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Amanda [17]
2 years ago
14

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

o 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?
Mathematics
2 answers:
prohojiy [21]2 years ago
8 0

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\%

adoni [48]2 years ago
7 0
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)

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