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The positive effect of water fluoridation on dental health is well documented. One study that validates this is described in the article "Impact of Water Fluoridation on Children's Dental Health: A Controlled Study of Two Pennsylvania Communities" (American Statistical Association Proceedings of the Social Statistics Section [1981]: 262-265). Two communities were compared. One had adopted fluoridation in 1966 , whereas the other had no such program. Of 143 randomly selected children from the town without fluoridated water, 106 had decayed reeth, and 67 of 119 randomly selected children from the town with fluoridated water had decayed teeth. Let \(p_{1}\) denote proportion of all children in the community with fluoridated water who have decayed teeth, and let \(p_{2}\) denote the analogous proportion for children in the community with unfluoridated water. Estimate \(p_{1}-p_{2}\) using a \(90 \%\) confidence interval. Does the interval contain 0 ? Interpret the interval.

Step by step solution

01

Calculate Sample Proportions

Firstly, calculate the sample proportions (\( \hat{p}_{1} \) and \( \hat{p}_{2} \)) for the two communities. The sample proportion is the ratio of the number of decayed teeth to the total number of observations. For the community with fluoridated water, \( \hat{p}_{1} \)= 67/119 and for the community without fluoridated water \( \hat{p}_{2} \) = 106/143.

02

Calculate The Difference Between The Two Sample Proportions

Now calculate the difference between these two sample proportions (i.e., \( \hat{p}_{1} - \hat{p}_{2} \)). This will give the point estimate for the difference between the two population proportions.

03

Calculate The Standard Error of The Difference

Next, calculate the standard error of the difference between the two sample proportions. The formula for this is \( \sqrt{\hat{p}_{1}(1-\hat{p}_{1})/n_{1} + \hat{p}_{2}(1-\hat{p}_{2})/n_{2} }\), where \( n_{1} \) and \( n_{2} \) are the sizes of the two samples (in this case 119 and 143).

04

Obtain The Critical Values and Calculate The Confidence Interval

The confidence interval can be calculated using the formula \((\hat{p}_{1} - \hat{p}_{2}) 卤 Z_{\alpha/2}鉁昐E\), where \(Z_{\alpha/2}\) is the critical value from the Z-distribution corresponding to the desired confidence level (in this case, 90% implies \(Z_{\alpha/2} = 1.645\)), and \(SE\) is the standard error calculated in the previous step.

05

Interpret The Interval

Finally, interpret the confidence interval. If the interval contains 0, it suggests no significant difference in the proportion of children with decayed teeth in the two communities, while if it does not contain 0, it indicates that a significant difference is present.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportions

Sample proportions play a crucial role in comparing different groups within a study, serving as a summary statistic for categorical data. Take the case of the fluoridation study, where the sample proportions are the ratio of children with decayed teeth to the total number of children examined in each community. Specifically,

for the fluoridated community, the sample proportion is calculated by dividing the number of affected children (67) by the total surveyed (119) to yield \(\frac{67}{119}\),

and for the non-fluoridated community, it鈥檚 \(\frac{106}{143}\).

These proportions serve as estimators for the true proportion of all children in each community who would have decayed teeth, giving us insight into the effectiveness of water fluoridation on dental health.

Standard Error Calculation

In statistics, the standard error measures the variability or uncertainty around a sample statistic, such as a sample proportion. Calculating the standard error helps us understand the precision of our sample statistic as an estimate of the population parameter.

The standard error of the difference between two independent sample proportions \(\frac{SE_{diff}}\) is determined by the formula \(SE_{diff} = \frac{sqrt{(\frac{\frac{{\frac p_1}{1-p_1}}}{n_1{\frac p_2}{1-p_2}}}{n_2{\frac}}}\).

For our dental health study, we plug in the sample proportions and sizes for each community, which then provides an estimate of how much the difference between the two sample proportions might vary due to random chance alone. This is a crucial step in constructing a confidence interval.

Statistical Significance

Statistical significance is the determination of whether an observed effect or difference is likely due to chance or represents a true effect in the population. It's commonly assessed using a p-value, which quantifies the probability of finding the observed, or more extreme, results when the null hypothesis鈥攁 default statement suggesting no effect or no difference鈥攊s true.

In the context of our study comparing communities with and without fluoridated water, statistical significance would mean that the difference in decayed teeth proportions between the two communities is likely not due to random variability, but rather is a result of the fluoridation. We typically use a significance level (commonly \(\frac05{0.05}{}\)鈥�), to decide on whether our results are statistically significant, which also relates to the confidence level when we are constructing confidence intervals.

Hypothesis Testing

Hypothesis testing is a formal procedure used by statisticians to test whether a certain hypothesis about a population parameter is plausible given the sample data. The process involves five steps: stating the null and alternative hypotheses, selecting an appropriate test statistic, determining the significance level and corresponding critical value, computing the test statistic and corresponding p-value, and finally making a decision to reject or not reject the null hypothesis.

The dental health study鈥檚 hypothesis test might be set up to compare the proportions of decayed teeth in fluoridated versus non-fluoridated communities.

If the calculated confidence interval for the difference does not include zero, it means we would reject the null hypothesis (that there is no difference), suggesting a statistically significant difference between the two communities attributable to water fluoridation.