BB108 Business Statistics

Question: Discuss about the Report for Business Statistics. Answer: ANSWER 1 Our group number is 7, hence we used the data for group 7 The data for our group is given as Group 7 7 7 6 4 5 1 2 7 7 2 2 5 2 1 5 4 5 7 2 3 2 7 2 6 2 5 3 4 4 5 4 2 6 4 2 2 3 4 1 1 1 4 2 5 7 1 6 2 7 3 5 7 7 7 5 2 4 6 4 3 1 7 6 7 The frequency distribution of Best T-shirts Ltd can be depicted as Size of the T-shirt Frequency Small 7 Medium 14 Large 5 XL 10 2XL 9 3XL 6 4XL 13 In excel we use the command COUNTIF. The syntax for the command is COUNTIF(range, criteria) for example is if our data is from cell no A2 to A65 and we have to count how many 1 are there then we write COUNTIF(A2:A65,1) The total COUNT of T-shirts is 64. In excel we use the command COUNT(range). To count the number of T-shirts we used COUNT(A2:A65) The cumulative frequency and relative frequency can be depicted as Cumulative Frequency Relative Frequency 7 0.1094 21 0.2188 26 0.0781 36 0.1563 45 0.1406 51 0.0938 64 0.2031 To calculate the cumulative frequency we add the present frequency of the variable with the cumulative frequency of till the last variable. In our example since F3 cell contained the data of the first variable it was copied as it is. The frequency of the second was at F4. Hence we add the frequency of F4 to the cell containing the value of F3. To calculate the relative frequency we divide the frequency by the total count. The Mode represents the central tendency of this data. We are trying to find the frequency and thus which is the most frequently used t-shirt size. The sales of the t-shirts can be summarized as The medium size t-shirt is in the highest demand. 14 t-shirts of Medium size were sold. The large size t-shirt is in least demand. Only 5 were sold. From last months data we find that 2XL size was purchased by 14.06% of the people 3XL was purchased by 9.38% and 4XL was purchased by 20.31% of the people. Thus individually these proportions of the customers will be affected. Overall 43.75% of the customers will be affected. The variable T-shirt size is an ordinal data. The data contained in the t-shirt size has a logical order even though the differences in the order may not be constant. The variables Length in inches and Width in inches are interval data, but the interval data is not constant. ANSWER 2 The average number of Hours spent watching TV is 24.8. We used the AVERAGE function. Since the data was contained in cells from B2 to K11 hence the command was AVERAGE(B2:K11) The standard deviation of the sample was 9.1905 hours. We used the function STDEV.S and the command was STDEV.S(B2:K11) We can use the t-distribution table for constructing the 96% confidence interval. We need to find the value of . determines the level of confidence. Since the sample size is 100 we can find the value of this from the t-distribution table which is given as 2.054 The standard normal distribution table can also be used. 96% CI means 4% level of significance. Hence a = 4/100 = 0.04 Hence the total area below the critical value is From the standard normal distribution table we find the z value corresponding to 0.98 which is less than 2.06 but more than 2.05. For a better value we used the t-table. The point estimate is a single value which describes the population. Thus the sample mean can be used as the point estimate of the population mean. Thus the point estimate is 24.8 hours. The margin of error for the population mean is given as E = where = the table value for 96% confidence interval = 2.054 s = sample standard deviation = 9.1905 n = size of the sample = 100 thus the margin of error = Thus the 96% confidence interval for the population mean is given as 24.8 8877. The lower limit of the interval is 22.9123 and the upper limit of the interval is 26.6877. ANSWER 3 The binomial probability distribution for the successful heart transplant surgery is 7.63611E-42 The command used was BINOMDIST(number_s, trials, probability_s, cumulative) where number_s is the number of heart transplants that took place = 6 trials is the total number of surgery that took place in the hospital = 65 probability_s is the probability of success of heart transplant = 0.85 cumulative is the probability of success of heart transplant given that a total of 65 surgery took place = FALSE The mean of the binomial distribution is given as Where m = mean N = the total number of surgery = 65 n = the probability of success of heart surgery = 0.85 Thus m = 65x0.85 = 55.25 (Mean) The variance of the binomial distribution Hence s2 = 65x0.85x(1 0.85) = 8.2875 (Variance) The standard deviation of a binomial distribution = = 2.8788 (standard deviation) The total number of different surgeries that took place in the hospital is 65. Of these 65 surgeries the number of heart surgery was 6, and the probability of success of these 6 heart surgery was 0.85. If all the 65 surgeries, were heart surgeries then the average number of successful heart surgery would be 55.25, with a standard deviation of 2.8788. A heart surgery cannot have a decimal part. It is either a success or a failure. Hence in other words if we consider repeated data from 65 heart surgeries we would find that the mean number of successful heart surgery to be 55.25 and the standard deviation to be 2.8788. For Binomial distribution the unusual values are more than two times standard deviation on either side of the mean. The mean mx = 55.25 The standard deviation sx = 2.8788 The values of x are 55.25 2x2.8788 i.e x 49.4924 and 55.25+2x2.8788 i.e., x 61.0076