Mathematics, 06.12.2019 03:31, lorelaistudent

Nielsen wants to estimate the percentage that are tuned t_q the tonight show. assume that they want 95% confidence that their sample percentage has a margin of error 2 percentage points a prior study found that 19% tune to the tonight show the number of households must nielsen survey is
a. 1479,
b. 2555
c, 1042
d. 633
e. 3034

Answers: 2

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Answers

answered: MidnightYT

a. 1479

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\hat p$ estimated proportion

n represent the sample size

Me =0.02 or 2% points represent the margin of error

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

We need to find a critical value in order to estimate the sample size required. In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$t_{\alpha/2}=-1.96, t_{1-\alpha/2}=1.96$

The confidence interval for the true proportion is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Wher the margin of error is given by:

$Me= z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

And we are interested in find n, solving for n we got:

$(\frac{Me}{z_{\alpha/2}})^2=\frac{\hat p (1-\hat p)}{n}$

$n=\frac{\hat p (1-\hat p)}{(\frac{Me}{z_{\alpha/2}})^2}$

And replacing the values that we have, we got:

$n=\frac{0.19 (1-0.19)}{(\frac{0.02}{1.96})^2}=1478.05$

And if we round up to th nearest integer we got that n=1479

answered: Guest

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.