Question

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 =0.05. (20 points, 4 points each)

1) Write down the hypothesis

2) What is the formula of test statistics?

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3) What is the critical value?

4) What is the test value? (show your work below)

5) What is the conclusion? What does the result mean?

1. 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 0=0.05. (20 points, 4 points each) 1) Write down the hypothesis 2) What is the formula of test statistics? 3) What is the critical value? 4) What is the test value? (show your work below) 5) What is the conclusion? What does the result mean?

Answer 1

s solution ceixen the information, n=40 uc population Meas PEF for childeren foee of asthmg = 306 n=40 X - scrmorple mean DEE - 2.79 = sample standaod deviation - Z) e level of significance - 0.05 1) TO test the claim mees PEF lower mean of PEE IO children with chronic bronchitis irea to test Hoiu = 300 Hile u < 306 (claim) vs 2) The foomula for test statisticis, t X-4 4) + slun = 279 - 306 711140 t - 2.4 D5 def=degree 33 coitical value=_toped-te) off freedom = n-1 dof. = 40-1 = 39 (612) = (0 05/2) = 0·025 t6-025,393 - 2:023 (From t-table] , coitical xolue = 2:023 coltical method Decision Relle using ro Reject to when I t-statistic) 7, lt-coitical l g fail to reject otherwise 97 % level of significance.

2.405 z 2.023 Reject Ho Qt 5% level of significance. conclusion > the say that there is sufficient evidence to suppost the clair lowey mean of PEE o children with cool chronic bromwitis at 5 % level of ficance of