Question

A recent poll of 85 randomly selected cable subscribers found that 39% would be willing to pay extra for a new nature channel. To the nearest percent, with a confidence level of 95% (z*-score 1.96), what is the confidence interval for the proportion of cable subscribers who would be willing to pay extra for the new nature channel?

Answer 1

Answer:

$29\%,49\%$

Step-by-step explanation:

A recent poll of 85 randomly selected cable subscribers found that 39% would be willing to pay extra for a new nature channel.

Here,

p = proportion= 39% = 0.39,

n = sample size = 85,

Now we have to construct a 95% confidence interval for the proportion.

Confidence interval can be calculated by,

$=p \pm Z_{critical}\cdot \sqrt{\dfrac{p(1-p)}{n}}$

Putting the values,

$=0.39 \pm 1.96\cdot \sqrt{\dfrac{0.39(1-0.39)}{85}}$

$=0.39 \pm 0.1037$

$=0.29,0.49$

$=29\%,49\%$

Answer 2

The mean is 85*0.39 = 33.15, while the standard error is sqrt(0.39*0.61/85) = 0.0529. Using the z-score of 1.96, the confidence interval is:33.15 +/- 1.96*0.0529 = (33.05, 33.25)By dividing by 85, this corresponds to a proportion of:(0.3888, 0.3912)