Question

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:

Answer 1

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.