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.


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