Question

A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy. We are interested in the population proportion of voters who feel the economy is the most important.

Construct a 90% confidence interval, and state the confidence interval and the error bound.

Short answer

The $90\%$confidence interval for the population proportion of voters who feel the economy is the most important issue is$[0.62735,0.67265]$ and the error bound is $0.022649$.

Step-by-step solution

Given information

Given in the question that

A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy.

Solution

The degree of uncertainty or assurance in a sampling procedure is calculated using confidence intervals. They can withstand a variety of probability thresholds, the most frequent of which are a $95$percent or 99 percent confidence level. Statistical tools such as the t-test are used to calculate confidence intervals.

tied to an error

The worst-case strategy for distinguishing between the assessed value of the procedure as supplied by the Taylor polynomial and the actual value of the function is shown by the Lagrange error bound of a Taylor polynomial.

The number of voters who believe the economy is the most important issue will be calculated here.

$0.65\times 1200=780$

After that, we need to calculate the confidant interval using the TI-83 calculator.

Press STAT and use the arrow (?) to TESTS.

Then press the arrow $(\downarrow )$to A: 1-PropZint and press enter.

Now press the arrow (?) and enter the inputs as shown below,

Then press the arrow (?) to calculator and press enter. The output is given below,

Solution

The $90\%$confidence interval for the population proportion of voters who feel the economy is the most important issue is $[0.62735,0.67265]$

Now we need to calculate the error bound

$EBM=z_{\frac{\alpha }{2}}\sqrt{\frac{p^{\mathrm{\prime }}{q}^{\mathrm{\prime }}}{n}}$

$EBM=z_{\frac{0.1}{2}}\sqrt{\frac{0.65\times (1-0.65)}{1200}}$

$=z_{0.05}\sqrt{\frac{0.65\times 0.35}{1200}}$

$=1.645\times 0.0137689$

$=0.022649$