A drug company manufactures antacid that is known to be successful in providing relief for 70% of people with heartburn. The company tests a new formula on a simple random sample of 150 people with heartburn. 114 of the subjects who try the new antacid report feeling some relief.

Conduct an appropriate hypothesis test, at the 10% level of significance, to determine if there is a difference in eeffectiveness between the old formula and the new formula.

The formula for the standard deviation of a "proportion" is:

sqrt(p*q*n)

your problem:

p = 114/150 = .76

q = 1-.76 = .24

n = 150

sd = (pluggin' and chuggin') = sqrt(27.36) = 5.23067872....

Also, usually you don't need to calculate "s" for hypothesis tests of proportions. Usually, you just calculate the standard error as:

sqrt(p*q / n) which = .0348711915...

Taking that as the denominator of :

(Observed - Expected) /(Standard Error) = (.76 - 0.7) / .0348711915 = 1.720618004...

(Which is your test statistic).

Since the critical T-value for a two-tailed test with 10% is only 1.644853626... you can reject the null.


Please follow the work so you know this for your exam to best prepare yourself.

in this case, can we use 70% as p? instead using 114/150