Suppose that a market research firm is hired to estimate

Ответ:

mazolethrin9632

26.01.2022

The 95% confidence interval is (0.81,0.874) and this can be determined by using the confidence interval formula and using the given data.

Given :

500 randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes – they own cell phones.95% confidence level.

The formula for the confidence interval is given by:

$\rm CI = p\pm z\sqrt{\dfrac{p(1-p)}{n}}$ --- (1)

where the value of p is given by:

$\rm p =\dfrac{421}{500}=0.842$

Now, the value of z for 95% confidence interval is given by:

$\rm p-value = 1-\dfrac{0.05}{2}=0.975$

So, the z value regarding the p-value 0.975 is 1.96.

Now, substitute the value of z, p, and n in the expression (1).

$\rm CI = 0.842\pm 1.96\sqrt{\dfrac{0.842(1-0.842)}{500}}$

The upper limit is 0.81 and the lower limit is 0.874 and this can be determined by simplifying the above expression.

So, the 95% confidence interval is (0.81,0.874).

For more information, refer to the link given below:

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Ответ:

jailinealvarado24

26.01.2022

Mathematics

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

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.