For the 2006 GSS a comparison of females and males on the nu

For the 2006 GSS, a comparison of females and males on the number of hours a day that the subject wached TV the following sample statistics: Among 870 men, the average number of hours was 2.86, the standard deviation was 2.22 and the standard error of the mean was 0.075. Among 1117 women, the average number of hours was 2.99, the standard deviation was 2.34 and the standard error of the mean was 0.070. It is not assumed that the two groups have equal variances.

1.What is the point estimate of the difference between men and women?

2.What is the absolute value of the t-statistic of a difference of means test?

3. Say that the null hypothesis being tested is that there is no difference in the number of hours spent by men and women watching TV and the alternative hypothesis is that men watch more TV than women. What is the P-value of the associated difference of means test?

Solution

1.

The point estimate of the difference is the difference of means Xd is,

Xd = Xmen - Xwomen = 2.86 - 2.99 = -0.13 [ANSWER]

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2. For the difference of means test,

t = Xd / sd

where

sd = sqrt(s1^2 / n1 + s2^2 / n2)

Thus,

sd = 0.102795

Thus,

t = 1.2646 [ANSWER]

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3.

As

df = n1 + n2 - 2 = 1985

tails = 1 [men watch more tv than women]

Using Excel, we type

=1 - tdist(t, df, number of tails)

There is a 1 - because it is a left tailed test with a negative t.

Thus,

p = 0.8969 [ANSWER]