The average repair cost of a random sample of 25 refrigerators is $150 with a standard...
The average repair cost of a random sample of 25 refrigerators is $150 with a standard deviation $15. What is the margin of error of the 90% confidence interval for the population mean cost constructed by using this sample? Enter a dollar amount, round to the nearest cents
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The average repair cost of a random sample of 25 refrigerators
is $150 with a standard...
The average repair cost of a random sample of 100 refrigerators is $150 with a standard...
The average repair cost of a random sample of 100 refrigerators is $150 with a standard deviation $15. If (a,b) is the 80% confidence interval for the population mean cost constructed by using this sample, then a=[a] enter a dollar amount, round to the nearest cents. b=[b] enter a dollar amount, round to the nearest cents.
In a random sample of 60 refrigerators, the mean repair cost was $140.00 and the population...
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...
In a random sample of four mobile devices, the mean repair cost was $60.00 and the...
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 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...
Libel In a random sample of four mobile devices, the mean repair cost was $60.00 and...
Libel In a random sample of four mobile devices, the mean repair cost was $60.00 and the standard deviation was $13.50. Assume the population is normally distrbuted and use at distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results ol The 90% confidence interval for the population mean pis (C. (Round to two decimal places as needed.) The margin of error iss (Round to two decimal places as needed)...
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 ?$90.00 and the...
In a random sample of six microwave? ovens, the mean repair cost was ?$90.00 and the standard deviation was ?$13.00 Assume the variable is normally distributed and use a? t-distribution to construct a 90?% confidence interval for the population mean ?. What is the margin of error of ???
to estimate the population mean repair cost of microwave ovens in macomb, a random sample of...
to estimate the population mean repair cost of microwave ovens in macomb, a random sample of 9 broken microwave ovens was selected and it is found that the sample mean repair cost was $80 with a sample standard deviation of $15. 1) what would be the “critical value” if you want to find a 95% confidence interval for the population mean? 2) what would be the “margin of error” if you want to find a 95% confidence interval for the...
A supervisor records the repair cost for 14 randomly selected refrigerators. A sample mean of $79.20...
A supervisor records the repair cost for 14 randomly selected refrigerators. A sample mean of $79.20 and standa deviation of $10.41 are subsequently computed. Determine the % confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval Round your answer to three decimal places Tables Keypad Answer How to enter your answer
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 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...
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...
Libel In a random sample of four mobile devices, the mean repair cost was $60.00 and the standard deviation was $13.50. Assume the population is normally distrbuted and use at distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results ol The 90% confidence interval for the population mean pis (C. (Round to two decimal places as needed.) The margin of error iss (Round to two decimal places as needed)...
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...
A supervisor records the repair cost for 14 randomly selected refrigerators. A sample mean of $79.20 and standa deviation of $10.41 are subsequently computed. Determine the % confidence interval for the mean repair cost for the refrigerators. Assume the population is approximately normal. Step 1 of 2: Find the critical value that should be used in constructing the confidence interval Round your answer to three decimal places Tables Keypad Answer How to enter your answer
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.)...
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