A. A random sample of 32 different juice drinks has a mean of 98
calories per serving and a standard deviation of 31.5 calories.
Construct a 99% confidence interval of the population mean number
of calories per serving, and interpret the 99% confidence interval
in 1 sentence:
B. A random sample of 50 standard hotel rooms in Philadelphia,
PA, has a mean nightly cost of $189.99 and a standard deviation of
$35.25. Construct a 95% confidence interval of the mean nightly
hotel cost, and interpret the 95% confidence interval in 1
sentence:
C. In a random sample of 45 refrigerators, the mean repair cost was $170.00 and the standard deviation was $18.50. Construct a 90% confidence interval for the population mean repair cost, and interpret the 90% confidence interval in 1 sentence:
A) At 99% confidence interval the critical value is t^{*} = 2.744
The 99% confidence interval for population mean is
+/- t^{*} * s/
= 98 +/- 2.744 * 31.5/
= 98 +/- 15.28
= 82.72, 113.28
We are 99% confident that the true population mean number of calories lies between the boundaries 82.72 and 113.28.
B) At 95% confidence interval the critical value is t^{*} = 2.01
The 95% confidence interval for population mean is
+/- t^{*} * s/
= 189.99 +/- 2.01 * 35.25/
= 189.99 +/- 10.02
= 179.97, 200.01
We are 95% confident that the true population mean nightly hotel cost lies between the boundaries 179.97 and 200.01.
C) At 90% confidence interval the critical value is t^{*} = 1.680
The 90% confidence interval for population mean is
+/- t^{*} * s/
= 170 +/- 1.680 * 18.5/
= 170 +/- 4.633
= 165.367, 174.633
We are 90% confident that the true population mean for repair cost of refrigerators lies between the boundaries 165.367 and 174.633.
A. A random sample of 32 different juice drinks has a mean of 98 calories per...
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...
6.2.19-T Question Help In a random sample of four microwave ovens, the mean repair cost was $85.00 and the standard deviation was $13.00. Assume the population is normally distributed and use a t-distribution to construct a 99% confidence interval for the population mean μ. What is the margin of error of μ? Interpret the results. The 99% confidence interval for the population mean μ is (DD (Round to two decimal places as needed.) 6.2.21-T Question Help In a random sample...
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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 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...
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 four 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 95% confidence interval for the population mean u what is the margin of error of u? interpret the results In a Round to two decimal places as needed.)
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 45 home theater systems has a mean price of $127.00. Assume the population standard deviation is $19.20. Construct a 90% confidence interval for the population mean. The 90% confidence interval...
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)...
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.