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# Class_29 - Example A hospital administrator claims that the...

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Example A hospital administrator claims that the average length of stay of patients admitted to the hospital is 5 days. An insurance company is investigating the possibility that the average length of stay is actually greater than 5 days. They select a random sample of 100 patients from last year and they find that the sample average length of stay of patients is 6 days. We assume that the population is approximately Normal with σ = 3 days. Does the data support the insurance company’s claim at level α = . 01? 0 Identify Parameter of Interest True mean time a person is hospitalized - μ TH 1 Identify test: n = 100 30 z-test for means 2 Formulate H 0 and H A : (Hint: start with H A ) H 0 : μ TH = 5 (innocent until proven guilty) H A : μ TH > 5 (company’s suspicion) Example 3 Compute test statistic (z-score) z obs = ¯ x - μ 0 ( s or σ ) / n = 6 - 5 3 / 100 3 . 33 4 Using Table A2 compute p-value: H A : μ TH > 5 p-value= P ( Z z obs ) = P ( Z 3 . 33) z obs = 3 . 33 Table A2 . 9996 (Area to the left of 3.33) p-value= P ( Z 3 . 33) = 1-.9996 = .0004 Interpretation: The probability of getting a sample mean of 6 days ( ¯ X ) or greater (see H A ), if the true mean hospitalization time was 5 days ( H 0 ), is .0004 or .04%. 5 Decision and Conclusion: level α = . 01 since p-value = .0004 < .01 = α , we reject the H 0 We have very strong evidence at α level .01 that the mean hospitalization time at the specific hospital is greater than 5 days. Example From extensive records it is known that the duration of treating a disease by standard therapy has a mean of 15 days. Some researchers think that a new therapy can reduce the treatment time. We consider a sample of 25 patients receiving this new therapy and their time to recovery are recorded. From this sample we observe an

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Class_29 - Example A hospital administrator claims that the...

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