# statistics/ Help!-thanks!

posted by .

The producers of a new toothpaste claim that it prevents more cavities than other brands of toothpaste. A random sample of 60 people use the new toothpaste for 6 months. The mean number of cavities at their next check up is 1.5. In the general population the mean # of cavities after a 6 month check up is 1.73 (@= 1.12).
a) is it 1 or 2 tailed test? (I think one- tailed)
b) What are the Ho and Ha for this study? I think:
Ho: µ = µ1 or µ cavities from new toothpaste= cavities any other brand toothpaste
Ho: µ0>µ1 or µ cavities from new toothpaste> µ cavities with any other brand toothpaste
Ha: µ0<µ1 or µ cavities from new toothpaste<µ cavities with any other brand toothpaste

c) Compute Zobt -I don't know how
d) What is Z cv?- Also don't know.
e) Should Ho be rejected (I think yes)
f) Determine the 95% confidence interval for the population mean, based on the sample mean.

• statistics/ Help!-thanks! -

I would phrase the hypotheses this way:

Ho: µ ≥ 1.73
Ha: µ < 1.73

This would be a one-tailed test since Ha is showing a specific direction (less than).

Use a z-test:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

Once you have the z-test statistic, compare to a critical value from a z-table. If the test statistic exceeds the critical value from the table, reject the null. If the test statistic does not exceed the critical value from the table, you cannot reject the null. You can draw your conclusions from there.

From looking at the data, it does appear Ho will be rejected.

Confidence interval formula:
CI95 = mean ± 1.96(sd/√n)
...where sd = standard deviation and n = sample size.

I hope these few hints will help get you started.