
LCL = 0.0745 - 0.078774
UCL = 0 ( SET TO ZERO )
The second p chart diagram is upload below;
NOTE; the red circle symbol on 28 denotes that the point is not used in computing the control limits
