How Do You Calculate a Z Score for Toothpaste Effectiveness Testing?

[altitus]

The problem is: A company wants to test the effectiveness of a new brand of toothpaste. They claim that this new brand of toothpaste will reduce the number of cavities that consumers will have as compared to the general population. The company tests a random sample population of 60 people for 6 months. At their next check-up, the sample population have a mean of 1.5 cavities. The general population (those who did not use the new brand of toothpaste), have a mean of 1.73 cavities at their 6 month check-up. The standard deviation for this problem is 1.12.

The question is: decide whether to use a z test or a t test, then find the z score or the t score.

I believe that this problem requires the use of a z test, but I cannot confirm this for sure, and cannot figure out what either the z obtained or t obtained would be.

I'd really appreciate some help. Thank you so very much.

 

[GJA]

Hi altitus,

Your intuition is correct, nicely done. Since we have a "large" sample size ($n >50 $) and the population standard deviation is known, we use the $z$-test. See the Z-test Wikipedia article - specifically, point 6 in the "Use in Location Testing" section. If you then scroll to the "Example" section, you will see a worked example very similar in spirit to the question you posted. See if you can follow their calculation and then apply it to your problem. If you're still stuck after that, feel free to post follow-up questions and I'd be happy to help more.