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Thread: hypothesis testing

  1. #1
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    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...
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  2. #2
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    Quote Originally Posted by Robb View Post
    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 $\displaystyle , \hat{p}=z*\sigma+\mu=(-1.645)(0.04)+0.2=0.1342$
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