Alocal board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. they randomly selected 850 residents in the community and contacted them by telephone. of the 850 residents surveyed, 410 supported the property tax levy. let p represent the proportion of residents in the community that support the property tax levy. a 90% confidence interval for p is (use decimal notation. give value to four decimal places and "z" value to three decimal a. 0.4489 to 0.5159.b. 0.4542 to 0.5105.c. 0.4487 to 0.5161.d. 0.4463 to 0.5185.

And the 90% confidence interval would be given (0.4542;0.5105).

So the correct option is:

B. 0.4542 to 0.5105

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

Solution to the problem

For this case the estimated proportion is given by:

The confidence interval would be given by this formula

For the 90% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.

And replacing into the confidence interval formula we got:

And the 90% confidence interval would be given (0.4542;0.5105).

So the correct option is:

B. 0.4542 to 0.5105

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step-by-step explanation:

i hope this : )