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.

Answer 1

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.