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): 5, 5, 6, 6, 6, 3, 4, 6, 6, 6
Males (Population 2): 7, 3, 4, 7, 4, 7, 3, 3, 7, 4

Give a 93% confidence interval for the difference between mean female and male drink consumption. (Assume that the population variances are equal.)

Confidence Interval =-0.76,+1.56

Step-by-step explanation:

The confidence interval can be determined by

(x1`- x2`) ± z∝/2 √s1²/n1 + s2²/n2

z∝/2 for 93% is 1.81

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

The mean x1`= ∑x/ n=  5+ 5+ 6+ 6+ 6+ 3+ 4+ 6+ 6+ 6

                                  = 53/10= 5.3

Standard deviation = s1`=  1.0049= 1.005

The mean x2`= ∑x/ n=  7+ 3+ 4 +7+ 4+ 7+ 3+ 3+ 7+ 4/10

                                  = 49/10= 4.9

Standard Deviation = s2= 1.75783= 1.76

Putting the values

(x1`- x2`) ± z∝/2   √s1²/n1 + s2²/n2

(5.3-4.9 ) ± 1.81 √1.005²/10 +1.76²/10

0.4 ± 1.81 √ 1.010025/10 + 3.0976/10

-0.76,+1.56

The confidence interval for 93% two tailed test is [-0.76,+1.56]

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step-by-step explanation:

because i said it is

no he doesnt because "-16' is negatiive

step-by-step explanation: