The critical value for Z at the 05 level is Peak expiratory

The critical value for Z at the .05 level is . Peak expiratory flow (PEF) is a measure of a patient’s 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 306. An investigator wants to test whether children with chronic bronchitis have restricted PEF. A sample of 40 children with chronic bronchitis are studied and their mean PEF is 279 with a standard deviation of 71. Is there statistical evidence of a lower mean PEF in children with chronic bronchitis? Run the appropriate test at a=0.05. What is the null hypothesis(2pts) H0 > 306 HO 306 H1<306 H1-=306 When would we reject HO?(2pts) When the answer = 0 If the alternative hypothesis is greater If Z = 1.645 If Z is < or = -1.645 Would we reject or accept the HO?(2pts) Accept Reject There is not enough information

Solution

Answer:

Mean X=279

standard devistion(SD)=71

sample size n = 40

Hypotheses:
Null hypothesis: H0: = 306
Alternate hypothesis: H1: < 306

Level of significance a = 0.05

Use z-test.
Z-statistic formula = (X - ) / (SD / sqrt( n))
Z = (279 - 306) / (71/sqrt(40)) = -2.405

Z= -2.405 (calculated value of Z )

The p-value for -2.41 is .00798 and the p-value for -2.40 is .00820. So the p-value for -2.405 is between .00798 and .00820. This is .00809.
If the p-value is less than .05, we reject the null hypothesis! .00809 < .05, so we reject the null hypothesis that the mean PEF in children with chronic bronchitis is equal to the mean PEF for children free of it. There is evidence that suggests