Question

In: Statistics and Probability

In a sample of 1901 patients who underwent a certain type of surgery, 22% experienced complications....

In a sample of 1901 patients who underwent a certain type of surgery, 22% experienced complications. Find a 90% confidence interval for the proportion of all those undergoing this surgery who experience complications.

Expert Solution

Solution :

Given that,

Point estimate = sample proportion = = 0.22

1 - = 1- 0.22 =0.78

At 90% confidence level

= 1 - 90%

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E Z/2 * $\sqrt$ (( * (1 - )) / n)

= 1.645 * $\sqrt$ ((0.22*0.78) / 1901)

E = 0.016

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.22-0.016 < p < 0.22+0.016

0.204< p < 0.236

orchestra answered 2 years ago

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