Question

A study measured how long men and women stared at the new 141" HD HGTV. Of the 48 men, the average was 19.6 minutes with a standard deviation of 3.4 minutes. Of the 11 women, the average was 6.8 minutes with a standard deviation of 1.7 minutes. The average difference was 12.8 minutes, and the standard deviation of the differences was 0.45. Find a 87% confidence interval for the average difference between the two genders.

Answer #1

Given:

For Males:
= 19.6, s_{1} = 3.4, n_{1} = 48

For Females:
= 6.8, s_{2} = 1.7, n_{2} = 11

= 0.13

Since s1 / s2 = 2, we use unequal variances.

The degrees of freedom for unequal variances is given by:

The tcritical (2 tail) for = 0.13, df = 32, is 1.554

The Confidence Interval is given by (- ) ME, where

(- ) = 19.6 – 6.8 = 12.8

The Lower Limit = 12.8 - 1.103 = 11.697

The Upper Limit = 12.8 + 1.103 = 13.903

The 87% Confidence Interval is **(11.697 ,
13.903)**

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