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Question Four a) The mean monthly sale of an insurance company is Kshs.81, 590. The manager designed a program to improve sales. Ten agents were selected randomlyto participate in the program. Their monthly sales were recorded in thousands as follows: By testing hypothesis, does this data provide enough evidence to indicate that the program was a success? b) A sample of 150 students selected randomly from a population of students in college. Their agesare presented in the table below: Age 19-23 24-28 29-33 34-38 39-43 44-48 No. of students 18 26 49 34 16 7 The diameter of shafts produced in a factory conforms to normal distribution. 31% of the shafts have a diameter less than 45 mm and 8 percent have more than 64 mm. Find the mean and standard deviation of the diameter of shafts. c) F10 Prtsc F11 F12 6
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a) Let X be the monthly sale in thousands We calculate mean and standard deviation of the sample data using Excel functions Average and S 10 81.6 12.2038 81.59 0.05 Sample Size Sample Mean Sample Standard Deviation Population Mean Assume 5% level of significance Since population standard deviation is unknown and sample size is small we use t-distributior Hence the null and alternative hypothesis are Ho : H 81.59 Ha : μ > 81.59 This is a one tailed right sided test To check if sales have improved 81.6 - 81.59 12.2038-0.0026 10 t = 0.0026 df Degrees of freedom 10 1 10 -1 9 α-0.05 we find critical value of t using Excel function T.INV(α, df) such that t-crit T.INV(0.95,9) t-crit- 1.8331 Decision rule for Rejecting the null hypothesis is Reject Ho if calculated t> 1.8831 Rounded to 3 decimals Since the test is right tailed we take 1 0.0026 <1.83311 Thus t-value >t-crit Hence we do not reject Ho Conclusion There does not exist enough statistical evidence at 0.05 level of significance to show that the program was a success (that is the sales did not increase) b) Note: Question is not clear.Let X be the diameter of shafts Let μ be the mean and σ be the standard deviation Given P(X < 45) 0.31 Converting to z-scores P(Z < z) = 0.31 where c) 45-μ From Excel function NORM.S.INV we find z NORM.S.INV(0.31) z0.4959 45-μ 0.4959- μ-0.4959ơ 45 Also P(X> 64) 0.08 1-P(X < 64) = 0.08 P(X < 64) 0.92 Converting to z-scores P(Z <z) 0.92 where-, 64-μ From Excel function NORM.S.INV we find z = NORMS. INV(0.92) z = 1.4051 1.4051 + 1.405|ơ 64 Solving (1) and (2) we get Mean μ-49.9564 Standard Deviation σ: 9.9947

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