Question

Try this now, Exercise: Suppose our hospital is concerned about the delay in transferring older patients to long-term institution after hospitalization. We are concerned that the delay in discharges from McAllen hospital is longer than Mission Hospital. To Evaluate our concern, we randomly select 15 recent discharges from each hospital and measured, in days the length of the delay. The result of the two sample as follow Mcallen Hospital: 15, 13, 16, 19, 10, 22, 23, 19, 18, 17, 15, 19, 16, 12, 8 Mission Hospital: 10, 13, 16, 18, 12,9,18, 15, 19, 11, 13, 9, 19, 15, 18, 20 if α-oos, use the data to evaluate the claim. Assume the Population variances are not equal.

Answer 1

(a) The Computation Table 10 13 1.13333 1284437 -3.13333 9.817757 0.133330.017777 2.866678.217797 -6.13333 37.61774 5.86667 34.41782 6.8666747.15116 2.86667 8.217797 1.86667 3.484457 0.866670.751117 1.13333 1.284437 2.86667 8.217797 0.13333 0.017777 413333 17.08442 813333 66.15106 -4.6875 21.97266 1.68752.847656 1.31251.722656 3.3125 10.97266 -2.6875 7.222656 -5.6875 32.34766 .3125 10.97266 0.3125 0.097656 43125 18.59766 3.6875 13.59766 1.6875 2.847656 -5.6875 32.34766 .3125 18.59766 0.3125 0.097656 3.3125 10.97266 5.3125 28.22266 15 13 16 18 12 10 23 19 15 19 13 15 19 19 15 12 20 16 243.7333 213.4375 The formula for calculating the point estimate of the population mean is 15 (242) 16.13333 16 13333 = Therefore, the point estimate of the population mean is X16 13333 The formula for calculating the sample standard deviation is VI5-(243.7333) 243.7333 14 17.40952 -4.172472 4172472 Therefore, the sample stan dard deviation is s, . 172472

The formula for calculating the point estimate of the sample mean is 16 (235) 16 14.6875 14.6875 Therefore, the point estimate of the sample mean is Y 14.6875 The formula for calculating the sample standard deviation is (213.4375) 16-1 213.4375 = 14.229 17 3.772157 S 3.772157 Therefore, the sample standard deviation is s, 3.772157

(b) Let A4 and約bethe population means And let x1 and x2 be the means ofthe respective samples The summary stati stics is Sample1 m, 15, 16.13333, s 4.172472. Sample2m-16,-14.6875, s -3.772157 The level of significance is 0.05 Step1: The degrees of freedom for the unequal variances uneual 1 n2 4.172472) 3.772157) 15 16 (4.172472) (3.772157) 15 15 15-1 15-1 17.40952 14.22917 15 16 17.40952 15 15-1 14.22917 15 15-1 (1.160635+0.889323) (1.160635)', (0.889323) 15-1 16-1 (2.049958)2 (1.160635)', (0.889323) 14 15 (2.049958) 0.09622+0.052726 4.202327 0.096220.052726 4.202327 0.148946 28.21379 28 unequal Step 2: State the null and altemative hypotheses We are to test for the difference in the two populati on means The null and alternative hypotheses are, respectively Ho:4(There is no significant difference between the populati on means) H H> (The populati on mean 1 is greater than the populati on mean2)

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Step3 16.13333-14.6875 17.40952 14.22917 4207) 1.445833 1.160635+0.889323) 1.445833 2.049958 1.445833 1.431767 -1.009824 Step4 Since, on using Excel2010 function -tinv(probability,df) The t-table value is tons (28) -1.701 Step5 Make a decisi orn Since, the value of the test statistic is t-1.01 <005,.701 forz^-2 falls in the non-rejecti on region, we do not reject the null hypothesis H0 There is not sufficient evidence to conclude that the delay in discharges from McAllen hospital islonger than Mission hospital