Question

A researcher is interested in the mean amount of time it takes people to complete a personality questionnaire. He selects 40 people at random and calculates the mean amount of time to be 20.4 min with a variance of 17.64 min2 .

a) Define the parameter of interest.

b) Define the random variable of interest.

c) Name the distribution required to calculate confidence intervals. (Check the relevant criteria.)

d) Construct a 98% confidence interval for the true mean amount of time.

e) Interpret your confidence interval.

Answer 1

Total sample size = 40

mean amount of time = = 20.4 min

variance =s² = 17.64 min²

Here parameter of interest is the average amount of time taken to complete a personality questionnaire.

(b) Here random variable of interest is amount of time taken by a random person to complete a personality questionnaire.

(c) Here the distribution required to calculate confidence interval is t distribution with degree of freedom = dF = n-1 = 40 -1 = 39

(d) Here 98% confidence interval = +- t_critical se_o

Here t_critical = TINV(0.02, 39) = 2.4258

standard error of sample mean = sqrt [s²/n] = sqrt [17.64/40] = 0.6641

98% confidence interval = 20.4 +- 2.4258 * 0.6641 = (18.8 min, 22.0 min)

(e) Here we interpret the confidence interval as if repeatedly samples of size 40 is taken than there is 98% chance that mean of these samples are in between 18.8 min and 22.0 min.