In: Statistics and Probability
105. 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 nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.
a. i. x = __________ ii. sx = __________ iii. n = __________ iv. n – 1 = __________
b. Define the random variable X in words.
c. Define the random variable X in words. d. Which distribution should you use for this problem? Explain your choice.
e. Construct a 95% confidence interval for the population mean length of time.
i. State the confidence interval.
ii. Sketch the graph.
iii. Calculate the error bound.
f. What does it mean to be “95% confident” in this problem?
a) x ( mean ) = sum(X_i)/n = 2.511
sx ( standard deviation of X_i ) = 0.31798
n = 9
n-1 = 8
b) X : effective period of the tranquilizer (in hours)
c) X : effective period of the tranquilizer (in hours)
d) Since, the population variance is normal and the sample size, n < 30. we will assume it to be t - distribution
e) 95% confidence interval for the population mean length of time = mean +/- t_critical*sx/sqrt(n)
95% confidence interval : it is an interval where, the probability of a sample sample lie within the interval is 0.95
degrees of freedom = n - 1 = 9 - 1 = 8
t_critical = 2.306
mean +/- t_critical*sx/sqrt(n) = 2.511 +/- 2.306*0.31798/sqrt(9) = 2.511 +/- 0.2444 = ( 2.2666, 2.7554 )
error bound = 2*t_critical*sx/sqrt(n) = 2*2.306*0.31798/sqrt(9) = 2*0.2444 = 0.4888
f) “95% confident” means that confidence that in the long-run 95% of the confidence interval will include the population mean.