Answer:
There is insufficient evidence at the 5% level of significance to reject H₀ in favour of the alternative hypothesis that a person who regularly went to the dentist had fewer cavities compared to normal.
Step-by-step explanation:
The average adult has an average (μ) of 3.28 decayed or missing teeth, with a standard deviation (σ) of 1.1.
Researchers investigating whether regular dental visits will lower the number of cavities found that, in a sample of 13 adults who had been regularly going to the dentist since they were ten years old, the average number of decayed or missing teeth was 2.97.
To determine whether the group that regularly went to the dentist had fewer cavities compared to normal, we need to perform a hypothesis test to assess whether the sample average of 2.97 is significantly lower than the population average of 3.28.
Let μ be the average number of decayed of missing teeth per adult.
We are testing to see if the number of decayed or missing teeth has decreased, so this is a one-tailed test. Therefore, the hypotheses are:
Null hypothesis H₀: μ = 3.28
Alternative hypothesis H₁: μ < 3.28
We have not been given a significance level, so we will assume it is 5%. Therefore, α = 0.05.
[tex]\boxed{\begin{array}{v}\textsf{If $X \sim \text{N}(\mu,\sigma^2)$, then $\overline{X} \sim \text{N}\left(\mu,\dfrac{\sigma^2}{n}\right) \implies Z=\dfrac{\overline{X}-\mu}{\sigma / \sqrt{n}} \sim \text{N}(0,1)$}\\\\\\\textsf{Then the value of the test statistic will be:\;\;$z=\dfrac{\overline{x}-\mu}{\sigma / \sqrt{n}}$}\end{array}}[/tex]
The sample mean [tex]\overline{x}[/tex] is 2.97 when n = 13. Therefore, the test statistic is:
[tex]z=\dfrac{2.97-3.28}{1.1/ \sqrt{13}}=\dfrac{-0.31}{0.3050851...}\approx -1.01610990..[/tex]
This is a one-tailed test and values more likely to occur under H₁ are at the lower end of the distribution. Therefore, we need to find the critical value, z, such that P(Z < z) = 0.05.
Using a calculator, P(Z ≤ z) = 0.05 for z = -1.645.
So the critical region is Z < -1.645.
Since -1.016 > -1.645, the result does not lie in the critical region and is not significant, so we fail to reject the null hypothesis.
Therefore, there is insufficient evidence at the 5% level of significance to reject H₀ in favour of the alternative hypothesis that a person who regularly went to the dentist had fewer cavities compared to normal.
In other words, this means that, based on the sample data, we do not have enough evidence to conclude that regular dental visits are associated with a lower number of cavities compared to the general population.
