n Ewekansige steekproef van 150 werknemers het getoon dat 90 tevrede is met die nuwe mediese-skema. Vind 'n 99% vertrouensinterval vir alle tevrede werknemers. / A ramdom sample of 150 workers showed that 90 workers are satisfied with the new medical scheme. Find a 99% confidence interval for all satisfied workers.

Question

[0.5216; 0.6784]


[0.4970 ; 0.703]


[0.5342 ; 0.6658]


[0.703 ; 0.4970]

Answer 1

To find the 99% confidence interval for the proportion of satisfied workers, we can follow these steps:

1. Calculate the sample proportion (p-hat): \[ \hat{p} = \frac{x}{n} = \frac{90}{150} = 0.6 \]

2. Calculate the standard error (SE): \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.6 \cdot (1 - 0.6)}{150}} = \sqrt{\frac{0.6 \cdot 0.4}{150}} = \sqrt{\frac{0.24}{150}} \approx 0.04 \]

3. Find the z-value for a 99% confidence level: For a 99% confidence level, the z-value (from the standard normal distribution) is approximately 2.576.

4. Calculate the margin of error (ME): \[ ME = z \cdot SE = 2.576 \cdot 0.04 \approx 0.1023 \]

5. Calculate the confidence interval: \[ \text{Confidence interval} = \hat{p} \pm ME = 0.6 \pm 0.1023 \] This gives us: \[ [0.6 - 0.1023, 0.6 + 0.1023] = [0.4977, 0.7023] \]

Now, rounding the endpoints, the 99% confidence interval for the proportion of satisfied workers is approximately: \[ [0.497, 0.702] \]

Thus, the closest answer choice to our calculation is: [0.4970 ; 0.703].