In a sample of 17 hip surgeries of a certain type, the sample mean surgery time was 136.9 minutes witha sample standard deviation s=22.6 minutes. Find a 95% confidence interval for the mean surgery time for this procedure. The CI is:

The 95% confidence interval = [ 125.28, 148.52]

Step-by-step explanation:

In a sample of 17 hip surgeries of a certain type, the sample mean surgery time was 136.9 minutes witha sample standard deviation s=22.6 minutes. Find a 95% confidence interval for the mean surgery time for this procedure. The CI is:

The formula for Confidence Interval =

Cl = Mean ± t × standard deviation/√n

We are using t score because our sample size is less than 30, hence it is small

Degrees of freedom = n - 1 = 17 - 1 = 16

T score for 95 % confidence interval = 2.120

CI = 136.9 ± 2.120 × 22.6/√17

CI = 136.9 ± 2.120 × 5.4813051258

CI = 136.9 ± 11.620366867

CI

= 136.9 - 11.620366867

= 125.27963313

≈ 125.28

136.9 + 11.620366867

= 148.52036687

≈ 148.52

The 95% confidence interval = [ 125.28, 148.52]

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )

11

step-by-step explanation:

32.50-7.50=25

225/25=11