# math-statistics help

A pharmaceutical manufacturer forms tablets by compressing a granular material that contains the active ingredient and var. fillers. The hardness of a sample from each lot of tablets produced is measured to control compression. The target values for the hardness are mean=11.5, standard dev=0.2. The hardness data for sample of 20 tablets:

11.627, 11.613, 11.493, 11.602, 11.630, 11.374, 11.592, 11.458, 11.552, 11.463, 11.383, 11.715, 11.485, 11.509, 11.429, 11.477, 11.570, 11.623, 11.472, 11.531

Is there signifcant evidence at the 5% level that the mean hardness of tablets is different from the target value?

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Okay, I actually solved most of the problem up to the part where you have to find the P-value. I have no idea how to do it....like how do you know when to multiply by 2P? Also, using Ti-83 calculator, how do you use normalcdf function to find out the percentage that you need to interpret the results to see whether it rejects the null hypothesis? What do you put into the normal cdf function to get that percentage? Please help asap thanks

If you are doing a one-sample t-test, then the value you found for the test statistic can be used to find your p-value. The p-value is the actual level of the test statistic. It represents the probability of getting a result as extreme as the test statistic itself. Also, if you are just looking for a significant difference at the 5% level, you can check a t-table for the cutoff points at both ends of the distribution (divide 5% in half for a two-tailed test). If your test statistic exceeds the positive or negative critical value (or cutoff point), then the null is rejected in favor of the alternate or alternative hypothesis. If your test statistic does not exceed the positive or negative critical value, then the null is not rejected.

Someone else may be able to help you with the calculator functions.

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