HOMEWORKOFTHEBUSINESSSTATISTICS应用统计学课后习题.doc

HOMEWORK OF THE BUSINESS STATISTICSMAJOR: MANAGEMENT SCIENCENAME: MENG ZEHUASTUDENT ID: 20090123618.54SOLUTION:(a). From the question we can know that the sample mean is =9.7 days and the sample standard deviation is S=4.0 days. Using the row for 24 degrees of freedom, for 95% confidence, we can find =2.0639 from the table. Since n=25, using Equation,We are 95% confident that the mean number of absences for clerical workers during the year is between 8.04888 and 11.35112 days .Although the true mean may or may not be in this interval, 95% of intervals formed in this manner will contain the true mean.(b).Since X =12 5 , n-X=13 5 , using Equation,P=0.48, and with a 95% level of confidence = 1.96,We are 95% confident that the population proportion of clerical workers absent more than 10 days during the year is between 0.284 and 0.676. Although the interval from 0.284 to 0.676 may or may not contain the true proportion, 95% of intervals formed from samples of size 25 in this manner will contain the true proportion.(c).Using Equation and e=1.5, s=4.5,and =1.96 for 95% confidence,=35Therefore, we should select a sample size of 35 clerical workers because the general rule for determining sample size is to always round up to the next integer value in order to slightly oversatisfy the criteria desired.(d).Because no information is available from past data, assume that =0.50.Using Equation and e=0.075, =0.50,and =1.645 for 90% confidence,Therefore you need a sample of 121 clerical workers to estimate the population proportion to within 0.075 with 90% confidence.(e).If a single sample were to be selected for both purposes, the larger of the two sample sizes (n=121) should be used.8.56SOLUTION:(a).From the question we can know that the sample mean is =$38.54 and the sample standard deviation is S=$7.26. Using the row for 59 degrees of freedom, for 95% confidence, we can find =2.0010 from the table. Since n=60, using Equation,We are 95% confident that the population mean amount spent per customer in the restaurant is between $36.665 and $40.415.Although the true mean may or may not be in this interval, 95% of intervals formed in this manner will contain the true mean.(b). Since X =18 5 , n-X=42 5 , using Equation,P=0.3, and with a 90% level of confidence = 1.645,We are 90% confident that the population proportion of customers who purchase dessert is between 0.2027 and 0.3973. Although the interval from 0.2027 to 0.3973 may or may not contain the true proportion, 90% of intervals formed from samples of size 60 in this manner will contain the true proportion.(c).Using Equation and e=1.50, s=8, and =1.96 for 95% confidence,=110Therefore, we should select a sample size of 110 customers because the general rule for determining sample size is to always round up to the next integer value in order to slightly oversatisfy the criteria desired.(d).Because no information is available from past data, assume that =0.50.Using Equation and e=0.04, =0.50,and =1.645 for 90% confidence,Therefore you need a sample of 423 customers to estimate the population proportion to within 0.04 with 90% confidence.(e).If a single sample were to be selected for both purposes, the larger of the two sample sizes (n=423) should be used.8.58SOLUTION:(a)&(b). Using the calculator we can know that the sample mean is =8.421 inches and the sample standard deviation is S=0.046 inches. Using the row for 48 degrees of freedom, for 95% confidence, we can find =2.0106 from the table. Since n=49, using Equation,We are 95% confident that the mean width of the troughs is between 8.4078 and 8.4342 inches. Although the true mean may o