Question

You are interested in finding a 95% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 13 randomly selected college students. Round answers to 3 decimal places where possible. 10 5 2 4 1 2 1 0 1 9 9 10 5 a. To compute the confidence interval use a distribution. b. With 95% confidence the population mean number of days of class that college students miss is between and days. c. If many groups of 13 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean number of missed class days and about percent will not contain the true population mean number of missed class days.

Answer 1

Sample mean using excel function AVERAGE(), x̅ = 4.5385

Sample standard deviation using excel function STDEV.S, s = 3.7775

Sample size, n = 13

a) To compute the confidence interval use a t - distribution.

b) 95% Confidence interval :  

At α = 0.05 and df = n-1 = 12, two tailed critical value, t-crit = T.INV.2T(0.05, 12) =    2.179

Lower Bound = x̅ - t-crit*s/√n = 4.5385 - 2.179 * 3.7775/√13 =    2.256

Upper Bound = x̅ + t-crit*s/√n = 4.5385 + 2.179 * 3.7775/√13 =    6.821

2.256 < µ < 6.821

With 95% confidence the population mean number of days of class that college students miss is between 2.256 and 6.821 days.

c) If many groups of 13 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About 95 percent of these confidence intervals will contain the true population mean number of missed class days and about 5 percent will not contain the true population mean number of missed class days.