A health professional obtains records for 38 patients who have undergone a certain surgical procedure and finds the average length of stay is 5.23 days with a standard deviation of 1.76 days. Using α=0.10α=0.10, is there evidence to suggest the actual average length of stay for this surgical procedure is 5 days?What are the hypotheses?H0:μ=5.23H0:μ=5.23 vs H1:μ≠5.23H1:μ≠5.23H0:¯x=5.23H0:x¯=5.23 vs H1:¯x≠5.23H1:x¯≠5.23H0:μ=5H0:μ=5 vs H1:μ≠5H1:μ≠5H0:¯x=5H0:x¯=5 vs H1:¯x≠5H1:x¯≠5Calculate the test statistic using the correct letter.t=0.81t=0.81z=−0.81z=-0.81z=0.81z=0.81t=−0.81t=-0.81Determine the critical values.±1.305±1.687±1.685±1.645±1.304±1.686±1.282State your decision.Accept H0Fail to reject H0Reject H0The correct conclusion is that we conclude that the average length of stay of surgeries of this type days.What level of concern do you have for the validity of your results?Low −- our sample size is near 30, so while that's likely sufficient for ¯X¯ to be approximately normally distributed, seeing that the original data isn't extremely skewed would make us feel more comfortableModerate −- the original data is not normally distributed which is an assumptions violationExtreme −- we should not use any part of what we've done to make a conclusionNone −- our sample size is larger than 30 so there should be no problem with the conclusion

Hey there! Thank you for posting the question. Since your question has more than 3 parts, we are…

Answered: A health professional obtains records…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Question

A health professional obtains records for 38 patients who have undergone a certain surgical procedure and finds the average length of stay is 5.23 days with a standard deviation of 1.76 days. Using α=0.10α=0.10, is there evidence to suggest the actual average length of stay for this surgical procedure is 5 days?

What are the hypotheses?

H0:μ=5.23H0:μ=5.23 vs H1:μ≠5.23H1:μ≠5.23
H0:¯x=5.23H0:x¯=5.23 vs H1:¯x≠5.23H1:x¯≠5.23
H0:μ=5H0:μ=5 vs H1:μ≠5H1:μ≠5
H0:¯x=5H0:x¯=5 vs H1:¯x≠5H1:x¯≠5

Calculate the test statistic using the correct letter.

t=0.81t=0.81
z=−0.81z=-0.81
z=0.81z=0.81
t=−0.81t=-0.81

Determine the critical values.

±1.305
±1.687
±1.685
±1.645
±1.304
±1.686
±1.282

State your decision.

Accept H0
Fail to reject H0
Reject H0

The correct conclusion is that we conclude that the average length of stay of surgeries of this type days.

What level of concern do you have for the validity of your results?

Low −- our sample size is near 30, so while that's likely sufficient for ¯X¯ to be approximately normally distributed, seeing that the original data isn't extremely skewed would make us feel more comfortable
Moderate −- the original data is not normally distributed which is an assumptions violation
Extreme −- we should not use any part of what we've done to make a conclusion
None −- our sample size is larger than 30 so there should be no problem with the conclusion

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution