In March of 2018, a survey asked 801 US adults - id201880111201233, AgarioEdit

Ответ:

kingofguns2826

17.03.2021

The 95% confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service is (0.5362, 0.6048).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of the 801 participants in the survey, 457 indicated they subscribed to at least one video-streaming service.

This means that $n = 801, \pi = \frac{457}{801} = 0.5705$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so Z = 1.96 .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5705 - 1.96\sqrt{\frac{0.5705*0.4295}{801}} = 0.5362$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5705 + 1.96\sqrt{\frac{0.5705*0.4295}{801}} = 0.6048$

The 95% confidence interval for the proportion of all US adults who would say they subscribe to a video-streaming service is (0.5362, 0.6048).

0,0(0 оценок)

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question