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Question (1): A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the .10 significance level, is there a difference in the population means? Solution: Let 1 is the mean of the population I and 2 be the mean of population II. We set up the Null and Alternate hypothesis as below: Null Hypothesis : H0 : 1 = 2 ( there is no difference in population means) Alternate Hypothesis : H1 : 1 ≠ 2 ( there is a difference in population means ) We tabulate the given data as below: Sample Mean Sample Standard Deviation Sample Size Population - I 350 ( x1) 12 (s1) 15 (n1) Population - II 342 (x2 ) 15 (s2 ) 17 (n2) An unbiased estimate of the common population standard deviation is given by (n1 1) s1 (n2 1) s 2 n1 n2 2 2 S= = 2 (15 1)(12) 2 (17 1)(15) 2 15 17 2 2016 3600 30 5616 = = 13.6822 . S = 13.68 30 The standard error of the difference between the two means is given by: 1 1 S.E. = S n1 n2 = 1 1 17 15 32 13.68 13.68 15 17 255 255 = 13.68 × 0.3542 = 4.8455 Therefore the standard error (S.E.) = 4.8455 Now we compute the t-value. x x2 350 342 t= 1 = 1.6510 S .E. 4.8455 = 13.68 Degree of freedom = n1 + n2 – 2 = 15 + 17 – 2 = 30 At 30 degrees of freedom and 0.10 significance level the table value of t is 1.310. Since the computed value of t ( 1.6510 ) is greater than the table value of t (1.310), hence we reject the Null hypothesis and accept the Alternate hypothesis. Therefore, we conclude that there is a significant difference in population means at 0.10 level. Question (2): Ms. Lisa Monnin is the budget director for Nexus Media, Inc. She would like to compare the daily travel expenses for the sales staff and the audit staff. She collected the following sample information. Sales($) 131 135 146 165 136 142 Audit($) 130 102 129 143 149 120 139 At the .10 significance level, can she conclude that the mean daily expenses are greater for the sales staff than the audit staff? What is the p-value? Solution : Let 1 represent the mean daily expenses of sales staff and 2 represent the mean daily expenses of audit staff. We set up the Null and Alternate hypothesis as below : H0 : 1 = 2 Null Hypothesis : Alternate Hypothesis : H1 : 1 > 2 Sales (x1) 131 135 146 165 136 142 x1 = 855 ( x1 x1 ) -11.5 -7.5 3.5 22.5 -6.5 -0.5 ( x1 x1 ) 2 132.25 56.25 12.25 506.25 42.25 0.25 749.5 Audit (x2) 130 102 129 143 149 120 139 x2 = 912 ( x2 x2 ) - 0.3 -28.3 - 1.3 12.7 18.7 -10.3 8.6 ( x2 x 2 ) 2 0.09 800.89 1.69 161.29 349.69 106.09 75.69 1495.43 x1 = 855 , n1 = 6 , x 1 ( x1 x1 ) 2 = 749.5 855 912 = 142.5 , x2 = 912 , n2 = 7 , x 2 = 130.3 6 7 , ( x2 x 2 ) 2 = 1495.43 ( x1 x1 ) 2 749.5 Sales sample variance = s12 = = 124.92 n1 6 ( x 2 x 2 ) 2 1495.43 Audit sample variance = s2 = = 213.63 n2 7 An unbiased estimate of the common population standard deviation is given by 2 (n1 1) s1 (n2 1) s 2 n1 n2 2 2 S= 2 (6 1)(124.92) (7 1)(213.63) 672 624.6 1281.78 = 13.15 11 S = 13.15 = Standard error ( S.E.) = S. 1 1 n1 n2 1 1 6 7 13 = 13.15 42 = 13.15 × 0.5563 = 7.315 we have S.E. = 7.315 Now we compute the value of t. = 13.15 x1 x 2 142.5 130.3 t = = 1.6678 S.E. 7.315 the computed value of t = 1.6678 degrees of freedom = n1 + n2 – 2 = 6 + 7 – 2 = 11 At 11 degrees of freedom and at 0.10 significance level the computed value of t ( 1.6678) is greater than the table value of t ( 1.363 ), therefore , the difference between the two means is significant. Hence we reject the Null hypothesis and accept the Alternate Hypothesis. Hence it is concluded that the mean daily expenses of sales staff is more than the mean daily expenses of the audit staff. The p value of the test is P( Z ≥ 1.66) = 0.399 . Therefore the probability that the mean daily expenses of sales staff being more than the mean daily expenses of the audit staff = 0.399. Question (3): The management of Discount Furniture, a chain of discount furniture stores in the Northeast, designed an incentive plan for salespeople. To evaluate this innovative plan, 12 salespeople were selected at random and their weekly incomes before and after the plan were recorded. Sales person Before ( in $) After ( in $) Sid Mahone 320 340 Carol Quick 290 285 Tom Jackson 421 475 Andy Jones 510 510 Jean Sloan 210 210 Jack Walker 402 500 Peg Mancuso 625 631 Anita Loma 560 560 John Cuso 360 365 Carl Utz 431 431 A.S.Kushner 506 525 Fern Lawton 505 619 Was there a significant increase in the typical salesperson's weekly income due to the innovative incentive plan? Use the .05 significance level. Estimate the p-value, and interpret it. Solution: Let 1 and 2 represent the typical sales person’s weekly income before and after the innovative incentive plan respectively. We set up the Null and Alternate Hypothesis as below: H0 : 1 = 2 ( No increase in weekly income ) H1 : 2 > 1 ( there is an increase in weekly income ) d n Here we apply the t-test difference formula : t= s Sales Person Before (in $) After (in $) Difference (d) Sid Mahone 320 340 20 Carol Quick 290 285 -5 Tom Jackson 421 475 54 Andy Jones 510 510 0 Jean Sloan 210 210 0 Jack Walker 402 500 98 Peg Mancuso 625 631 6 Anita Loma 560 560 0 John Cuso 360 365 5 d2 400 25 2916 0 0 9604 36 0 25 Carl Utz A.S.Kushner Fern Lawton 431 506 505 431 525 619 We have n = 12 , d = 311 d = d 2 n (d ) 2 S= n 1 s = 40.75 0 0 19 361 114 12996 d = 311 d2 = 26363 d 311 = 25.92 n 12 26363 12(25.92) 2 = 12 1 18300.8 = 40.75 11 d n Now we find the value of t using the above stated formula : t = s 25.92 12 t= = 0.6361 × 3.4641 = 2.2035 40.75 Therefore the computed value of t = 2.2035 Degrees of freedom = n – 1 = 12 – 1 = 11 At 11 degrees of freedom, at 0.05 significance level the calculated value of t (2.2035) is greater than the table value of t (1.796). Therefore, we reject the Null hypothesis and accept the Alternate hypothesis. Hence, we conclude that there is a significant increase in weekly income after the innovative plan. The p-value of the test is P(Z ≥ 2.203) = 0.5 – 0.4871 = 0.0129