This document provides study material and solved assignments for the subject of Quantitative Methods for Business. It includes topics like simple random sampling, quota sampling, sample frame, cluster sampling, regression coefficients, and probability calculations.
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Quantitative Methods For Business
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Question 1: (a) Range Mid value (X)Frequency (F)FXx̅ - X( x̅ - X)2 F( x̅ - X)2 30 - 3532.517552.5-11.9583143.00172431.03 35 - 4037.524900-6.9583348.41841162.042 40 - 4542.519807.5-1.958333.83506972.86632 45 - 5047.52813303.0416679.251736259.0486 50 - 5552.519997.58.04166764.66841228.7 55 - 6057.513747.513.04167170.08512211.106 Total12053357364.792 I.Mean ( x̅ )=Ʃ FX / Ʃ F =44.45833 II.Standard deviation = =42.909 III. Result I show that on an average 44.45 seconds taken by computer disc to run particular program. On the other hand; result II shows that 42.909 seconds variations to its mean average occur by disc to run particular program. (b) I.Simple random sampling: Simple random sampling is the most basic and common type of sampling method commonly used in quantitative social science research and scientific research. The main advantage of simple random sampling is that each member of the population has an equal chance of being selected for study. This means that it guarantees that the selected sample is representative of the population
and that the sample is chosen in an unbiased manner. In turn, the statistical conclusions drawn from the analysis of the sample would be valid. II.Quota sampling: A quota is a type of sample that is a non-probability sample, in which the researcher selects people according to some set standard. That is, units are selected based on predetermined characteristics in a sample so that only one distribution of the assumed characteristics present in the total sample population is selected. III.Sample frame: Sample frame, known as "sample structure", refers to the scope of the overall sample and the sample to determine the composition of the composite unit can be selected as roster lists or sorting numbers. A sample frame design, by voting or units. The required number for drawing can be through a random number table. If there is no sampling frame, you can calculate the probability of the sample units. Cannot be calculated, and thus probability cannot be sampled. IV.Cluster Sampling: Cluster sampling refers to a sampling method in which members of a population are randomly selected, from naturally occurring clusters known as 'clusters'. The totality of each unit is a number of non-overlapping will merge, and the group called collection is not copied, so a sample group is a sample unit sample. The application of cluster samples, that is, group differences in different units within larger, inter-group differences, requires a better representation of the group. (c) RangeFrequency ( F) Cumulative Frequency (CF) 0 - 733 7 - 1469 14 - 211120 21 - 282646 28 - 35450
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Question 2: (a) The meaning of four regression coefficients is as follows: The value 0.794x and 0.427x shows positive slope; which indicates that any increase in x value which is independent value or distance travelled will simultaneously increase Y variable which is running cost of the car and vice a verse.On the other hand; values like 2.650 and 5.585 shows the intersection of two regression line at single point. (b) New travelling distance of car F = 12.1 * 1.5 = 18.15 New travelling distance of car L = 12.3 * 1.5 = 18.45 Choices: Total cost of car F: Y = 2.650 + 0.794X = 2.650 + 0.794 (18.15) =1,698.8 pounds Total cost of car L: Y = 5.585 + 0.427X = 5.585 + 0.427 (18.45) =1,346.5 pounds Hence, the cost of car L is low and therefore this option will be chosen for this region.
(c) Car L Distance Travelled Running Costs 10% increase XY 3.56.97.59 4.67.68.36 5.37.98.69 68.39.13 7.28.89.68 8.49.210.12 10.19.610.56 11.110.311.33 11.510.111.11 12.311.312.43 Mean =89.9 Expected total running costs for 5 of car L in this new region next year: =8×9.9×5 =39,600 pounds Question 3 (a) Probability of Chris not solving the problem P (C’) =1−1 6=5 6 Probability of Albert not solving the problem P (A’) =1−1 8=7 8 Probability of John not solving the problem P (J’) =1−1 3=2 3 I. Probability of none of them solves the problem =P (C’). P (A’). P (J’)
¿5 6×7 8×2 3=70 144 II. Probability of at least one of them solves the problem: ¿1−70 144=74 144 III. Probability of only one of them will solve the problem: = P(C)⋅P(A’)⋅P(J’)+P(A)⋅P(C’)⋅P(J’)+P(J)⋅P(A’)⋅P(C’) ¿{1 6×7 8×2 3}+{1 8×5 6×2 3}+{1 3×7 8×5 6}=59 144 (b) I. Probability of one is green and other is white: ¿4G 9+3W 5=47 45 II. Probability of both ball have same color: ¿2B 5+5B 9=18 45