Question

Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 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. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adults residents of this city who have cell phones.

Answer 1

Answer:

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.

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

In which

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

For this problem, we have that:

$n = 500, \pi = \frac{421}{500} = 0.842$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 - 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.81$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 + 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.874$

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

Answer 2

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:

https://brainly.com/question/23044118