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The management is interested to know the percentage of idle time of equipment. The trial study showed that percentage of idle time would be 20%. The number of random observations necessary for 95% level of confidence and ± 5% accuracy is:
- a)6400
- b)1600
- c)640
- d)160
Correct answer is option 'A'. Can you explain this answer?
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To determine the number of random observations necessary to estimate the percentage of idle time of equipment with a 95% level of confidence and 5% accuracy, we can use the formula for sample size calculation for proportions.
Sample size calculation for proportions:
n = (Z² * p * (1-p)) / E²
Where:
n = Sample size
Z = Z-value for the desired level of confidence
p = Estimated proportion
E = Margin of error
Given:
Estimated proportion (p) = 0.20 (20%)
Level of confidence = 95%
Accuracy (Margin of error, E) = 0.05 (5%)
To find the Z-value for a 95% level of confidence, we can refer to the standard normal distribution table, which gives the Z-value corresponding to the desired level of confidence. For a 95% level of confidence, the Z-value is approximately 1.96.
Calculating the sample size:
n = (1.96² * 0.20 * (1-0.20)) / 0.05²
n = (3.8416 * 0.16) / 0.0025
n = 0.614656 / 0.0025
n = 245.86
Since the sample size must be a whole number, we round up the calculated value to the nearest whole number:
Sample size (n) = 246
Therefore, the number of random observations necessary for a 95% level of confidence and 5% accuracy is 246.
However, it is important to note that the correct answer provided in the question is option 'A' (6400). This implies that there may be an error in either the question or the provided answer. The calculation performed above is correct, and the result obtained is 246, not 6400.
Sample size calculation for proportions:
n = (Z² * p * (1-p)) / E²
Where:
n = Sample size
Z = Z-value for the desired level of confidence
p = Estimated proportion
E = Margin of error
Given:
Estimated proportion (p) = 0.20 (20%)
Level of confidence = 95%
Accuracy (Margin of error, E) = 0.05 (5%)
To find the Z-value for a 95% level of confidence, we can refer to the standard normal distribution table, which gives the Z-value corresponding to the desired level of confidence. For a 95% level of confidence, the Z-value is approximately 1.96.
Calculating the sample size:
n = (1.96² * 0.20 * (1-0.20)) / 0.05²
n = (3.8416 * 0.16) / 0.0025
n = 0.614656 / 0.0025
n = 245.86
Since the sample size must be a whole number, we round up the calculated value to the nearest whole number:
Sample size (n) = 246
Therefore, the number of random observations necessary for a 95% level of confidence and 5% accuracy is 246.
However, it is important to note that the correct answer provided in the question is option 'A' (6400). This implies that there may be an error in either the question or the provided answer. The calculation performed above is correct, and the result obtained is 246, not 6400.
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The management is interested to know the percentage of idle time of equipment. The trial study showed that percentage of idle time would be 20%. The number of random observations necessary for 95% level of confidence and ± 5% accuracy is:a)6400 b)1600 c)640 d)160Correct answer is option 'A'. Can you explain this answer?
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