Question

A drug company is marketing a new pain relief formula in 800 mg. tablets. The company claims that on the average 2.6 of the tablets will eliminate a migraine headache. A neurologist asks 36 of her patients with migraines to keep track of how many of the new tablets are needed to get relief from their next migraine. Her patients have an average of 3.1 tablets with a standard deviation of 1.1. Can the neurologist conclude at the 0.05 level of significance that the new pain relief formula is not as potent as the drug company claims?

I have problems finding the p value using tcdf on ti-84

Answer 1

Ho: μ = 2.6

Ha: μ ≠ 2.6

t=(​−) / (s/) ​​

=(3.1−2.6) /(1.1/​

t = 2.727

Degree of Freedom= n-1

Degree of Freedom=36-1

Degree of Freedom=35

Now following steps gives you P-value

Press button 2ND then press button VARS , scroll down to the tcdf  and press button Enter.

Syntax is tcdf( Test statistic value, larger value, degrees of freedom).

Then enter values as follows

tcdf(2.727, 9999, 35) then press button Enter

you will gate 0.0049603984

this value is for one tail but we want for two tail so just multiply it by 2.

P-value = (2* 0.0049603984)

P-value = 0.0099207969

Steps for TI-83 plus and TI-84 Calculator are same to use tcdf() function.

Following Image show the output of TI-83 plus Calculator.