according to a recent pew research center report, many american adults have made money by selling something online. in a random sample of 4579 american adults, 914 reported that they earned money by selling something online in the previous year. assume the conditions for inference are met. construct a 98% confidence interval for the proportion of all american adults who would report having earned money by selling something online in the previous year.

As per the 98% confidence interval for the proportion of all American adults who would report having earned money by selling something online in the previous year is (0.1859,0.2133).

Confidence interval:

In statistics, confidence interval is an estimate of an interval in statistics that may contain a population parameter.

Given,

According to a recent pew research center report, many American adults have made money by selling something online. in a random sample of 4579 American adults, 914 reported that they earned money by selling something online in the previous year. assume the conditions for inference are met.

Here we need to construct a 98% confidence interval for the proportion of all American adults who would report having earned money by selling something online in the previous year.

As per the given question, the value of

Confidence interval = 98%

Number of samples = 4579

Reported numbers = 914

Therefore, the number of failure is calculated as,

=> 4579 - 914

=> 3665

Here the sample proportion is the number of successes divided by the sample size:

=> 941/4579

=> 0.1996

Here for confidence level 1 − α = 0.98,

We have to determine zα / 2 = z0.01 using the normal probability table in the appendix, we get

zα/2 = 2.33

Then the margin of error is calculated as,

E = 2.33 x √[0.1996 x (1 - 0.1996)]/4579

E = 0.0137

Then the boundaries of the confidence interval are then:

p⁻ˣ = 0.1996−0.0137 = 0.1859

p⁺ˣ = 0.1996+0.0137 = 0.2133

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