Question at position 5 Leah is the operations manager at a chair company. She is getting concerned about the number of wrong customer account numbers recorded. She decides to take measures into her own hands. She conducts a random sample every week of 489 deposits and records the number of incorrect account numbers for 8 weeks. The results for the 8 weeks are listed below. She needs your help to do the first step in her research. Please calculate the p for every week and come up with the average p for the 8 weeks. (Please round your answer to 3 decimal places) [table]
Week | Number defective | Calculate P
1 | 10 | 2 | 12 | 3 | 5 | 4 | 8 | 5 | 22 | 6 | 11 | 7 | 9 | 8 | 7 |
[/table] Answer Leah is the operations manager at a chair company. She is getting concerned about the number of wrong customer account numbers recorded. She decides to take measures into her own hands. She conducts a random sample every week of 489 deposits and records the number of incorrect account numbers for 8 weeks. The results for the 8 weeks are listed below. She needs your help to do the first step in her research. Please calculate the p for every week and come up with the average p for the 8 weeks. (Please round your answer to 3 decimal places) [table]
Week | Number defective | Calculate P
1 | 10 | 2 | 12 | 3 | 5 | 4 | 8 | 5 | 22 | 6 | 11 | 7 | 9 | 8 | 7 |
[/table] [input]

Question

Question at position 5  Leah is the operations manager at a chair company. She is getting concerned about the number of wrong customer account numbers recorded. She decides to take measures into her own hands. She conducts a random sample every week of 489 deposits and records the number of incorrect account numbers for 8 weeks. The results for the 8 weeks are listed below. She needs your help to do the first step in her research. Please calculate the p for every week and come up with the average p for the 8 weeks. (Please round your answer to 3 decimal places) [table]
Week | Number defective | Calculate P
1 | 10 |  2 | 12 |  3 | 5 | 4 | 8 | 5 | 22 |  6 | 11 |  7 | 9 | 8 | 7 |
[/table] Answer Leah is the operations manager at a chair company. She is getting concerned about the number of wrong customer account numbers recorded. She decides to take measures into her own hands. She conducts a random sample every week of 489 deposits and records the number of incorrect account numbers for 8 weeks. The results for the 8 weeks are listed below. She needs your help to do the first step in her research. Please calculate the p for every week and come up with the average p for the 8 weeks. (Please round your answer to 3 decimal places) [table]
Week | Number defective | Calculate P
1 | 10 |  2 | 12 |  3 | 5 | 4 | 8 | 5 | 22 |  6 | 11 |  7 | 9 | 8 | 7 |
[/table] [input]

BlackTom AI · Accepted Answer

First, I will restate the task to ensure clarity: for each week, calculate p = (number of defective deposits) / 489, then compute the average of these eight p-values, rounding each p to three decimals as requested.

Week 1: defective = 10. Compute p = 10 / 489 ≈ 0.020436... Rounding to three decimals gives 0.020.
Week 2: defective = 12. p = 12 / 489 ≈ 0.024548... Rounds to 0.025.
Week 3: defective = 5. p = 5 / 489 ≈ 0.010226... Rounds to 0.010.
Week 4: defective = 8. p = 8 / 489 ≈ 0.016355... Rounds to 0.016.
Week 5: defective = 22. p = 22 / 489 ≈ 0.044991... Rounds to 0.045.
Week 6: defective = 11. p = 11 / 489 ≈ 0.022492... Rounds to 0.022.
Week 7: defective = 9. p = 9 / 489 ≈ 0.018394... Rounds to 0.018.
Week 8: defective = 7. p = 7 / 489 ≈ 0.014331... Rounds to 0.014.

Next, I’ll sum the eight rounded p-values to obtain the total, then divide by 8 to get the average p.

Sum of rounded p-values: 0.020 + 0.025 + 0.010 + 0.016 + 0.045 + 0.022 + 0.018 + 0.014 = 0.170.
Average p = 0.170 / 8 = 0.02125. Rounding to three decimals yields 0.021.

Thus, the eight weekly p-values (to three decimals) are: 0.020, 0.025, 0.010, 0.016, 0.045, 0.022, 0.018, 0.014, and the average p across the eight weeks is 0.021.

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The number of defectives in 10 different groups of 50 observations each is the following: 5, 1, 1, 2, 3, 3, 1, 4, 2, 3. A larger sample of 500 observations is obtained by combining the 10 groups. Based on this combined sample, what is the sample proportion of defective items?

From a sample of 300 items, 15 are defective. The point estimate of the population proportion defective will be:

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