Question

Suppose you are a researcher in a hospital. You are experimenting with a new tranquilizer. You...

Suppose you are a researcher in a hospital. You are experimenting with a new tranquilizer. You collect data from a random sample of 12 patients. The period of effectiveness of the tranquilizer for each patient (in hours) is as follows: Duration 3 2.7 2.6 2 2.9 2.2 2.3 2.3 2.4 2.9 2.1 2 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? t-distribution normal distribution What must be true in order to construct a confidence interval in this situation? The population mean must be known The population standard deviation must be known The population must be approximately normal The sample size must be greater than 30 Construct a 95% 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 "95% confident" in this problem? 95% of all simple random samples of size 12 from this population will result in confidence intervals that contain the population mean The confidence interval contains 95% of all samples There is a 95% 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? No Yes Is it likely? Yes No

Homework Answers

Answer #1

The point estimate for population mean is = (3 + 2.7 + 2.6 + 2 + 2.9 + 2.2 + 2.3 + 2.3 + 2.4 + 2.9 + 2.1 + 2)/12 = 2.4500

sample standard deviation(s) = sqrt(((3 - 2.45)^2 + (2.7 - 2.45)^2 + (2.6 - 2.45)^2 + (2 - 2.45)^2 + (2.9 - 2.45)^2 + (2.2 - 2.45)^2 + (2.3 - 2.45)^2 + (2.3 - 2.45)^2 + (2.4 - 2.45)^2 + (2.9 - 2.45)^2 + (2.1 - 2.45)^2 + (2 - 2.45)^2)/11) = 0.361

We will use t-distribution.

The population must be approximately normal.

At 95% confidence interval the critical value is t0.025, 11 = 2.201

The 95% confidence interval is

+/- t0.025, 9 * s/

= 2.45 +/- 2.201 * 0.361/

= 2.45 +/- 0.23

= 2.22, 2.68

There is a 95% chance that the confidence interval cointains the population mean.

Since 2 does not lie between the confidence interval, so it is not possible that the mean time for all patients is 2 hours.

No it is not likely.

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