Answer to Question #231620 in Statistics and Probability for maya

Question

The purpose of a study by Luglie et al. (A-5) was to investigate the oral status of a group of patients

diagnosed with thalassemia major (TM). One of the outcome measures was the decayed, missing, and

filled teeth index (DMFT). In a sample of 18 patients the mean DMFT index value was 10.3 with a

standard deviation of 7.3. Is this sufficient evidence to allow us to conclude that the mean DMFT

index is greater than 9.0 in a population of similar subjects? Let a ¼ :10.

Answer 1

Total problem

Sample size, n=18

Sample mean, $\bar{x}=10.3$

Sample standard deviation, s =7.3

We have to test whether the mean DMFT index is greater than 9 is a population of similar subjects.

The hypothesis are:

$H_0: \mu ≤9 \\ H_1: \mu >9$

This is a right tailed test.

Since the population standard deviation $\sigma$ is unknown, we will use t-test.

Test-statistic:

$t=\frac{\bar{x}-\mu}{s/ \sqrt{n}} \\ t= \frac{10.3-9}{7.3 / \sqrt{18}} \\ = 0.7555$

Degree of freedom df=n-1=18-1=17

P-value = P($t_{df}$ > test statistic) (since it is a right tailed test)

$= P(t_{17}>0.7555)$

= 0.23 (using t table, P-value is between 0.20 and 0.25)

a) answer: test statistic = 0.7555

p-value = 0.23

b) If p-value < α => Reject H₀.

If p-value > α => Fail to reject H₀.

Here p-value > α

(0.23>0.10)

We accept H₀.

Answer: accept H₀.

Conclusion: There is not enough evidence to support the claim that $\mu>9.0.$