A sample of 8 giraffes spent an average of 61 hours per week

A sample of 8 giraffes spent an average of 6.1 hours per week exercising, std dev of 4.6 hours. a sample of 14 turtles spent and average of 4 hours per week exercising, std dev of 1.5 hours.

A) test whether variances in exercise time differ between the two species. Alpha = .05

B) based on the above, construct a 95% confidence interval for the difference in means between the two. is there evidence that they differ?

Solution

A) test whether variances in exercise time differ between the two species. Alpha = .05

Let o1^2 be the varaince for giraffes

Let o2^2 be the variance for turtles

The test hypothesis:

Ho: o1^2=o2^2 (i.e. null hypothesis)

Ha: o1^2 not equal to o2^2 (i.e. alternative hypothesis)

The test statistc is

F=s1^2/s2^2

=4.6^2/1.5^2

=9.40

So the P-value = P(F with df1=n1-1=8-1=7, df2=n2-1=14-1=13 >9.40) =0.9997 (from F table)

Since the p-value is larger than 0.05, we do not reject the null hypothesis.

So we can not conclude that variances in exercise time differ between the two species

-------------------------------------------------------------------------------------------------------------

B) based on the above, construct a 95% confidence interval for the difference in means between the two. is there evidence that they differ?

The degree of freedom =n1+n2-2=8+14-2=20

Given a=1-0.95=0.05, t(0.025, df=20) =2.09 (from student t table)

So the lower bound is

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

=(6.1-4)-2.09*sqrt(4.6^2/8+1.5^2/14)

=-1.400806

So the upper bound is

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

=(6.1-4)+2.09*sqrt(4.6^2/8+1.5^2/14)

=5.600806

Since the interval includes 0, we can not conclude that they differ