Hypothesis Testing for Dental Program: MATH-125 Assignment Solution

This assignment solution addresses a hypothesis testing problem related to a dental program's impact on the average time to tooth decay in Yanomami tribesmen. The solution demonstrates the five-step hypothesis testing process, including formulating null and alternative hypotheses, identifying the appropriate test statistic (t-test), determining the critical value, and drawing conclusions based on the test statistic and alpha level. The assignment also includes the identification of the p-value associated with the test statistic. The solution assumes a single-sample t-test with known standard deviation. The analysis aims to determine if the dental program increased the average time to tooth decay, providing a practical application of statistical methods.

The average time it takes for the teeth of all Yanomami (Amazonian tribesmen) to decay is 15.8 years.
This is probably because little or no preventive treatments are available. A longitudinal program that
brought dental services to a group of 14 Yanomami tribesmen ran from age 4 to age 21. The result of this
sample was a mean time to tooth decay of 18.9 years, with a sample standard deviation of 1.3 years.
Assuming α = .05, test the claim that the dental program increased the average time to tooth decay.
(a) Show Steps 1 – 5 of the HT process. You do not have to draw a graph in Step 2, just state which
distribution you are using and the CV(s). Step 3 is given as TS = 2.13.
(b) Identify the p-value of this test assume TS = 2.13.
Solution:
Part (a)
Step-1
The assumption of the test is single sample t-test with known standard deviation.
Null hypothesis (H0): The dental program is not increased the average time to tooth
decay.
i.e. μ ≤15.8
Alternative hypothesis: The dental program is increased the average time to tooth
decay.
i.e. μ > 15.8
Step-2
Normal distribution with known standard deviation.
tcv = 1.75
Step -3
Test statistic (t) = 2.13
Step- 4
Alpha = 0.05 (at 5% significance level)
Step-5
Conclusion: test statistic > tcv
Hence reject the null hypothesis and accept the alternative hypothesis.
Thus the dental program is increased the average time to tooth decay.
Part (b)
Given test statistic = 2.13
P-value = 0.049