Newt Dimswitch owns a small construction company in Chesapeake, Virginia. He often uses more than one construction crew, and has had a difficult time maintaining records on profit levels, cost controls, and other variables that are necessary in making business decisions. Newt somehow feels that the use of one of his construction crews is not proving as profitable as other crews when sent out on a job. To make some kind of intelligent estimate of profit levels Newt collects cost data for the crew in question for the last 28 jobs it completed.
Completion times, in hours, for the these jobs were 79, 83, 77, 81, 65, 92, 72, 71, 90, 79, 73, 70, 64, 91, 77, 80, 65, 85, 65, 89, 80, 65, 73, 78, 68, 88, 82, and 69. Newt must pay his crew a total of $112 per hour while it is on the job. This crew has bought in revenues on these jobs of $11200, 11700, 11500, 11100, 8300, 11000, 9900, 9400, 11500, 9200, 10200, 9100, 8400, 10300, 10900, 10900, 8300, 9700, 8600, 11200, 11200, 8600, 9200, 10900, 8700, 11100, 10400, and 8400.
Now that Newt has these data, he isn't sure what to do with them. He heard something about confidence intervals could be used to make estimates. Your job is to help poor Newt decide if this crew is profitable.
Note: Profit = Revenue - Cost
1. Enter the completion time data in one column. Use Time as your header. Select Stat > Basic Statistics > 1 - sample t... The variable is Time. This will yield a 95% confidence interval for the mean job completion time for this crew.
2. Determine the 95% confidence interval for cost of a job by this crew. (Hint: use interval from question 1) Show calculations
3. Enter the revenue data in one column. Use Revenue as your header. Select Stat > Basic Statistics > 1 - sample t... The variable is Revenue. this will yield a 95% confidence interval for the mean revenue for this crew.
4. Now, is this crew making a profit? Determine the 95% confidence interval for
profit. (Hint: use intervals from questions 2 & 3). Show calculations.
This solution is comprised of a detailed explanation of confidence interval using excel. In this solution, step-by-step explanation of this complicated topic provides students with a clear perspective of confidence interval using excel.
Statistics: Tweens, Rainfall, Body Temperature
When it comes to advertising, "'tweens" are not ready for the hard line messages that advertisers often use to reach teenagers. The Geppeto Group study found that 78% of 'tweens understand and enjoy ads that are silly in nature. Suppose that the study involved n = 1030 'tweens.
a. Construct a 90% confidence interval for the proportion of 'tweens who understand and enjoy ads that are silly in nature.
b. Do you think that "more than 75%" of all 'tweens enjoy ads that are silly in nature? Why?
What is the normal body temperature for healthy humans? A random sample of 130 healthy human body temperatures provided by Allen Shoemaker yielded x (with a line over it) =98.25 degrees and standard deviation 0.73 degrees.
a. Give a 99% confidence interval for the average body temperature of healthy people.
b.Does the confidence interval obtained in part (a) contain the value 98.6 degrees, the accepted average temperature cited by physicians and others? What conclusions can you draw?
Refer to Exercise 1. How many 'tweens should have been interviewed in order to estimate the proportion of 'tweens who understand and enjoy ads that are silly in nature, correct to within .02, with probability .99? Use the proportion from the previous sample in approximating the standard error of the estimate.
Suppose that you want to estimate the mean pH of rainfalls in an area that suffers from heavy pollution due to the discharge of smoke from a power plant. Assume that - is in the neighborhood of .5 pH and that you want your estimate to lie within .1 of ? with probability near .95. Approximately how many rainfalls must be included in your sample (one pH reading per rainfall)? Would it be valid to select all of your water specimens from a single rainfall? Explain.
Although there are many treatments for bulimia nervosa, some subjects fail to benefit from treatment. In a study to determine which factors predict who will benefit from treatment, Wendy Baell and E. H. Wertheim found that self-esteem was one of the important predictors. The mean and standard deviation of post treatment self esteem scores for n = 21 subjects were x (with a line over)= 26.6 and s = 7.4, respectively. Find a 95% confidence interval for the true post treatment self esteem scores.