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In a random sample of four mobile devices, the mean repair cost was \$60.00 and the standard deviation was \$14.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results. The 90% confidence interval for the population mean is (DO (Round to two decimal places as needed.) The margin of error is \$ (Round to two decimal places as needed.) Interpret...

• ### In a random sample of six mobile devices, the mean repair cost was \$70.00 and the...

In a random sample of six mobile devices, the mean repair cost was \$70.00 and the standard deviation was \$11.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 95% confidence interval forte population mean. Interpret the results. The 95% confidence interval for the population m ean μ is (DO). Round to two decimal places as needed.) The margin of error is s (Round to two decimal places as needed.)...

• ### In a random sample of six microwave​ ovens, the mean repair cost was ​\$65.00 and the...

In a random sample of six microwave​ ovens, the mean repair cost was ​\$65.00 and the standard deviation was ​\$12.50. Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results. The 90​% confidence interval for the population mean mu is ​( nothing​, nothing​). ​(Round to two decimal places as​ needed.) The margin of error is nothing. ​(Round to...

• ### 6.2.20-T Question Help In a random sample of four mobile devices, the mean repair cost was...

6.2.20-T Question Help In a random sample of four mobile devices, the mean repair cost was \$90.00 and the standard deviation was \$13.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results.

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In a random sample of six mobile​ devices, the mean repair cost was ​\$80.00 and the standard deviation was ​\$12.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results. The 95​% confidence interval for the population mean mu is

• ### Question Help In a random sample of live microwave ovens, the mean repair cost was \$60.00...

Question Help In a random sample of live microwave ovens, the mean repair cost was \$60.00 and the standard deviation was \$12.00. Assume the population is normally distributed and use at distribution to construct a 90% confidence interval for the population mean . What is the margin of error of ? Interpret the results The 90% confidence interval for the population mean (Round to two decimal places as needed.) is C D Enter your answer in the edit fields and...

• ### in a random sample of six microwave ovens, the mean repair cost was \$60.00 and the...

in a random sample of six microwave ovens, the mean repair cost was \$60.00 and the standard deviation was \$11.00 Round to two decimal places as needed) (Round to two decimal places as needed.) C. It can be said that 99% of microwaves have a repair cost between the bounds of the c interval DOLL

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In a random sample of 24 people, the mean commute time to work was 32 minutes and t andard deviation was 7 minutes. Assume the population is normally debuted and use a distribution to constructa 99% confidence interval for the population mean what is the margin of error of Interpret the results The concerteval for the population means Round tone decimal places needed) The margin of error of Round to ore decimal place as needed) Interpret the results DAW concertante...

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In a random sample of 60 refrigerators, the mean repair cost was \$140.00 and the population standard deviation is \$17.30 Construct a 95% confidence interval for the population mean repair cost. Interpret the results. Construct a 95% confidence interval for the population mean repair cost The 95% confidence interval is ( 0 0 (Round to two decimal places as needed ) interpret you results Choose the correct answer below A, with 95% confidence, it can be said that the confidence...

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In a random sample of 25 ​people, the mean commute time to work was 30.2 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 80​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results. The confidence interval for the population mean mu is left parenthesis nothing comma nothing right parenthesis . ​(Round to one decimal place as​ needed.) The...