Question

A hospital wants to determine if the type of treatment for pneumonia is a factor in recovery time? The table below shows the number of days to recovery for several randomly selected pneumonia patients that had various types of treatment.

Sent Home With Medicine	A Few Hours in the Hospital	Overnight Hospital Stay
30	25	20
21	12	12
19	15	20
16	17	8
26	11	16
26	11	15
13	20	6
25
22

Assume that all distributions are normal, the three population standard deviations are all the same, and the data was collected independently and randomly. Use a level of significance of α=0.01α=0.01.

H0: μ1=μ2=μ3H0: μ1=μ2=μ3
H1:H1: At least two of the means differ from each other.

1. For this study, we should use Select an answer TInterval χ²GOF-Test χ²-Test 2-PropZInt 1-PropZTest 2-SampTTest ANOVA T-Test 2-SampTInt 1-PropZInt 2-PropZTest
2. The test-statistic for this data =  (Please show your answer to 3 decimal places.)
3. The p-value for this sample =  (Please show your answer to 4 decimal places.)
4. The p-value is = Select an answer greater than alpha less than (or equal to) alpha  αα
5. Base on this, we should Select an answer *reject the null hypothesis *accept the null hypothesis* fail to reject the null hypothesis  hypothesis
6. As such, the final conclusion is that...
   • There is sufficient evidence to support the claim that treatment type is a factor in recovery time for pneumonia.
   • There is insufficient evidence to support the claim that treatment type is a factor in recovery time for pneumonia.

Answer 1

Date: 21/07/2019 Answer To test the hypothesis is that treatment type is a factor in recovery time for pneumonia at 5% level of significance The null and alternative hypothesis is, -- Ha at leasttwo H, is unequal For the study, ANOVA test should be used due to two or more means and assume the three population standard deviations are all the same. The F-test statistics is, By using Excel software, find F-test statistics with the help of following steps: 1) 2) 3) 4) 5) Import the data. Choose ANOVA: Single Factor from Data Analysis in Data menu. Choose Input Range and Colunn in Groups by Choose Output Range Click Ok

ANOVA Source of Variation Between Groups Within Groups Total F crit df 2 146.7516 5.08042 0.016439 3.492828| P-value Ss MS 293.5031056 577.7142857 20 28.88571 22 871.2173913 From the Excel output, the F-test statistics is 5.080 The p-value for this test is, From the Excel output, the p value for this test is 0.0164 Decision The conclusion is that p-value in this context is less than 0.05 which is 0.0164, hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that treatment type is a factor in recovery time for pneumonia. The result is statistically significant. so the null