A ramdom sample of 150 workers showed that 90 workers are satisfied with the new medical scheme. Find a 90% confidence interval for all satisfied workers.Give the upper limit.
To find a 90% confidence interval, we first calculate the sample proportion of workers satisfied with the new medical scheme:
Sample proportion (P̂) = x/n
P̂ = 90/150
P̂ = 0.6
Next, we need to determine the margin of error, which is given by:
Margin of error = Z * sqrt((P̂(1-P̂))/n)
For a 90% confidence level, the Z-value corresponding to a two-tailed test is 1.645.
Plugging in the values:
Margin of error = 1.645 * sqrt((0.6 * (1-0.6))/150)
Margin of error ≈ 0.050
Finally, we can calculate the confidence interval by subtracting and adding the margin of error from the sample proportion:
Confidence interval = P̂ ± Margin of error
Confidence interval = 0.6 ± 0.050
The upper limit of the confidence interval is 0.6 + 0.050 = 0.65.
Therefore, the 90% confidence interval for all satisfied workers is (0.55, 0.65), and the upper limit is 0.65.
To find the 90% confidence interval for all satisfied workers, we can use the formula for calculating confidence intervals for proportions.
The formula is given as:
CI = p̂ ± Z * √(p̂(1-p̂)/n)
Where:
CI = Confidence Interval
p̂ = Proportion of sample (satisfied workers)
Z = Z-Score for the desired confidence level (90% confidence corresponds to a Z-Score of approximately 1.645)
n = Sample size
First, calculate the proportion of satisfied workers in the sample:
p̂ = 90/150 = 0.6
Next, calculate the margin of error using the formula:
ME = Z * √(p̂(1-p̂)/n)
ME = 1.645 * √(0.6*(1-0.6)/150)
ME ≈ 0.065
Finally, calculate the upper limit of the confidence interval:
Upper Limit = p̂ + ME
Upper Limit = 0.6 + 0.065
Upper Limit ≈ 0.665
Therefore, the upper limit of the 90% confidence interval for all satisfied workers is approximately 0.665.