Mathematics, 23.07.2021 02:00 mochoa4

A certain test preparation course is designed to help students improve their scores on the LSAT exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 6 students' scores on the exam after completing the course: 6,16,19,12,15,14. Using these data, construct a 90% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Find the critical value that should be used in constructing the confidence interval.

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Answer from: katelyn03anderson

The critical value is T_c = 2.5706 .

The 90% confidence interval for the average net change in a student's score after completing the course is (9.04, 18.30).

Step-by-step explanation:

Before building the confidence interval, we need to find the sample mean and the sample standard deviation:

Sample mean:

$\overline{x} = \frac{6+16+19+12+15+14}{6} = 13.67$

Sample standard deviation:

$s = \sqrt{\frac{(6-13.67)^2+(16-13.67)^2+(19-13.67)^2+(12-13.67)^2+(15-13.67)^2+(14-13.67)^2}{5}} = 4.4121$

Confidence interval:

We have the standard deviation for the sample, so the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 6 - 1 = 5

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 2.5706, that is, the critical value is T_c = 2.5706

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.5706\frac{4.4121}{\sqrt{6}} = 4.63$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 13.67 - 4.63 = 9.04.

The upper end of the interval is the sample mean added to M. So it is 13.67 + 4.63 = 18.30.

The 90% confidence interval for the average net change in a student's score after completing the course is (9.04, 18.30).

Answer from: Quest

rick has $1400 and sally has $1500

step-by-step explanation:

to find amount of money rick has after 10 years :

principal = $1000 , time = 10 years , rate = 4% annually

$\text{simple interest} = \frac{principal\times rate\times time}{100} \\\\\text{simple interest} = \frac{1000\times 4\times 10}{100}=\$400\\\\amount = principal + interest\\amount = 1000 + 400 = \$1400$

to find amount of money sally has after 10 years :

her friend gives to pay her $50 per year, therefore money she has after 10 years = 50 × 10 = $500

also, sally gets $1000 after 10 years,

therefore, total amount of money sally has after 10 years = 500 + 1000 = $1500

so, after 10 years : rick has $1400 and sally has $1500

Answer from: Quest

10.416 meters per second

step-by-step explanation:

distance=1 00 meters

time= 9.6 seconds in decimal form or 9 3/5 as a mixed number

average speed= 100/9.6

=10.416 meters per second or 10 52/125 as a mixed number.

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