Question

During a flu vaccine shortage in the United States, it was believed that 45 percent of vaccine-eligible people received flu vaccine. The results of a survey given to a random sample of 2,350 vaccine-eligible people indicated that 978 of the 2,350 people had received flu vaccine. Construct a 99 percent confidence interval for the proportion of vaccine-eligible people who had received flu vaccine. Use your confidence interval to comment on the belief that 45 percent of the vaccine-eligible people had received flu vaccine. Suppose a similar survey will be given to vaccine-eligible people in Canada by Canadian health officials. A 99 percent confidence interval for the proportion of people who will have received flu vaccine is to be constructed. What is the smallest sample size that can be used to guarantee that the margin of error will be less than or equal to 0.02

Answer 1

Answer:

Part A: 99% CI is given by : (0.38998, 0.44236)

Part B:

The given proportion 45 % or 0.45 is not included in the 99% confidence interval which is 0.39- 0.44 . So it is not consistent with the statement that 45 % got vaccinated.

Part C:

There should be at least 4148 vaccine eligible people in Canada to guarantee that the margin of error will be less than 0.02

Explanation:

Part A:

The test statistic p^ ±z sqrt (p`q`/n) will be used because

It is a random sample and both np^≥10 also nq^≥10 so the large sample interval procedure is used.

Here

p= proportion of vaccine- eligible people who received flu vaccine

Sample size= n= 2350

Sample success= p^=978/2350= 0.4161

Sample failure= q^= 1-p`= 1- 0.4161= 0.5838

For 99% CI the z- value is 2.576

Using the test static

p^ ±z sqrt (p^q^/n)

0.4161± 2.576 sqrt( 0.4161*0.5838/2350)

0.4161± 2.576 * 0.010167

0.4161± 0.02619

(0.38998, 0.44236)

Part B:

The given proportion 45 % or 0.45 is not included in the 99% confidence interval which is 0.39- 0.44 . So it is not consistent with the statement that 45 % got vaccinated.

Part C:

z* sqrt ( p^q^/n) ≤ 0.02

sqrt ( p^q^/n) ≤ 0.02/z

Taking square of both sides

( p^q^/n) ≤ [0.02/z]²

Reciprocating both sides

n/p^q^ ≥[ z/ 0.02]²

n≥[ z/ 0.02]²p^q^

n≥ z²p^q^/[ 0.02]²

Putting the values

n ≥(2.576)² (0.5)(0.5)/ 0.2²

n≥ [2.576 /2 (0.02) ]²

n≥4,147.36

There should be at least 4148 vaccine eligible people in Canada to guarantee that the margin of error will be less than 0.02