ID: A 3. A small stock brokerage firm wants to determine the average daily sales
ID: 3317508 • Letter: I
Question
ID: A 3. A small stock brokerage firm wants to determine the average daily sales (in dollars) of stocks to their clients. A sample of the sales for 40 days revealed average daily sales of $200,000. Assume that the standard deviation of the population is known to be $18,000. Provide a 96% confidence interval estimate for the average daily sale. 4. In order to determine how many hours per week freshmen college students watch television, a random sample of 300 students was selected. It was determined that the students in the sample spent an average of 15 hours with a standard deviation of 3.2 hours watching TV per week. Provide a 99% confidence interval estimate for the average number of hours that all college freshmen spend watching TV per week.Explanation / Answer
Mean is 200000 and s is 18000
for sample size of 40, the standard error SE=s/sqrt(N)=18000/sqrt(40)=2846.0499
for 96% confidence, the z value can be calculated as the value for (1-0.96)/2 +0.96 or for 0.98 which is 2.05 from normal distribution table
thus lower bound is mean-z*SE=200000-2.05*2846.0499=194165.5977
upper bound is mean+z*SE=200000+2.05*2846.0499=205834.4023