Question

A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of 9 patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.6; 2.9; 3.1; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

(i) Round your answer to two decimal places.

x =

(ii) Round your answer to two decimal places.

sx =

(iii) Enter an exact number as an integer, fraction, or decimal.

n =

(iii) Enter an exact number as an integer, fraction, or decimal.

n − 1 =

2. Which distribution should you use for this problem? (Enter your answer in the form z or tdf where df is the degrees of freedom.)

3. Construct a 95% confidence interval for the population mean length of time.

(i) State the confidence interval. (Round your answers to two decimal places.)

(ii) Sketch the graph.

(iii) Calculate the error bound. (Round your answer to two decimal places.)

Answer 1

i. Calculate the sample mean (x) as 2.51. ii. Calculate the sample standard deviation (Sx) as 0.318. iii. Use a Student's t-distribution with 8 degrees of freedom. iv. The 95% confidence interval is (2.27, 2.76). v. The error bound is 0.25.

Step-by-step explanation:

i. Calculate the sample mean:

x = (2.6 + 2.9 + 3.1 + 2.3 + 2.3 + 2.2 + 2.8 + 2.1 + 2.4) / 9 = 2.51

ii. Calculate the sample standard deviation:

Sx = √[((2.6 - 2.51)^2 + (2.9 - 2.51)^2 + (3.1 - 2.51)^2 + (2.3 - 2.51)^2 + (2.3 - 2.51)^2 + (2.2 - 2.51)^2 + (2.8 - 2.51)^2 + (2.1 - 2.51)^2 + (2.4 - 2.51)^2) / (9 - 1)] ≈ 0.318

iii. Given that there are 9 patients in the sample, n = 9.

iv. We need to use a Student's t-distribution because the population standard deviation is unknown.

3. Construct a 95% confidence interval for the population mean length of time:

i. The confidence interval is given by x ± t * (Sx / √n), where t is the critical value for a 95% confidence interval with n - 1 degrees of freedom. From the t-distribution table, t = 2.306.

Therefore, the confidence interval is (2.51 - (2.306 * (0.318 / √9)), 2.51 + (2.306 * (0.318 / √9))), which simplifies to (2.27, 2.76).

ii. Sketching a graph of the confidence interval would require plotting the sample mean x, the lower endpoint of the interval (2.27), and the upper endpoint of the interval (2.76) on a number line.

iii. Calculate the error bound by finding the difference between the sample mean x and the lower endpoint of the interval, or the difference between the upper endpoint of the interval and the sample mean x. The error bound in this case is (2.76 - 2.51) / 2 = 0.25.