Question

Suppose you are a researcher in a hospital. You are experimenting with a new tranquilizer. You collect data from a random sample of 13 patients. The period of effectiveness of the tranquilizer for each patient (in hours) is as follows: Duration 2.7 2.9 2.5 3 2.2 2 2.3 2.8 2.3 2.4 2.7 3 2.6 What is a point estimate for the population mean length of time. (Round answer to 4 decimal places) Which distribution should you use for this problem? normal distribution t-distribution What must be true in order to construct a confidence interval in this situation? The population mean must be known The population must be approximately normal The population standard deviation must be known The sample size must be greater than 30 Construct a 90% confidence interval for the population mean length of time. Enter your answer as an open-interval (i.e., parentheses) Round upper and lower bounds to two decimal places. What does it mean to be "90% confident" in this problem? 90% of all simple random samples of size 13 from this population will result in confidence intervals that contain the population mean The confidence interval contains 90% of all samples There is a 90% chance that the confidence interval contains the population mean Suppose that the company releases a statement that the mean time for all patients is 2 hours. Is this possible? Yes No Is it likely? Yes No

Answer 1

The point estimate for population mean() = (2.7 + 2.9 + 2.5 + 3 + 2.2 + 2 + 2.3 + 2.8 + 2.3 + 2.4 + 2.7 + 3 + 2.6)/13 = 2.5692

s = sqrt(((2.7 - 2.5692)^2 + (2.9 - 2.5692)^2 + (2.5 - 2.5692)^2 + (3 - 2.5692)^2 + (2.2 - 2.5692)^2 + (2 - 2.5692)^2 + (2.3 - 2.5692)^2 + (2.8 - 2.5692)^2 + (2.3 - 2.5692)^2 + (2.4 - 2.5692)^2 + (2.7 - 2.5692)^2 + (3 - 2.5692)^2 + (2.6 - 2.5692)^2)/12) = 0.3172

We will use t-distribution.

The population must be approximately normal.

At 90% confidence interval the critical value is t_{0.05, 12} = 1.782

The 90% confidence interval for population mean is

+/- t_{0.05, 12} * s/

= 2.5692 +/- 1.782 * 0.3172/

= 2.5692 +/- 0.1568

= 2.4124, 2.7260

= 2.41, 2.73

There is a 90% chance that the confidence interval cointains the population mean .

Since 2 does not lie in the confidence interval, so it is not possible.

No, it is not possible.