Question at position 4 Riverdale Hospital is concerned over the number of times patients must wait more than 30 minutes beyond their scheduled appointments. The quality team decided to take random samples of 67 patients to see how many in each sample had to wait more than 30 minutes. Each instance is considered a defect in the clinic process. The table below contains the data for 15 samples. [table]
Sample | Number of defects 1 | 5
2 | 2 3 | 1
4 | 3 5 | 1
6 | 5 7 | 2
8 | 3 9 | 6
10 | 3 11 | 9 12 | 9 13 | 5 14 | 2 15 | 3
[/table] σ\sigmap = ______ (enter response rounded to 3 decimal places). Riverdale Hospital is concerned over the number of times patients must wait more than 30 minutes beyond their scheduled appointments. The quality team decided to take random samples of 67 patients to see how many in each sample had to wait more than 30 minutes. Each instance is considered a defect in the clinic process. The table below contains the data for 15 samples. [table]
Sample | Number of defects 1 | 5
2 | 2 3 | 1
4 | 3 5 | 1
6 | 5 7 | 2
8 | 3 9 | 6
10 | 3 11 | 9 12 | 9 13 | 5 14 | 2 15 | 3
[/table] σ\sigmap = ______ (enter response rounded to 3 decimal places). [input]题目解析

Question

Question at position 4  Riverdale Hospital is concerned over the number of times patients must wait more than 30 minutes beyond their scheduled appointments. The quality team decided to take random samples of 67 patients to see how many in each sample had to wait more than 30 minutes. Each instance is considered a defect in the clinic process. The table below contains the data for 15 samples. [table]
Sample | Number of defects 1 | 5
2 | 2 3 | 1
4 | 3 5 | 1
6 | 5 7 | 2
8 | 3 9 | 6
10 | 3 11 | 9 12 | 9 13 | 5 14 | 2 15 | 3
[/table] σ\sigmap = ______ (enter response rounded to 3 decimal places). Riverdale Hospital is concerned over the number of times patients must wait more than 30 minutes beyond their scheduled appointments. The quality team decided to take random samples of 67 patients to see how many in each sample had to wait more than 30 minutes. Each instance is considered a defect in the clinic process. The table below contains the data for 15 samples. [table]
Sample | Number of defects 1 | 5
2 | 2 3 | 1
4 | 3 5 | 1
6 | 5 7 | 2
8 | 3 9 | 6
10 | 3 11 | 9 12 | 9 13 | 5 14 | 2 15 | 3
[/table] σ\sigmap = ______ (enter response rounded to 3 decimal places). [input]题目解析

BlackTom AI · Accepted Answer

First, I restate the scenario and data to ensure clarity: Riverdale Hospital collected 15 samples, each consisting of 67 patients. For each sample, the number of defects (patients waiting more than 30 minutes past their appointment) is given. The total across all samples is 59 defects (sum of: 5, 2, 1, 3, 1, 5, 2, 3, 6, 3, 9, 9, 5, 2, 3). The total number of observations across all samples is 15 samples × 67 patients per sample = 1005 observations.

Step 1: Estimate the overall defect probability p̂.
- p̂ is the total number of defects divided by the total number of observations: p̂ = 59 / 1005 ≈ 0.0587.
- This means about 5.87% of all patients waited more than 30 minutes beyond their scheduled time, on average, across the combined data.

Step 2: Identify the standard deviation (standard error) of the sample proportion p̂ for a single sample of size n = 67.
- The standard error for a sample proportion in a binomial context is SE = sqrt[ p̂ (1 − p̂) / n ].
- Compute p̂(1 − p̂): 0.0587 × (1 − 0.0587) ≈ 0.0587 × 0.9413 ≈ 0.0552.
- Divide by n = 67: 0.0552 / 67 ≈ 0.000824.
- Take the square root: SE ≈ sqrt(0.000824) ≈ 0.0287.

Step 3: Round to three decimal places as required.
- SE ≈ 0.0287 rounds to 0.029 (to three decimals).

Additional notes on potential alternative interpretations: Some calculations might attempt to use slightly different p̂ values or a different n (for example, using a pooled proportion across all samples versus the per-sample estimation). However, the standard approach here uses the overall proportion p̂ = 59/1005 and the per-sample size n = 67, leading to SE ≈ 0.029. If a different convention were used (e.g., treating each sample as an independent trial and computing a standard error of the mean proportion across the 15 samples), the numeric result would differ, but the conventional calculation for a single-sample proportion with n = 67 yields 0.029.

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