A random sample of 900 voters in Cincinnati found that 43% favored a half-percent increase in the...

Question:

A random sample of {eq}900 {/eq} voters in Cincinnati found that {eq}43 \% {/eq} favored a half-percent increase in the sales tax fund mass transit. Construct a {eq}90 \% {/eq} confidence interval for the parameter.

Confidence Interval

A confidence interval is an interval of values estimated using the sample statistics which, with a certain degree of confidence is likely to contain the value of the true paramerter.

The formula for confidence interval for proportion is:

{eq}\hat p \pm {z_{1 - \dfrac{\alpha }{2}}} \cdot \sqrt {\dfrac{{\hat p\left( {1 - \hat p} \right)}}{n}} {/eq}

Answer and Explanation:

Given Information:

Of n = 900 selected voters, 0.43 percent favored a half-percent increase in the sales tax fund mass transit.

The sample proportion is:

{eq}\hat p = 0.43. {/eq}

The lower limit of 90% confidence interval for the proportion is:

{eq}\begin{align*} \hat p - {z_{0.95}} \cdot \sqrt {\dfrac{{\hat p\left( {1 - \hat p} \right)}}{n}} &= 0.43 - \left( {1.645} \right) \cdot \sqrt {\dfrac{{0.43\left( {1 - 0.43} \right)}}{{900}}} \\ &= 0.402853 \end{align*} {/eq}

The upper limit of 90% confidence interval for the proportion is:

{eq}\begin{align*} \hat p + {z_{0.95}} \cdot \sqrt {\dfrac{{\hat p\left( {1 - \hat p} \right)}}{n}} &= 0.43 + \left( {1.645} \right) \cdot \sqrt {\dfrac{{0.43\left( {1 - 0.43} \right)}}{{900}}} \\ &= 0.457147 \end{align*} {/eq}

The z-value is obtained using the z-tables.

Therefore, the 90% confidence interval for the true proportion of the people who favored a half-percent increase in the sales tax fund mass transit is {eq}\left( {0.402853,0.457147} \right). {/eq}


Learn more about this topic:

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Finding Confidence Intervals for Proportions: Formula & Example

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 8
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