Question

the prices in dollars for a particular phone are shown below for 7 online retailers. estimate...

the prices in dollars for a particular phone are shown below for 7 online retailers. estimate and interpret the true mean price for this particular phone with 95% confidence.

1015   1022   977   1000   985   950   998

Homework Answers

Answer #1

From the sample data,

Sample mean = 992.4286

Sample standard deviation = S = 24.3780

t critical value at 0.05 level with 6 df = 2.447

95% Confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

992.4286 - 2.447 * 24.3780 / sqrt(7) < < 992.4286 + 2.447 * 24.3780 / sqrt(7)

969.88 < < 1014.98

95% CI is ( 969.88 , 1014.98 )

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The prices (in dollars) for a graphing calculator are shown below for 8 online vendors. 121...
The prices (in dollars) for a graphing calculator are shown below for 8 online vendors. 121 125 151 129 127 133 121 125 (1) Estimate the true mean price for this particular calculator with 95% confidence. (2) Test at α = 0.05 to see if the mean price for this particular calculator is higher than $122.
Digital Camera Prices The prices (in dollars) for a particular model of digital camera with 6...
Digital Camera Prices The prices (in dollars) for a particular model of digital camera with 6 megapixels and an optical 3X zoom lens are shown below for 9 online retailers. Round sample statistics and final answers to at least one decimal place. 221 180 200 187 265 222 197 195 181 Estimate the true mean price for this particular model with 90% confidence. Assume the variable is normally distributed.
The following data represent the prices for a 4 GB flash memory card from online retailers....
The following data represent the prices for a 4 GB flash memory card from online retailers. Treat these data as a simple random sample of all online retailers. It was verified that the data are normally distributed and that s = 2.343 dollars. 12.25 13.49 13.76 14.75 14.99 14.99 15.27 15.95 17.88 20.49 Determine the chi-square critical values using a 95% confidence. Construct a 95% confidence interval for the standard deviation price of a 4GB flash memory card. Interpret the...
Two online retailers offer a particular style of webcam for sale. The prices of these cameras...
Two online retailers offer a particular style of webcam for sale. The prices of these cameras vary day by day. Consider two groups: Group #1 consists of consumers who purchased a camera from retailer #1 during the past three months and Group #2 consistes of consumers who purchased a camera from retailer #2 during the same period of time. A 99% confidence interval for ?1−?2μ1−μ2, the difference in population mean sale prices, is 30 to 60 dollars. (b) The confidence...
A marketing specialist wants to estimate the average amount spent by visitors to an online retailers...
A marketing specialist wants to estimate the average amount spent by visitors to an online retailers newly-designed website. From the data in a preliminary study she guesses that the standard deviation of the amount spent is about 14 dollars. Question 1. How large a sample should she take to estimate the mean amount spent to within 5 dollars with 90% confidence? (Round your answer up to the next largest integer).
Part A: A manufacturer of a brand of designer jeans realises that many retailers charge less...
Part A: A manufacturer of a brand of designer jeans realises that many retailers charge less than the suggested retail price of $100. The manufacturer provides you with the information she collects from a random sample of 12 retailers and asks you to perform some statistical analysis to have some statistical evidence on this realisation. You find that the mean of prices of the jeans is $70 and find the 95% confidence interval for the mean retail price of jeans...
Question 1. Online shopping statistics are routinely reported by www.shop.org. Of interest to many online retailers...
Question 1. Online shopping statistics are routinely reported by www.shop.org. Of interest to many online retailers are gender-based differences in shopping preferences and behaviours. The summary data of monthly online expenditures for a sample of male and female online shoppers are shown in the following table: Male Female n 45 45 Mean $352 $310 StDev s $95 $80 Find the 90% confidence interval for the difference in mean monthly online expenditures between males and females.   lower bound of confidence interval...
A random sample of the closing stock prices in dollars for a company in a recent...
A random sample of the closing stock prices in dollars for a company in a recent year is listed below. Assume that sigma is ​$2.29. Construct the 90​% and 99​% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. 18.16 17.37 20.79 21.54 16.89 19.22 22.91 18.71 15.42 15.21 20.67 20.85 18.53 22.85 18.47 17.13 The​ 90% confidence interval is left parenthesis $ nothing comma $ nothing right parenthesis . ​(Round to...
The random sample shown below was selected from a normal distribution. 9​, 5​, 3​, 4​, 7​,...
The random sample shown below was selected from a normal distribution. 9​, 5​, 3​, 4​, 7​, 8 a. Construct a 95​% confidence interval for the population mean mu.
The table below lists weights​ (carats) and prices​ (dollars) of randomly selected diamonds. Find the​ (a)...
The table below lists weights​ (carats) and prices​ (dollars) of randomly selected diamonds. Find the​ (a) explained​ variation, (b) unexplained​ variation, and​ (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear​ correlation, so it is reasonable to use the regression equation when making predictions. For the prediction​ interval, use a​ 95% confidence level with a diamond that weighs 0.8 carats. Weight 0.3 0.4 0.5 0.5 1.0 0.7    Price ​$516 ​$1175 ​$1333 ​$1416 ​$5673...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT