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i get:

\hat p=493/880 the test statistic is {\hat p -.5\over \sqrt{(.5)(.5)\over 880}}

I put \hat p=493/880 into the formula:

{(443/880) -.5\over \sqrt{(.5)(.5)\over \880}}=3.573

is...

OK, here is the work I've got so far:

\bar X+-t_{n-1,.025}s/\sqrt{n}

where can i find the t stat

880 randomly chosen people surveyed. 493 of them admitted to chewing gum. If we use a signifagant level of 0.01, can we conclude that a majority of people chew gum?

39 nuts are gathered. They have an average size of 3.34 cent and a standard deviation of .45 cent. Find the 95% confidence interval for the mean size of all the nuts