Solutions

2. The producers of a new toothpaste claim that it prevents more cavities than other brands of toothpaste. A random sample of 60 people used the new toothpaste for 6 months. The mean number of cavities at their next checkup is 1.5. In the general population, the mean number of cavities at a 6-month checkup is 1.73 (' = 1.12).
a. Is this a one- or two-tailed test?
Solution:
It is one side test, because the producer of new toothpaste claim that it will prevents more cavities. Thus it will be a left sided test
b. What are H0 and Ha for this study?
Solution:
Ho: New toothpaste doesn’t lead to reduction in the cavities
H1: New toothpaste leads to reduction in the cavities
c. Compute zobt.
Solution:
Below is the calculation of zobt.

= (1.73-1.5)/(1.12/square root of 60)
= (0.23)/(0.1445)
= 1.59

d. What is zcv?
Solution:
The Z critical value at 5% level is -1.645

e. Should H0 be rejected? What should the researcher conclude?
Solution:
The computed Z-value is less than the critical Z-value, thus we can reject the null hypothesis.
And conclude that the new toothpaste leads to reduction in cavities.

f. Determine the 95% confidence interval for the population mean, based on the sample mean
Solution:
Number in the sample n = 60
Population standard deviation σ = 1.12
Mean of the sample = 1.59
Confidence Interval 95% means that α = 1 - .95 = .05, so α = 0.05
It is conventional, however, to state confidence intervals with 95%, not 97.5%, confidence. We can easily create a one-sided 95% confidence interval. To do this, we simply compute a 90% two-sided confidence interval instead of 95%.
The critical number corresponding to a proportion to .5 is approximately 1.645

Thus, the 90% confidence interval for the population mean is:

Where E=zσn=1.6451.1260≅0.2378
1.59-0.2378<μ<1.59+0.2378