I could really use some help with this!!!

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. Whe

kherson [118]

Answer:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

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".

The population proportion have the following distribution

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

Solution to the problem

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:

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

The margin of error for the proportion interval is given by this formula:

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

And on this case we have that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We don't have a prior estimation for the proportion $\hat p$ so we can use 0.5 as an approximation for this case

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

5 0

3 years ago

it takes 89 pounds of seed to completely plant a 12 acre field how many pounds of seed are needed per acre

yanalaym [24]

Answer:

7.41 rounded answer

Step-by-step explanation:

3 0

3 years ago

5.3.4 Journal: Two-Variable Systems Elimination <br><br> I need urgent help

max2010maxim [7]

Answer:

1) X stands for individual acts and y, group acts. 2) Each scenario describes a different period in minutes, but each one respecting their different amounts (individual and group acts). 3) $S=\left \{ 5,10 \right \}$

Step-by-step explanation:

Completing with what was found:

<em> 1) Here is a summary of the scenario your classmate presented for the talent show:Main show The main show will last two hours and will include twelve individual acts and six group acts.Final show The final show will last 30 minutes and will include the top four individual acts and the top group act.The equations he came up with are: 12x+ 6y= 120, 4x+ y= 30</em>

1. What do x and y represent in this situation?

X stands for individual acts and y, group acts.

Besides that, In the system of equation, they represent the time for x, and the time for y.

2. Do you agree that your classmate set up the equations correctly? Explain why or why not.

Yes, that's right. Each scenario describes a different period in minutes, but each one respecting their different amounts (individual and group acts). Either for 120 minutes or 30 minutes length. And their sum totalizing the whole period.

3. Solving the system by Elimination

$\left\{\begin{matrix}12x+ 6y= 120\\ 4x+ y= 30\end{matrix}\right.\\\left\{\begin{matrix}12x+ 6y= 120\\ 4x+ y= 30\:*(-3)\end{matrix}\right.\\\left\{\begin{matrix}12x+ 6y= 120\\ -12x+ -3y= -90\end{matrix}\right.\\3y=30\Rightarrow y=10\\4x+(10)=30\Rightarrow 4x=20\Rightarrow x=5\\S=\left \{ 5,10 \right \}$

8 0

3 years ago

Rectangle with side lengths of 1/2 ft. and 3/4 ft. What is the area and perimeter?

deff fn [24]

Perimeter: all the sides added together, so that would be 1/2+1/2+3/4+3/4, which is 2 and a half feet

Area: the two side lengths multiplied, so 1/2*3/4, which is 3/8 feet squared

4 0

3 years ago

In a survey of 1,003 adults concerning complaints about restaurants, 732 complained about dirty or ill-equipped bathrooms and 38

Ganezh [65]

Answer:

a

$0.716 < p < 0.744$

b

$0.3498 < p < 0.4089$

c

With the result obtained from a and b the manager can be 95 % confidence that the proportion of the population that complained about dirty or ill-equipped bathrooms are within the interval obtained at a

and that

the proportion of the population that complained about loud or distracting diners at other tables are within the interval obtained at b

Step-by-step explanation:

From the question we are told that

The sample size is $n = 1003$