Question

8 points 2. An Insurance Company reports that the mean cost to process claim is $50. An industry comparison showed this amount to be larger than most other insurance companies, so they instituted cost-cutting measures. To evaluate the effect of cost- cutting measure, the company selected a random sample of claims processed last month and determine the cost to process these selected claims. The sample information is reported as follows: 45, 48, 52, 58, 38, 42. At the 0.05 significance level is it reasonable to conclude that the mean cost to produce a claim is now less than $50?* Your answer

Answer 1

Solution Given data 45, 48,52, 58, 38, 42 n=6 x=0.05 * Now to find the sample mean (x) X= 2x 45+48 +52 +58 + 38+42 X= 6 X= 283 6 deviation X= 47.167 Now to find the stan sample standard {(xi-x) ² S=1 (45-47.167) ?+(48-47.167) +(52-47.167) + (58 - 47.167)2+ S=1 (38-47.16-) + (42-47.167) ? 6-1 256.8333 S= 5 S= V 51.3667

S= 7.1671 Now to state the mall and alternative hypothesis Null hynthesis Ho: 4 =50 Altesi native hypothesis H :d <50 This is "left-tailed' test. * Now to find the test statistic We know that The formula of test statistic is given by Test - statistic, t = X-M 2 47.667-50 71678/VG -2.833 t= 2.9260 * t = -0.968 Test statistic, t= -0.968 Now to find the degrees of freedom degrees of freedom - n-1 df = 6-1 df =

* Now to find the Pvalue 3 P-value = 0.1888 Decision P-value > 0.1888 >0.05 Since the P-value is greater than significance level. Therefore We do not reject the null hypothesis (H) Conclusion : There is not sufficient evidence to conclude that the Cost to Produce a claim claim is now less than $50 mean