I have done the first part of the question already I just need help on the second part.

You read in a U.S Census Bureau that a 99% confidence interval for the mean income in 2005 of American households headed by a college educated person at least 25 years old was 100,272±1651. Based on this interval, can you reject the null hypothesis that the mean income in this group is $95,000? What is the alternate hypothesis of the test? What is its significance level?

I answered that you can reject the null hypothesis at the 0.01 level as confidence interval is between 98621 and 101923, and 95,000 is not in that in that range.

I am confused about the alternate hypothesis is it Ha>95,000 or is it Ha does not equal 95,000. Also How do I get the signicance level of the null hypothesis?

Thank you very much everyone.


1 Answer 1


The alternative hypothesis is $H_1: \mu \neq \,\$95,000.$ The significance level is the same as the error rate of the confidence interval, $\alpha = 0.01$. You cannot recover an observed significance level (OSL, i.e. $p$-value) directly from a confidence interval. You can backsolve to find the standard error of the mean and compute $Z$ from that. From $Z$ you can find the OSL for one- or two-tailed alternatives.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.