( -3,12),(14,6) slope

The Scholastic Aptitude Test (SAT) is a standardized test for college admissions in the U.S. Scores on the SAT can range from 60

kodGreya [7K]

Answer:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

Step-by-step explanation:

Notation and previous concepts

$n_1$ represent the sample after the preparation

$n_2$ represent the sample without preparation

$\bar x_1 =678$ represent the mean sample after preparation

$\bar x_2 =1837$ represent the mean sample without preparation

$s_1 =197$ represent the sample deviation after preparation

$s_2 =328$ represent the sample deviation without preparation

$\alpha=0.1$ represent the significance level

Confidence =90% or 0.90

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}$ (1)

The point of estimate for $\mu_1 -\mu_2$

The appropiate degrees of freedom are $df=n_1+ n_2 -2$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,df)

The standard error is given by the following formula:

$SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}$

After replace in the formula for the confidence interval we got this:

$3.0 < \mu_1 -\mu_2$

And we need to interpret this result:

A. The PrepIt! claim of statistically significant differences is valid. PrepIt! classes produce improvements in SAT scores that are 3% to 13% higher than improvements seen in the comparison group.

False, We conduct a confidence interval associated to the difference of scores with additional preparation and without preparation. And we can't conclude that the results are related to a % of higher improvements.

B. Compared to the control group, the PrepIt! course produces statistically significant improvements in SAT scores. But the gains are too small to be of practical importance in college admissions.

Correct, since we net gain is between 3.0 and 13 with 90% of confidence and if we see tha range for the SAT exam is between 600 to 2400 and this gain is lower compared to this range of values.

C. We are 90% confident that between 3% and 13% of students will improve their SAT scores after taking PrepIt! This is not very impressive, as we can see by looking at the small p-value.

False, we not conduct a confidence interval for the difference of proportions. So we can't conclude in terms of a proportion of a percentage.

5 0

3 years ago

I need help with question b

vodomira [7]

Answer:

A triangular pyramid, otherwise known as a tetrahedron.

Step-by-step explanation:

I learned this in math class a long time ago. My major in college revolves around math and I tutor in geometry, so I memorized this type of stuff.

Glad to see that my hard work is paying off.

6 0

3 years ago

What is the slope of the line between (−4, 4) and (−1, −2)? (1 point)

Ivanshal [37]

Answer:

-2

Step-by-step explanation:

if u evee need help on questions like this try desmos.com. its a free graphing calculator that is really easy to use. you just plug in numbers and it does the work for you!

i hope u have a good day

6 0

4 years ago

Is median a measure of center or a measure of variation

Marianna [84]

Answer:

a

Step-by-step explanation:

7 0

2 years ago

A normal distribution has a mean of μ = 100 with LaTeX: \sigmaσ = 20. If one score is randomly selected from this distribution,

alisha [4.7K]

Answer:

$P(100$

And we can find this probability with this difference:

$P(0$

And in order to find these probabilities we can using tables for the normal standard distribution, excel or a calculator.

$P(0$

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(100,20)$