Question

A company says that its product has a 95% positive approval rating. A polling company found that 89% of 500 people surveyed approved of the product. Construct a 90% proportion confidence interval based on this data and determine whether the data supports the 95% positive approval rating claim.

Start by using the binomial distribution (p $N1ncatQAAAABJRU5ErkJggg==$ + q)ⁿ = 1, and x $HHv7mUDho+Bd6rUpbgAAAAASUVORK5CYII=$ = np $N1ncatQAAAABJRU5ErkJggg==$ .

Use Summary 5b, Table 1, Column 3.

1. What is n?
2. What is p $N1ncatQAAAABJRU5ErkJggg==$ ?
3. What is q?
4. What is x $HHv7mUDho+Bd6rUpbgAAAAASUVORK5CYII=$ ?
5. What is p? (Note: p is based on our belief about the population. p is based on the sampl)
6. Let σ_p = sqrt(pqn). What is σ_p ?
7. We want to make a confidence interval by using the formula

z = k/σ_p

Should we use a 1 tail or 2 tail z-score?

8. What value of z corresponds to the 90% confidence interval?
1. What is k?
10. Construct the confidence interval

(p $N1ncatQAAAABJRU5ErkJggg==$ – k) < p_true < (p $N1ncatQAAAABJRU5ErkJggg==$ + k)

11. Is your confidence intervalconsistent with the belief that 95% approval rating claim? (yes or no, and explain your answer using your computed confidence interval).

Answer 1

(a)

n = 500

(b)

$\hat{p}$ = 0.89

(c)

q=1-p=0.11

(d)

$\bar{x}_{p}$ = 0.89

(e)

p = 0.95

(f)

Op = 0.89 x 0.11 = 500 = 0.0140

(g)

We should use 2 tail Z test

(a)

value of z corresponds to the 90% confidence interval = 1.645

(b)

k = 1.645 X 0.0140 = 0.0230

(c)

Confidence Interval:

0.89 - 0.0230 < ptrue < 0.89 + 0.0230

i.e.,

0.8670 < ptrue < 0.9130

(a)

Correct opton:

No

Explanation:
Since 0.95 is not included in the confidence interval 0.8670 < ptrue < 0.9130, our confidence interval is not consistent with the belief that 95% approval rating claim