Question

A local 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 p represent the proportion of residents in the community that support the property tax levy. A 90% confidence interval for p p is_______________.
A.0.4489 to 0.5159.
B. 0.4542 to 0.5105.
C.0.4487 to 0.5161.
D.0.4463 to 0.5185.

Answer 1

Answer:

$0.482 - 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.454$  

$0.482 + 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.510$  

And the 90% confidence interval would be given (0.454;0.510).  

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".  

Solution to the problem

The estimated proportion of residents in the community that support the property tax levis is given by:

$\hat p =\frac{x}{n}= \frac{410}{850}= 0.482$

The confidence interval for a proportion is given by this formula  

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$  

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.64$  

And replacing into the confidence interval formula we got:  

$0.482 - 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.4542$  

$0.482 + 1.64 \sqrt{\frac{0.482(1-0.482)}{850}}=0.5105$  

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

B. 0.4542 to 0.5105.

Answer 2

Answer:

90% confidence interval for p is [0.4542 , 0.5105] .

Step-by-step explanation:

We are given that a local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. Of the 850 residents surveyed, 410 supported the property tax levy.

Let p = proportion of residents in the community that support the property tax levy

$\hat p$ = proportion of residents in the community that support the property tax levy in a survey of 850 residents = $\frac{410}{850}$ = $\frac{41}{85}$

The pivotal quantity that will be used here population proportion p is;

         P.Q. = $\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

So, 90% confidence interval for p is given by;

P(-1.6449 < N(0,1) < 1.6449) = 0.90 {At 10% significance level the z table give

                                                            critical value of 1.6449)

P(-1.6449 < $\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.6449) = 0.90

P( $-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p - p}$ < $1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.90

P( $\hat p -1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p +1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.90

90% confidence interval for p = [ $\hat p -1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p +1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

                                       = [ $\frac{41}{85} -1.6449 \times {\sqrt{\frac{\frac{41}{85} (1-\frac{41}{85} )}{850} }$ , $\frac{41}{85} +1.6449 \times {\sqrt{\frac{\frac{41}{85} (1-\frac{41}{85} )}{850} }$ ]

                                       = [ 0.4542 , 0.5105 ]

Therefore, 90% confidence interval for p is [0.4542 , 0.5105] .