Question

Pages Pages 2698 2028 Mean 2408.8 2474 Standard Error 96.2684672 2395 Median 2387 2372 Mode #N/A...

Pages Pages
2698
2028 Mean 2408.8
2474 Standard Error 96.2684672
2395 Median 2387
2372 Mode #N/A
2475 Standard Deviation 304.4276232
1927 Sample Variance 92676.17778
3006 Kurtosis 0.976147469
2334 Skewness 0.345521688
2379 Range 1079
Minimum 1927
Maximum 3006
Sum 24088
Count 10
Confidence Level(95.0%) 217.7744026

Please answer these questions by using the information above.

  1. What is the point estimate?
  1. What is the confidence interval (using the information you generated)?
  1. What is the approximate probability that the true mean number of pages printed by the new ink-jet printer is equal to or less than 2,191?
  1. Finally, verify your results by manually calculating the mean of the sample.   Go on to determine the standard deviation and the standard error of the mean. Then compute a confidence interval based on your manually created numbers. If your interval happens to differ from the one generated by excel, please try to explain why.

Homework Answers

Answer #1

i. Point Estimate = mean = 2408.8

ii. Confidence interval = 2408.8 217.7744 = (2191.0256 , 2626.5744)

iii. Probability that the true mean number of pages printed by the new ink-jet printer is equal to or less than 2,191:

Manual calculations:

Mean:

Standard deviation:

Standard error:

Confidence interval

Both confidence interval exact same only the difference between decimal places. This occur due to round of decimal places.

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
Current Dividend in $ Mean    0.14 Standard Error    0.01 Median    0.12 Mode   ...
Current Dividend in $ Mean    0.14 Standard Error    0.01 Median    0.12 Mode    0.05 Standard Deviation    0.12 Sample Variance    0.01 Kurtosis    1.19 Skewness    1.28 Range    0.51 Minimum    0.01 Maximum    0.52 Sum    12.07 Count    84 Use the above summary measures to compute a 95% confidence interval of the mean Dividend of all Companies that pay a dividend. How would you interpret this interval? A reputable financial advisor recently made...
Interpret the mean, median and mode and standard deviation for each variable according to the statistics....
Interpret the mean, median and mode and standard deviation for each variable according to the statistics. Statistics Advertising Sales N Valid 36 36 Missing 0 0 Mean 24.25    28 28.5278 Median 24.25 23.0000 Mode 15.50a 21.00 Std. Deviation 6.18142 18.77763 Variance 38.210 352.599 Skewness .044 .488 Std. Error of Skewness .393 .393 Kurtosis -.618     -.715 Std. Error of Kurtosis .768 .768 Minimum 12.00 1.00 Maximum 36.50 65.00 a. Multiple modes exist. The smallest value is shown
The cost of ink cartridges for inkjet printers can be substantial over the life of a...
The cost of ink cartridges for inkjet printers can be substantial over the life of a printer. Printer manufacturers publish the number of pages that can be printed from an ink cartridge in an effort to attract customers. A company claims that its black ink cartridge will yield an average 2 comma 387 pages. To test this​ claim, an independent lab measured the page count of 62 cartridges and found the average page count to be 2 comma 211.4. Assume...
The cost of ink cartridges for inkjet printers can be substantial over the life of a...
The cost of ink cartridges for inkjet printers can be substantial over the life of a printer. Printer manufacturers publish the number of pages that can be printed from an ink cartridge in an effort to attract customers. A company claims that its black ink cartridge will yield 495 pages. To test this​ claim, an independent lab measured the page count of 47 cartridges and found the average page count to be 485.8 Assume the standard deviation for this population...
Number -5 6 8 -2 6 9 8 2 1 6 2 1 Find the following...
Number -5 6 8 -2 6 9 8 2 1 6 2 1 Find the following : Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count Confidence Level(95.0%) Give three decimals for all answers. Please make sure that the meaning of three decimals is the same as this formate(111.xxx). Example: suppose you got the answer like 111.98764, the answer using just 3 decimals is the same this formating (111.988).
A simple random sample with n=54 provided a sample mean of 24.0 and a sample standard...
A simple random sample with n=54 provided a sample mean of 24.0 and a sample standard deviation of 4.3. a. Develop a confidence interval for the population mean (to 1 decimal). , b. Develop a confidence interval for the population mean (to 1 decimal). , c. Develop a confidence interval for the population mean (to 1 decimal). , d. What happens to the margin of error and the confidence interval as the confidence level is increased? - Select your answer...
Salary Hrly Rate Yrs of Svc Ed Age Mean 111379.5 53.54783 6.890181 15.45161 35.83333 Standard Error...
Salary Hrly Rate Yrs of Svc Ed Age Mean 111379.5 53.54783 6.890181 15.45161 35.83333 Standard Error 1677.28 0.806385 0.211498 0.092706 0.502823 Median 103853 49.92933 5.730556 16 35 Mode 86918 41.7875 15.79444 16 32 Standard Deviation 32350.21 15.55298 4.079219 1.788047 9.698093 Sample Variance 1.05E+09 241.8953 16.64003 3.197113 94.05301 Kurtosis 4.149903 4.149903 -0.39339 0.032418 -0.71449 Skewness 1.470986 1.470986 0.730403 0.17764 0.310279 Range 231063 111.088 15.40556 8 44 Minimum 59147 28.43606 0.388889 12 18 Maximum 290210 139.524 15.79444 20 62 Sum 41433167 19919.79...
34. The following data show the number of hours per day 12 STAT 3309 students spent...
34. The following data show the number of hours per day 12 STAT 3309 students spent in front of screens watching​television-related content. Use the printout below to answer the following​ question: HOURS 1.2 Mean 4.666666667 4.8 Standard Error 0.695802712 4.1 Median 4.75 4.7 Mode ​#N/A 7.9 Standard Deviation 2.410331299 7.5 Sample Variance 5.80969697 5.1 Kurtosis ​-1.182783193 2.3 Skewness 0.141232566 5.8 Range 7.1 1.9 Minimum 1.2 2.4 Maximum 8.3 8.3 Sum 56 Count 12 Confidence​ Level(95.0%) 1.531451444 What is the​ 95%...
A simple random sample with n=50 provided a sample mean of 22.5 and a sample standard...
A simple random sample with n=50 provided a sample mean of 22.5 and a sample standard deviation of 4.2 . a. Develop a 90% confidence interval for the population mean (to 1 decimal). ( , ) b. Develop a 95% confidence interval for the population mean (to 1 decimal). ( , ) c. Develop a 99% confidence interval for the population mean (to 1 decimal). ( , ) d. What happens to the margin of error and the confidence interval...
A simple random sample with n=50provided a sample mean of 23.5 and a sample standard deviation...
A simple random sample with n=50provided a sample mean of 23.5 and a sample standard deviation of 4.3 a. Develop a 90% confidence interval for the population mean (to 1 decimal).   ,   b. Develop a 95% confidence interval for the population mean (to 1 decimal).   ,   c. Develop a 99% confidence interval for the population mean (to 1 decimal).   ,   d. What happens to the margin of error and the confidence interval as the confidence level is increased?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT