An intervention program to improve dietary behavior was test

An intervention program to improve dietary behavior was tested for effectivness by comparing the intervention group with a control group through a dietary knowledge test. The size of each group, and their average test scores and standard deviations are given in the table below.

Group n xhat s

Intervention 101 4.10 1.19

Control 129 3.67 1.12

a) Carry out a signifigance test for the effectivenss of the program. State your null and alternative hypotheses, and report the test statistic and the P-value along with a summary of your conclusion.

b) Find a 95% confidence interval for the difference between two means

Solution

(a) Let mu1 be the mean for Intervention

Let mu2 be the mean for Control

Null hypohtesis: mu1=mu2

Alternative hypothesis: mu1 not equal to mu2

The test statistic is

Z=(xbar1-xbar2)/sqrt(s1^2/n1+s2^2/n2)

=(4.10-3.67)/sqrt(1.19^2/101+1.12^2/129)

=2.79

It is a two-tailed test.

So the p-value= P(Z>2.79) =0.0026 (from standard normal table)

Assume that the significant level a=0.05

Since the p-value is less than 0.05, we reject Ho.

So we can conclude that there is the difference between two means.

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

(b) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

So the lower bound is

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

=(4.10-3.67)-1.96*sqrt(1.19^2/101+1.12^2/129)

=0.1279766

So the upper bound is

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

=(4.10-3.67)+1.96*sqrt(1.19^2/101+1.12^2/129)

=0.7320234