Question

It is generally believed that only 7% of the population has type O-negative blood (the universal donor). A blood bank is expecting 1000 donors at its upcoming drive. In our drive of 1000 donors, we found 8.5% of the donors were O-negative. Construct a 90% confidence interval for the number of O-negative donors we would expect to see based on our sample. Does this support or refute the estimate of 7%? Assume conditions are met (so don't check them)!

Conditions:

Calculation:

Conclusion:

Answer 1

Conditions:

1. We assume that the sample of 1000 donors is a random sample.

np(1-p) = 1000 * 0.085 * (1 - 0.085) = 77.775

2. Since np(1-p) > 10, the sample size is large enough to approximate the sampling distribution of proportion as normal distribution

3. The sample size of 1000 donors can be assumed to be less than or equal to 5% of the population size.

Calculation:

Sample proportion, = 0.085

Standard error of sample proportion, SE =

= 0.008819014

Z value for 90% confidence interval is 1.645

Margin of Error, E = Z * SE = 1.645 * 0.008819014 = 0.0145

90% confidence interval for the number of O-negative donors is,

(0.085 - 0.0145 ,  0.085 + 0.0145)

(0.0705 , 0.0995)

(7.05% , 9.95%)

Conclusion:

Since the 90% confidence interval for the number of O-negative donors does not contain the value 7%, we refute the estimate of 7% at 90% confidence level.