Use limits to find the area of the shaded region in the graph y = x2.

Suppose that a researcher is designing a survey to estimate the proportion of adults in your state who oppose a proposed law tha

irinina [24]

Answer:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

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

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

Solution to the problem

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)

If solve n from equation (a) we got:

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

The margin of error desired for this case is $ME= \pm 0.02$ equivalent to 2% points

For this case we need to assume a confidence level, let's assume 95%. And since we don't have prior estimation for the population proportion of interest the best value to do an approximation is $\hat p =0.5$

In order to find the critical value we need to take in count that we are finding the margin of error 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:

$z_{\alpha/2}=\pm 1.96$

Now we have all the values needed and if we replace into equation (b) we got:

$n=\frac{0.5 (1-0.5)}{(\frac{0.02}{1.96})^2}= 2401$

So without prior estimation for the population proportion, using a confidence level of 95% if we want a margin of error about 2% we need al least a sample size of 2401.

5 0

3 years ago

The conditional relative frequency table below was generated by column using frequency table data comparing the number of calori

inessss [21]

Answer:

B. There is an association because the value 0.15 is not similar to the value 0.55

Step-by-step explanation:

Based on the above picture, for the nutritionist to determine whether there is an association between where food is prepared and the number of calories the food contains, there must be an association between two categorical variables.

The conditions that satisfy whether there exists an association between conditional relative frequencies are:

1. When there is a bigger difference in the conditional relative frequencies, the stronger the association between the variables.

2. When the conditional relative frequencies are nearly equal for all categories, there may be no association between the variables.

For the given conditional relative frequency, we can see that there exists a significant difference between the columns of the table in the picture because 0.15 is significantly different from 0.55 and 0.85 is significantly different from 0.45

We can conclude that there is an association because the value 0.15 is not similar to the value 0.55

3 0

3 years ago

Read 2 more answers

What is the slope of (-2,2) (-4,-4)

Flauer [41]

Answer:

3

Step-by-step explanation:

slope = (y2-y1)/(x2-x1)

(-4-2)/(-4--2)

-6/-2