Question

a) A random sample of 100 housewives shows that 60 prefer detergent A. Compute a 99% confidence interval for the fraction of total housewives favoring detergent A. Give an interpretation of a confidence interval.

b) b) A poll of 1000 randomly selected voters was taken to estimate the proportion of population that were in support of the president's current policy on foreign aid. If 555 were favorable, did we find sufficient evidence to support the president's claim that he has the support of the (simple) majority of all voters? Use Q=0.05.

Answer 1

Solution:

a)Given:

n = 100

x = 60

We have to find 99% confidence interval for population proportion.

Sample propotion:

The critical value for α=0.01 is zc​=z1−α/2​=2.576

we are 99% confident that the true population proportion p is contained by the interval (0.474, 0.726).

b)

Given:

Sample size = n = 1000

x= 555

This corresponding to a right-tailed test, for which a z-test for one population proportion needs to be used.

Using the P-value approach: The p-value is p = 0.0003 …Using standard normal table

And α = 0.05

Since p = 0.0003<0.01, it is concluded that the Null Hypothesis is rejected.
Conclusion: It is concluded that the null hypothesis is rejected. Therefore, there is sufficient evidence to support the president's claim that he has the support of the (simple) majority of all voters at 0.05 significance level.

Done