# hypothesis testing

• Nov 13th 2009, 09:58 PM
Robb
hypothesis testing
A manufacturer claimed that at least 20% of the public preferred her product. A sample of 100 persons is taken to check her claim. With $\displaystyle \alpha=0.05$ how small would the sample percentage need to be before the claim could legitimately be refuted? (notice that this would invovle a one-tailed test of the hypothesis)

The book mentions that the answer is $\displaystyle \hat{p}<.1342$ but im not sure how they calculated this.
I tried using
$\displaystyle \frac{\hat{p}-.2}{\sqrt{\frac{\hat{p}\cdot (1-\hat{p}}{100}}}=z_{0.05}$ but got really stuck and couldn't calculate it out, so not sure if im on the right track...
• Nov 14th 2009, 06:21 AM
novice
Quote:

Originally Posted by Robb
A manufacturer claimed that at least 20% of the public preferred her product. A sample of 100 persons is taken to check her claim. With $\displaystyle \alpha=0.05$ how small would the sample percentage need to be before the claim could legitimately be refuted? (notice that this would invovle a one-tailed test of the hypothesis)

The book mentions that the answer is $\displaystyle \hat{p}<.1342$ but im not sure how they calculated this.
I tried using
$\displaystyle \frac{\hat{p}-.2}{\sqrt{\frac{\hat{p}\cdot (1-\hat{p}}{100}}}=z_{0.05}$ but got really stuck and couldn't calculate it out, so not sure if im on the right track...

Adopt decision rule using one-tailed test as follows:

Accept, if $\displaystyle p > 0.2$ or the standard mean,$\displaystyle z*$ greater than $\displaystyle -1.645$;

Reject otherwise.

Sample size, $\displaystyle n=100$
Proportion, $\displaystyle p=0.2$
Sample mean, $\displaystyle \mu = p = 0.2$
Standard deviation $\displaystyle \sigma = \sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{(0.2)(0.8)}{100}}=0.04$

Minimum required sample proportion $\displaystyle , \hat{p}=z*\sigma+\mu=(-1.645)(0.04)+0.2=0.1342$