Consider the following data Use Table 2 Construct a 99 confi

Consider the following data: Use Table 2.

Construct a 99% confidence interval for the difference between the population means. Assume the population variances are unknown but equal. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)

Specify the competing hypotheses in order to determine whether or not the population means differ.

= ?18.1	= ?17.0
s12 = 8.2	s22 = 8.7
n1 = 13	n2 = 13

Solution

(a) The degree of freedom =n1+n2-2=13+13-2 =24

Given a=1-0.99=0.01, t(0.005, df=24) =2.797 (from student t table)

So the lower bound is

(xbar1-xbar2) -t*sqrt(s1^2/n1+s2^2/n2)

=(-18.1+17)-2.797*sqrt(8.2/13+8.7/13)

=-4.29

So the upper bound is

(xbar1-xbar2) +t*sqrt(s1^2/n1+s2^2/n2)

=(-18.1+17)+2.797*sqrt(8.2/13+8.7/13)

=2.09

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(b)H₀: ?₁ ? ?₂ = 0; H_A: ?₁ ? ?₂ ? 0

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(c)Yes, since the confidence interval does not include the hypothesized value of 0.