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In the article "Fluoridation Brushed Off by Utah" (Associated Press, August 24,1998 ), it was reported that a small but vocal minority in Utah has been successful in keeping fluoride out of Utah water supplies despite evidence that fluoridation reduces tooth decay and despite the fact that a clear majority of Utah residents favor fluoridation. To support this statement, the article included the result of a survey of Utah residents that found \(65 \%\) to be in favor of fluoridation. Suppose that this result was based on a random sample of 150 Utah residents. Construct and interpret a \(90 \%\) confidence interval for \(\pi\), the true proportion of Utah residents who favor fluoridation. Is this interval consistent with the statement that fluoridation is favored by a clear majority of residents?

Step by step solution

01

Compute sample proportion and standard error

Firstly, calculate the sample proportion which is given as 0.65. Then, we need to calculate the standard error of the proportion. We use the formula for the standard error, \( SE = \sqrt{ \frac{p(1 - p)}{n} } \). So, \( SE = \sqrt{ \frac{0.65(1 - 0.65)}{150} } \).

02

Find the confidence interval

We were asked to construct a 90% confidence interval. The Z value for a 90% confidence interval is 1.645 (This can be looked up in a standard Z table). The confidence interval is given by \(p \pm Z * SE\). So, it's \(0.65 \pm 1.645 * SE\).

03

Interpret the confidence interval

The interval obtained in the previous step gives us an estimate of the true population proportion (\pi). It's interpreted as 'We are 90% confident that the true proportion of Utah residents who favor fluoridation is between the two values obtained.'

04

Check the statement

To verify if the confidence interval supports the statement that fluoridation is favored by a clear majority, we check if the values in the interval are greater than 50%. If they are, then the statement is supported by the confidence interval. If not, the statement is not supported by the confidence interval.

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