Defective Child Restraint Systems The Tracolyte Manufacturing ...

Question

Defective Child Restraint Systems The Tracolyte Manufacturing Company produces plastic frames used for child booster seats in cars. During each week of production, 120 frames are selected and tested for conformance to all regulations by the Department of Transportation. Frames are considered defective if they do not meet all requirements. Listed below are the numbers of defective frames among the 120 that are tested each week. Use a control chart for p to verify that the process is within statistical control. If it is not in control, explain why it is not.

3 2 4 6 5 9 7 10 12 15

3 2 4 6 5 9 7 10 12 15

Accepted Answer

All right. Hello, everyone. So this question says, a company manufactures safety helmets and inspects 150 helmets each week for defects. The number of defective helmets found in each of 10 consecutive weeks is as follows. We have 586479, 1113, 10, and 14. Using a P chart to determine whether the process is in statistical control. If not, explain why. And here we have 4 different answer choices labeled A through D. So here Right, the first step of this process is to calculate the proportion defective or P for each week, which is why I have labeled here P1 through 10. Now for the record, P is equal to the number of defective helmets. Divided by the total inspected, which in this case would be 150 for all of them. So for P1, for example, we saw 5 defective helmets out of the 150. Which equals 0.033? For week 2 that was 8/150. Which equals 0.053. For week 3, that was 6 out of 150. Which equals 0.040. For week 4, that was 4 out of 150. Which equals 0.027. For week 5 that was 7 out of 150. Which equals 0.047. For week 6, that was 9 out of 150. Which equals 0.060. For week 7 that was 11 out of 150. Which equals 0.073. Week 8 was 13 out of 150. Which equals 0.087. Week 9 was 10 out of 150. Which equals 0.067. And week 10 was 14 out of 150, which equals 0.093. So from here We would find P bar, which represents the average proportion defective. So here, you're going to sum up all of the P values you found earlier and divide that by 10. That eventually is going to equal 0.580. Which divided by 10. Equals 0.058. So now, using the value for P bar, you can find the standard deviation for the P chart. So here Sigma of P. Is equal to the square root of P bar. Multiplied by one subtracted by Par. And divided by Using the value calculated earlier, that's going to equal the square root of 0.058, multiplied by 1 subtracted by 0.058. And then divided by the sample size. Or excuse me, of N equal to 150. So after evaluating this expression, sigma of P is equal to approximately 0.019. And now you can use this value to calculate the control limits. We have the upper control limit and the lower control limit. The upper control limit is equal to P bar, added to 3 multiplied by sigma of P. And the lower control limit is equal to P bar, subtracted by 3 multiplied by Sigma P. So starting with the upper control limit first, that's equal to 0.058. Added to 3. Multiplied by 0.019. And this equals 0.115. The lower control limit would be equal to 0.058. Subtracted by 3 multiplied by 0.019. Which equals 0.001? Now, for the record, if the lower control limit happens to be negative, you would set it to 0. But now let's compare each week's p-value to the control limits. When you do so, it just so happens that all of the p values obtained are between. 0.001 and 0.115. And so since all of the points here are within the control limits, the process is in statistical control. Which means that the correct answer. Ultimately it is going to be option B in the multiple choice, which reads that the process is in control because all sample proportions are within the control limits. And there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Key Concepts

Control Chart for Proportion (p-chart)

Defective Rate Calculation

Statistical Control

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