You measure 28 turtle's weights, and find, they have a mean weight of 51 ounces. Assume the population standard deviation is 6.5 ounces. Based on this, what is the maximal margin of error associated...

You measure 28 turtle's weights, and find, they have a mean weight of 51 ounces. Assume the population standard deviation is 6.5 ounces. Based on this, what is the maximal margin of error associated with a 90% confidence interval for the true population mean turtle weight. Give your answer as a decimal, to two places +/- ______ ounces.

I need help on figuring out how to  input this in the calculator as I know that the mean is 51oz, standard deviation is 6.5 oz, confidence interval is % is 90 and the sample size is 28 turtles.

To set it up I know I need to find the  ounces but how I set it on on a TI-84 calculator. I need to know the steps please. This calculator is confusing. lol. I know that the confidence interval is x (with a line over the top) +/- t o  (= standard deviation) /2, n-1 , s/square root n.       So the required numbers to fill those in would be 51.0 -  ?   6.5/ square root 28, 51.0 + ? 65/ square root 28, I believe anyways. But I do not know to get the missing numbers