Please calculate and show the work for this problem Peak expiratory flow (PEF) i
ID: 3131426 • Letter: P
Question
Please calculate and show the work for this problem
Peak expiratory flow (PEF) is a measure of a patients ability to expel air from the lungs. Patients with asthma or other respiratory conditions often have restricted PEF. The mean PEF for children free of asthma is 500. An investigator wants to test whether children with chronic bronchitis have restricted PEF. A sample of 60 children with chronic bronchitis are studied and their mean PEF is 350 with a standard deviation of 95. Is there statistical evidence of a lower mean PEF in children with chronic bronchitis? Run the appropriate test at ?=0.05. Would we reject or accept the HO (show your work)?Explanation / Answer
Formulating the null and alternative hypotheses,
Ho: u >= 500
Ha: u < 500
As we can see, this is a left tailed test.
Thus, getting the critical z, as alpha = 0.05 ,
alpha = 0.05
zcrit = - 1.644853627
Getting the test statistic, as
X = sample mean = 350
uo = hypothesized mean = 500
n = sample size = 60
s = standard deviation = 95
Thus, z = (X - uo) * sqrt(n) / s = -12.23047372
Also, the p value is
p = 1.06844E-34
As |z| > 1.6449, and P < 0.05, we REJECT THE NULL HYPOTHESIS. [DECISION]
Hence, there is significant evidence that the mean PEF for children with chronic bronchitis is lower than the population at 0.05 level. [CONCLUSION]