Question

A Gallup survey of 2322 adults (at least 18 years old) in the U.S. found that 408 of them have donated blood in the past two years. Construct a 90% confidence interval for the population proportion of adults in the U.S. who have donated blood in the past two years. Round your answer to three decimal places. 10 points (0.163, 0.189) A. (0.160, 0.191) B. C. (0.156, 0.196) D. O (0.154, 0.198)

Answer 1

Answer: $(0.163,\ 0.189)$

Step-by-step explanation:

The confidence interval for population proportion(p) is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ = Sample proportion.

n= Sample size.

z* = Critical value.

Let p = Proportion of adults in the U.S. who have donated blood in the past two years.

As per given , we have

n= 2322

Sample proportion of adults in the U.S. who have donated blood in the past two years. : $\hat{p}=\dfrac{408}{2322}\approx0.1757$

By z-table , the critical value for 90% confidence : z*= 1.645

Then, the 90% confidence interval for the population proportion of adults in the U.S. who have donated blood in the past two years will be

$0.1757\pm ( 1.645)\sqrt{\dfrac{0.1757(1-0.1757)}{2322}}$

$0.1757\pm ( 1.645)\sqrt{0.0000623727433247}$

$0.1757\pm ( 1.645)(0.00789764163056)$

$0.1757\pm 0.013$

$=(0.1757- 0.013,\ 0.1757+ 0.013)$

$=(0.1627,\ 0.1887)\approx(0.163,\ 0.189)$

Hence, the 90% confidence interval for the population proportion of adults in the U.S. who have donated blood in the past two years. = $(0.163,\ 0.189)$