We are told that in a sample of 47 18-23 year-olds 37 are registered to vote, while the remaining 10 are not registered. Assuming that we are interested in the percentage of this age group who are registered to vote, the population proportion p is the percentage of 18-23 year old people who are registered to vote.
(1) We can create a confidence interval using this sample -- the sample size is adequate. (Typically we would want n>30.)
The point estimate we use is `hat(p)=37/47~~.7872 ` . Then for the confidence interval we add/subtract the margin of error given by the product of the confidence factor and the standard error. Assuming a 95% confidence we have:
`.7872 - 1.96sqrt((.7872 * .2128)/47)<p<.7872+1.96sqrt((.7872 * .2128)/47) `
`"or" .6701<p<.9042`
Thus we can say with 95% certainty that the population proportion lies in this interval.
(2) In order to run an hypothesis test, we need a null hypothesis. Since you have not provided this, I will provide an example so that you can redo with the correct given information.
Suppose that you are told that only 60% of people in the age range from 18 to 23 are registered to vote.
Further suppose that you believe the proportion is higher.
The null hypothesis is that p=.6, while the alternative hypothesis is that p>.6
The alternative test will be our claim.
This is a one-tailed test, so the critical value is 1.645. A test value greater than this lies in the critical region.
The test value is computed by taking the difference of the observed value from the sample and the expected value from the null hypothesis divided by the standard error:
` "tv"=(.7872 - .6)/sqrt((.6 * .4)/47)~~2.62`
Since this value is greater than the critical value we reject the null hypothesis. Also, the p-value method gives p approximately .004 which is less than the alpha of .05 so we reject the null.
We conclude that there is sufficient evidence to support the claim that p>.6
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