Question

You are interested in comparing the effectiveness of zinc and vitamin C for preventing the
common cold. Suppose that we randomly sampled 160 individuals who regularly take zinc and
204 individuals who regularly take vitamin C. Of the 160 zinc users sampled, 32 came down
with the cold, and of the 204 vitamin C users sampled, 51 likewise came down with the cold.
Test whether or not there is an equal proportion of cold sufferers across regular takers of zinc
and vitamin C (using the level of significance α = 0:10). Make sure to state the null and
alternative hypotheses, the test statistic, and the conclusion you draw (in complete, plain
language sentences) from the test. In addition, compute a 90% confidence interval for the
difference in the proportion

Answer 1

Solution:

Given

n1= 160 sample size of zinc user

n2 = 204 sample size of vitamin c user.

X1= 32 zinc user with the cold

X2 = 51 vitamin c user with the cold

sample proportion of zinc user with the cold.

sample proportion of vitamin c user with the cold.

pooled proportion

level of significance.

To test the hypothesis

Vs.

Test statistic

Z = - 1.1250897

Test statistic Z = -1.1251

The Z critical value at  is

from Z table

Decision:

1.1251 < 1.64

Fail to reject Ho.

Conclusion: fail to rejecte Ho there is insufficient evidence to conclude that there is not equal proportion of cold sufferes across regular takers of zinc and vitamin c.

The 90% confidence interval for the difference in the proportion is

At

from Z table

( -0.121844, 0.0218443)

( - 0.1218, 0.0218)

The 90% confidence interval for the difference in the proportion is

( -0.1218, 0.0218)