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 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 places.)

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:

B. 0.4542 to 0.5105

Step-by-step explanation:

A 90% confidence interval for p is calculated as:

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

This apply if n*p≥5 and n*(1-p)≥5

Where p is the proportion of sample, n is the size of the sample and $z_{\alpha /2}$ is equal to 1.645 for a 90% confidence.

Then, in this case p, n*p and n*(1-p) are calculated as:

$p=\frac{410}{850} =0.4824$

n*p = (850)(0.4824) = 410

n*(1-p) = (850)(1-0.4824) = 440

So, replacing values we get:

$0.4824-1.645\sqrt{\frac{0.4824(1-0.4824)}{850} }\leq p\leq 0.4824+1.645\sqrt{\frac{0.4824(1-0.4824)}{850} }$

$0.4824-0.0282\leq p\leq 0.4824+0.0282$

$0.4542\leq p\leq 0.5105$

It means that a  90% confidence interval for p is 0.4542 to 0.5105

Answer 2

Answer:

The correct answer in the option is;

B. 0.4542 to 0.5105.

Step-by-step explanation:

To solve the question, we note that

Total number of residents, n = 850

Number supporting property tax levy = 410

Proportion supporting tax levy, p = $\frac{410}{850}$ = 0.48235

The formula for confidence interval is

$p +/-z*\sqrt{\frac{p(1-p)}{n} }$

Where

z = z value

The z value from the tables at 90 % = 1.64

Therefore we have

The confidence interval given as

$0.48235 +/-1.64*\sqrt{\frac{0.48235(1-0.48235)}{850} }$ = 0.48235 ± 2.811 × 10⁻²

= 0.4542 to 0.5105

The confidence interval is 0.4542 to 0.5105.