Which of the following functions has a rate of change of 5?(2 points)

The city of Raleigh has 11700 registered voters. There are two candidates for city council in an upcoming election: Brown and Fe

Vesnalui [34]

Answer:

The sample statistic for the proportion of voters surveyed who said they'd vote for Brown is 0.308.

The expected numbers of voters who would vote for Brown is 3604.

Step-by-step explanation:

The sample statistic is a numerical quantity used to represent a characteristic of a sample. For example, sample mean is a sample statistic representing the average of the sample. Or, sample proportion is a sample statistic representing the proportion of a particular variable in the sample. Or sample variance is a sample statistic representing the variance of the sample.

The sample statistic can be used to estimate the population parameter.

They are known as the point estimate of the parameter.

The sample proportion of a variable X is given by:

$\hat p=\frac{X}{n}$

It is provided that of the n = 250 registered voters selected, the number of voters who said they would vote for Brown is X = 77.

Compute the sample statistic for the proportion of voters surveyed who said they'd vote for Brown as follows:

$\hat p=\frac{X}{n}$

$=\frac{77}{250}\\$

$=0.308$

Thus, the sample statistic for the proportion of voters surveyed who said they'd vote for Brown is 0.308.

Compute the expected number of people of the 11700 registered voters who would vote for Brown as follows:

$E(X)=N\times \hat p$

$=11700\times 0.308\\=3603.6\\\approx 3604$

Thus, the expected numbers of voters who would vote for Brown is 3604.

6 0

3 years ago

An SRS of 350 high school seniors gained an average of x¯=21 points in their second attempt at the SAT Mathematics exam. Assume

Lynna [10]

Answer:

a) $21-2.58\frac{52}{\sqrt{350}}=13.83$

$21+2.58\frac{52}{\sqrt{350}}=28.17$

So on this case the 99% confidence interval would be given by (13.83;28.17)

b) $ME=2.58\frac{52}{\sqrt{350}}=7.17$

c) $ME=2.58\frac{52}{\sqrt{100}}=13.42$

d) D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

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".

Part a

$\bar X=21$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=52$ represent the population standard deviation

n=350 represent the sample size

99% confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$