Statistics for business - Week 6 assignment - Assume that data contained in the BUSI1013-Numerical Data A file is from a simple random sample of

Statistics for business - Week 6 assignment - Assume that...

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Assume that data contained in the BUSI1013-Numerical Data A.xls file is from a simple random sample of customers of a Canadian retailer. Use this data to answer the following: ( 4 Points ) a. Develop a point estimate of the mean number of transactions for the retailer. b. Develop a point estimate of the population standard deviation of the number of transactions. 1. a) x = ∑xi / n - x = 1298 / 30 = 43.26 - Therefore the point estimate mean for the number of transactions = 43.26 b) s = (SQRT of ∑(xi – x)^2 / n-1 - s = (SQRT of 5115.87 / 29) - s = (SQ of 176.41) = 13.33 - Therefore the point estimate standard deviation for the number of transactions is 13.33 The average waiting time for customers at a fast food restaurant is 3.5 minutes. The population standard deviation is 2.2 minutes. ( 6 Points ) a. Assume that the waiting time of customers at this restaurant follow a normal distribution, what is the probability that a simple random sample of 60 customers will have a sample waiting time of less than 3 minutes? b. Do we need to make any assumption on the distribution of waiting times for the calculation in Part a? Why or why not? c. What is the probability that a simple random sample of 120 customers will have a sample waiting time of less than 3 minutes? 2. a) P(x < 3) - Ox = o / (SQRT of n) - Ox = 2.2 / 7.75 = 0.28 - z = x – u / o
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- z = 3 – 3.5 / 0.28 = -1.79 - Using Standard Table: - 1.79 = 0.0367 - Therefore the probability that the sample of 60 customers will wait less than 3 minutes = 0.0367 b) Yes, we had to assume that the population was infinite, making it appropriate to need to compute a new standard deviation for the given sample population - We also assumed that the restaurant follows a normal distribution, allowing a standard normal conversion and calculation to be made to find the probability. c) Ox = o / (SQRT of n) - Ox = 2.2 / 10.95 = 0.2 - z = 3 – 3.5 / 0.2 = -2.5 - Using the standard table: -2.5 = 0.0062 - Therefore the probability that the sample of 120 customer’s will wait less than 3 minutes = 0.0062 Traditionally, 30% of customers of a cell phone company leave after their term contract expires. If this is the true population proportion of churn (customers leaving after their term expired), answer the following questions: ( 6 Points ) a. If simple random samples of 100 customers are taken from the population of customers of this company, describe the sampling distribution of the sample proportion of churn.
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