Question

Random samples of female and male UVA undergraduates are asked to estimate the number of alcoholic drinks that each consumes on a typical weekend. The data is below:

Females (Population 1): 2, 2, 2, 5, 5, 5, 1, 2, 3, 5
Males (Population 2): 5, 4, 5, 7, 4, 7, 5, 4, 4, 8
Give a 97.6% confidence interval for the difference between mean female and male drink consumption. (Assume that the population variances are equal.)
Confidence Interval =

Answer 1

Confidence Interval =-3.4085 ,+0.7915

Explanation:

The confidence interval can be determined by

(x1`- x2`) ± 1.955 √s1²/n1 + s2²/n2

The mean and the standard deviations can be calculated using the calculator.

The mean x1`= ∑x/ n= 2+ 2+ 2+ 5+ 5+ 5+ 1+2+ 3+ 5/10

= 32/10= 3.2

Standard deviation = s1`= 1.536

The mean x2`= ∑x/ n= 5 +4 +5+ 7+ 4+ 7+ 5+ 4+ 4+ 8/10

= 53/10= 5.3

Standard Deviation = s2= 1.4177=1.42

Putting the values

(x1`- x2`) ± 1.955 √s1²/n1 + s2²/n2

(3.2-5.3 ) ± 1.955 √1.54²/10 +1.42²/10

-2.1 ± 1.955 √ 2.37/10 +2.11/10

-2.1 ± 1.3085

-3.4085 ,+0.7915