Respuesta :
Answer:
a) The 95% CI for the mean surgery time is (133.05, 140.75).
b) The 99.5% CI for the mean surgery time is (131.37, 142.43).
c) The level of confidence of the interval (133.9, 139.9) is 69%.
d) The sample size should be 219 surgeries.
e) The sample size should be 377 surgeries.
Step-by-step explanation:
We have a sample, of size n=132, for which the mean time was 136.9 minutes with a standard deviation of 22.6 minutes.
a) We have to find a 95% CI for the mean surgery time.
The critical value of z for a 95% CI is z=1.96.
The margin of error of the CI can be calculated as:
[tex]E=z\cdot s/\sqrt{n}=1.96*22.6/\sqrt{132}=44.296/11.489=3.85[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=\bar x-E=136.9-3.85=133.05\\\\UL=\bar x+E=136.9+3.85=140.75[/tex]
The 95% CI for the mean surgery time is (133.05, 140.75).
b) Now, we have to find a 99.5% CI for the mean surgery time.
The critical value of z for a 99.5% CI is z=2.81.
The margin of error of the CI can be calculated as:
[tex]E=z\cdot s/\sqrt{n}=2.81*22.6/\sqrt{132}=63.506/11.489=5.53[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=\bar x-E=136.9-5.53=131.37\\\\UL=\bar x+E=136.9+5.53=142.43[/tex]
The 99.5% CI for the mean surgery time is (131.37, 142.43).
c) We can calculate the level of confidence, calculating the z-score for the margin of error in that interval.
We know that the difference between the upper bound and lower bound is 2 times the margin of error:
[tex]UL-LL=2E\\\\E=\dfrac{UL-LL}{2}=\dfrac{139.9-133.9}{2}=\dfrac{6}{2}=3[/tex]
Then, we can write the equation for the margin of error to know the z-value.
[tex]E=z \cdot s/\sqrt{n}\\\\z= E\cdot \sqrt{n}/s=2*\sqrt{132}/22.6=2*11.5/22.6=1.018[/tex]
The confidence level for this interval is then equal to the probability that the absolute value of z is bigger than 1.018:
[tex]P(-|z|<Z<|z|)=P(-1.018<Z<1.018)=0.69[/tex]
The level of confidence of the interval (133.9, 139.9) is 69%.
d) We have to calculate the sample size n to have a margin of error, for a 95% CI, that is equal to 3.
The critical value for a 95% CI is z=1.96.
Then, the sample size can be calculated as:
[tex]E=z\cdot s/\sqrt{n}\\\\n=(\dfrac{z\cdot s}{E})^2=(\dfrac{1.96*22.6}{3})^2=14.77^2=218.015\approx 219[/tex]
The sample size should be 219 surgeries.
e) We have to calculate the sample size n to have a margin of error, for a 99% CI, that is equal to 3.
The critical value for a 99% CI is z=2.576.
Then, the sample size can be calculated as:
[tex]E=z\cdot s/\sqrt{n}\\\\n=(\dfrac{z\cdot s}{E})^2=(\dfrac{1.96*22.6}{3})^2=19.41^2=376.59\approx 377[/tex]
The sample size should be 377 surgeries.
