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,

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05