For the first six months of the year, an analyst records the monthly closing stock price (in $) for a firm as: 90, 95, 96, 99, 91, 93. (You may find it useful to reference the table.) 121 Click here for the Excel Data File a. Calculate the sample mean and the sample standard deviation. (Round final answer to 2 decimal places.) Answer is complete and correct. Sample mean Sample standard deviation 94.00 3.35 b. Calculate the 95% confidence interval for the mean stock price of the firm assuming that the stock price is normally distributed. (Round final answer to 2 decimal places.) Confidence interval Answer is complete but not entirely correct. 91.55 to 96.44

For the first six months of the year, an analyst records the monthly closing stock price (in $) for a firm as: 90, 95, 96, 99, 91, 93. (You may find it useful to reference the table.) 121 Click here for the Excel Data File a. Calculate the sample mean and the sample standard deviation. (Round final answer to 2 decimal places.) Answer is complete and correct. Sample mean Sample standard deviation 94.00 3.35 b. Calculate the 95% confidence interval for the mean stock price of the firm assuming that the stock price is normally distributed. (Round final answer to 2 decimal places.) Confidence interval Answer is complete but not entirely correct. 91.55 to 96.44

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

See similar textbooks

Similar questions

what is the unexplained variation? how would i find the indicated prediction interval?
A researcher collected sample data for 19 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 192.4, with a standard deviation of 8.2. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 90% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: Upper limit:
Express the confidence interval 10.8%±5.1%10.8%±5.1% in interval form.
What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.92 per 100 pounds. (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. Round your answers to two decimal places.)lower limit $ upper limit $ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.25 for the mean price per 100 pounds of watermelon. (Enter a number. Round up to the nearest whole number.) farming regions (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop (in dollars). What is the margin of error (in dollars)? Hint: 1 ton is 2000…
In the past, a chemical company produced a certain type of plastic. Now, using a newly developed and less expensive process, the mean daily yield of plastic for the first 50 days of production is 871 pounds; the standard deviation is 21 pounds. (b) Find a 90% confidence interval for the mean daily yield of plastic per day for the newly developed process.
What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of = $6.88 per 100 pounds of watermelon. Assume that σ is known to be $1.90 per 100 pounds. (a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop (in dollars). What is the margin of error (in dollars)? (For each answer, enter a number. Round your answers to two decimal places.)lower limit $ upper limit $ margin of error $ (b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.37 for the mean price per 100 pounds of watermelon. (Enter a number. Round up to the nearest whole number.) farming regions (c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop (in dollars). What is the margin of error (in dollars)? Hint: 1 ton is 2000…
A researcher collected sample data for 20 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 190.4, with a standard deviation of 9.7. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 99% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 99% confidence interval.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place Lower limit= Upper limit=
It is not taking (0.1404, 0.3596) as the confidence interval
Imagine that the confidence interval around a candidate's popularity score is [37.22,48.86],what is the point estimate?