Baron Asiimwe

2022-10-05

•The purpose of a study by Luglie was to investigate the oral status of a group of patients diagnosed with thalassemia major (TM) . One of the outcome measure s was the decayed , missing, filled teeth index (DMFT) . In a sample of 18 patients ,the mean DMFT index value was 10.3 with standard deviation of 7.3 . Is this sufficient evidence to allow us to conclude that the mean DMFT index is greater than 9 in a population of similar subjects? Let α =0.05

RizerMix

To determine if there is sufficient evidence to conclude that the mean DMFT index is greater than 9 in a population of similar subjects, we can perform a hypothesis test.
Let's set up the null and alternative hypotheses:
Null hypothesis (${H}_{0}$): The mean DMFT index is not greater than 9. μ ≤ 9
Alternative hypothesis (${H}_{1}$): The mean DMFT index is greater than 9. μ > 9
We will use a one-sample t-test to test these hypotheses. Given that the sample size is small (n = 18) and the standard deviation is known, we can calculate the test statistic as follows:
$t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$
where:
$\overline{x}$ is the sample mean (10.3),
μ is the hypothesized population mean (9),
s is the standard deviation of the sample (7.3), and
n is the sample size (18).
Using the given values, we can calculate the test statistic:
$t=\frac{10.3-9}{\frac{7.3}{\sqrt{18}}}$
Calculating this expression will give us the value of the test statistic.
To determine whether the evidence is significant at the α = 0.05 level, we need to compare the test statistic to the critical value from the t-distribution with n - 1 degrees of freedom.
If the test statistic is greater than the critical value, we can reject the null hypothesis and conclude that there is sufficient evidence to support the alternative hypothesis.

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