Statistics in Business - Assignment Example

Question µ = 20 minutes and σ = 3 minutes (a) P(X > 22) = ? The probability that an employee will take longer than 22 minutes to assemble one of these components is about 0.2514 (or 25.14%). (b) P(> 22) = ? , as population X is normally distributed. The probability that the average time for 15 employees exceeds 22 minutes is about 0.0049 (or 0.49%). (c) No, we do not need to call upon the central limit theorem to answer either of these questions. This is because the time it takes an employee to assemble the component (X) is normally distributed. Further, is normally distributed, since the population X is normally distributed. (d) P(X > 22) = 0.2525 {Excel Function = 1-NORMDIST(22,20,3,TRUE)} P(> 22) = 0.0049 {Excel Function = 1-NORMDIST(22,20,3/SQRT(15),TRUE)} Question 2 µ = 18 kg and σ = 5 kg Let X be the weights of suitcases taken by travellers on international journeys. (a) Total weight limit = 3500 kg and number of suitcases on board, n = 190, therefore kg is approximately normally distributed even though the population X is not normally distributes as n = 190 (> 30) (Central Limit Theorem). since n ≥ 190. = 0.1229 {Excel Function =1-NORMDIST(3500/190,18,5/SQRT(190),TRUE)} There is probability of about 0.123 that that weight limit of 3500 kg will be exceeded if the total number of suitcases on board is 190. (b) Yes, we would still be able to answer part (a) if we were now told that the weights of suitcases do follow a normal distribution. This is because we do not need to apply Central Limit Theorem, as is normally distributed, since the population of X is normally distributed for any n. Question 3 E = 0.5 units, α = 1 – 0.99 = 0.01 and σ = 1.75 Z0.01/2 = 2.575 = 82 (rounded up) The sample size required is 82. Question 4 (a) Step 1: = 15,500, s = 9950, n = 1000. The population is normally distributed. Step 2: 90% confidence  α = 0.1 and (The population variance is unknown) Step 3: The confidence interval is (b) Step 1: = 15,500, σ = 9950, n = 1000. The population is normally distributed. Step 2: 90% confidence  α = 0.1 and (The population variance is known) Step 3: The confidence interval is (c) The interval estimates in part (a) and (b) were virtually identical because the sample size of 1000 observations is very large. As the sample size (or d.f.) increases (d.f  ∞), the t-distribution approaches to the Z distribution. When the d.f. is infinitely large (i.e. d.f. = ∞), then the t distribution approaches the Z distribution. Question 5 (a) Step 1: = 5.125 hours, s = 3.310 hours, n = 200. The population is normally distributed. Excel command used for mean and sample standard deviation are “=AVERAGE(A2:A201)” and “=STDEV(A2:A201)”, respectively. Step 2: 95% confidence  α = 0.05 and (The population variance is unknown) Excel command used to determine t is “=TINV(0.05,199)”. Step 3: The confidence interval is We estimate (with 95% confidence) the mean amount of time all 15-year-old high school students devote per week to part-time jobs lies somewhere between 4.7 hours to 5.6 hours. Question 6 (a) The null and alternate hypotheses are (This is a two-tailed test (non-directional)) (b) Hypotheses Test Step 1: The null and alternate hypotheses Step 2: The test statistics Since, assumption made is that the population is normally distributed and the population variance σ2 is unknown, the test statistic is Step 3: Level of significance α α = 0.01 Step 4: Decision rule Reject H0 if |t| > tα/2,n – 1 = t0.005,60 – 1 = 2.662 {Excel command =TINV(0.01,59)} Step 5: Value of test statistic The value of t statistic Step 6: Decision Since t = 10.106 > 2.662 we reject H0 and conclude that there is enough statistical evidence to infer that the historical average price per square foot for apartments in Sydney has been changed from $30. Question 7 (a) Decision rule (Step 4) Reject H0 if z < -zα = -z0.05 = -1.645 (b) Value of the test statistic (Step 5) (c) Decision (Step 6) Since z = -1.040 > -1.645 we fail to reject H0 and conclude that there is not enough statistical evidence to infer that the mean is less than 24.4. Question 8 (a) In the context of this problem, it is important to use a one-tail test with right side because we want to test whether the metal stands can support average weight of more than 50 kg. If it can support average weight less than or equal to 50 kg than it will not pass safety test and the safety inspector will not certify the product for sale to the general public. (b) In the context of this problem, if inspector commits a Type I error (Rejecting the null hypothesis when it is in fact true) than he will wrongly certify the product for sale to the general public, although it not passes the safety test. (c) In the context of this problem, if inspector commits a Type II error (Not rejecting the null hypothesis when it is in fact false) than he will not certify the product for sale to the general public, although it passes the safety test. (d) If the company’s lawyers are worried about being sued for selling an unsafe product, they should decrease the value of α (e.g. from 0.05 to 0.01). By doing this, they will minimize the probability of making a Type I error and hence wrongly certifying the product for sale to the general public when actually the product fails the safety test. Question 9 Step 1: The null and alternate hypotheses Step 2: Table with observed and expected frequencies n = 180, k = 5 The expected frequencies ei are given below: For Brand A, eA = n(pA) = 180(0.20) = 36 For Brand B, eB = n(pB) = 180(0.20) = 36 For Brand C, eC= n(pC) = 180(0.20) = 36 For Brand D, eD = n(pD) = 180(0.20) = 36 For Brand E, eE = n(pE) = 180(0.20) = 36 The oi and ei are given below in table. Brand Observed Frequency (oi) Expected Frequency (ei) A 30 36 B 40 36 C 25 36 D 35 36 E 50 36 Total n 180 180 Step 3: Test statistic The statistic is d.f. = k – 1 Step 4: Level of significance α α = 0.05 Step 5: Decision Rule Reject H0 if Step 6: The value of test statistic Brand oi ei A 30 36 B 40 36 C 25 36 D 35 36 E 50 36 n 180 180 Sum = 10.278 The test statistics is = 10.278 Step 7: Decision Since = 10.278 > 9.488 we reject H0 and conclude that there is enough evidence that there is a preferred brand of orange juice. Read More