Arecent 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?

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,

Putting the values,

0

step-by-step explanation:

you can not divide by 0 so 4 ÷ 0 would be 0

0 × 9999 =0

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