A certain test preparation course is designed to help students improve their scores on the MCAT 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 7 students' scores on the exam after completing the course: 37,12,12,17,13,32,23 Using these data, construct a 80% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Step 1 of 4 : Calculate the sample mean for the given sample data. Round your answer to one decimal place.

Answers

Answer 1

Answer:

Answer:

The 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

Step-by-step explanation:

The net change in 7 students' scores on the exam after completing the course are:

S = {37 ,12 ,12 ,17 ,13 ,32 ,23}

Compute the sample mean and sample standard deviation as follows:

$\bar x=(1)/(n)\sum x=(1)/(7)* 146=20.857\n\ns=\sqrt{(1)/(n-1)\sum (x-\bar x)^(2)}}=\sqrt{(1)/(7)* 622.8571}=10.189$

As the population standard deviation is not known, a t-interval will be formed.

Compute the critical value of t for 80% confidence interval and 6 degrees of freedom as follows:

$t_(\alpha/2, (n-1))=t_(0.20/2, (7-1))=t_(0.10,6)=1.415$

*Use a t-table.

Compute the 80% confidence interval for the average net change in a student's score after completing the course as follows:

$CI=\bar x\pm t_(\alpha/2, (n-1))*(s)/(√(n))$

$=20.857\pm 1.415*(10.189)/(√(7))\n\n =20.857\pm 5.4493\n\n=(15.4077, 26.3063)\n\n\approx (15.4,26.3)$

Thus, the 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

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You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you survey? Assume that you want to be 99% confident that the sample percentage is within 5.5 percentage points of the true population percentage. Complete parts (a) and (b) below. a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. nequals 549 (Round up to the nearest integer.) b. Assume that a prior survey suggests that about 34% of air passengers prefer an aisle seat. nequals nothing (Round up to the nearest integer.)

Answers

Answer:

a) $n=(0.5(1-0.5))/(((0.055)/(2.58))^2)=550.116$

And rounded up we have that n=551

b) $n=(0.34(1-0.34))/(((0.055)/(2.58))^2)=493.78$

And rounded up we have that n=494

Step-by-step explanation:

Previous concept

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

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

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 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_(\alpha/2)=-2.58, t_(1-\alpha/2)=2.58$

Part a

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

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

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\hat p (1-\hat p))/(((ME)/(z))^2)$ (b)

We can assume that $\hat p =0.5$ since we don't know prior info. And replacing into equation (b) the values from part a we got:

$n=(0.5(1-0.5))/(((0.055)/(2.58))^2)=550.116$

And rounded up we have that n=551

Part b

$n=(0.34(1-0.34))/(((0.055)/(2.58))^2)=493.78$

And rounded up we have that n=494

Final answer:

To determine the required sample size for the survey, we can use a sample size formula based on the desired confidence level and margin of error. If nothing is known about the passenger preferences, a sample size of 549 would be needed. If a prior survey suggests a certain proportion, the sample size can be calculated using the known proportion.

Explanation:

In order to determine the number of randomly selected air passengers that must be surveyed, we need to calculate the required sample size for a desired confidence level and margin of error.

a. If nothing is known about the percentage of passengers who prefer aisle seats, we can use a sample size formula given by n = (Z^2 * p * q) / E^2, where Z is the z-score corresponding to the desired confidence level, p and q are the estimated proportions for aisle seat preference and non-aisle seat preference respectively, and E is the desired margin of error. Since a confidence level of 99% and a margin of error of 5.5% are specified, we can round up the sample size to 549.

b. If a prior survey suggests that about 34% of air passengers prefer an aisle seat, we can use the same sample size formula but with the known proportion p = 0.34. We do not have information about the non-aisle seat preference, so we cannot determine the required sample size.

Learn more about Sample size calculation here:

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Find the arc length of the curve below on the given interval. y equals one third (x squared plus 2 )Superscript 3 divided by 2y= 1 3x2+23/2 on [00,66]

Answers

Answer

$\int_(0)^(6)√(1+12x^4+8x^2)dx$

Step-by-step explanation:

We are given that

$y=(1)/(3)(3x^2+2)^{(3)/(2)}$

Interval=[0,6]

a=0 and b=6

Differentiate w.r. t x

$(dy)/(dx)=(1)/(3)(3x^2+2)^{(1)/(2)}* 6x=2x(3x^2+2)^{(1)/(2)}$

By using the formula ; $(dx^n)/(dx)=nx^(n-1)$

We know that arc length of curve

$s=\int_(a)^(b)\sqrt{1+((dy)/(dx))^2}dx$

Substitute the values

$s=\int_(0)^(6)\sqrt{1+(2x(3x^2+2)^{(1)/(2)})^2}dx$

$s=\int_(0)^(6)√(1+4x^2(3x^2+2))dx$

$s=\int_(0)^(6)√(1+12x^4+8x^2)dx$

Length of curve,= $s=\int_(0)^(6)√(1+12x^4+8x^2)dx$

What'd the greatest common factor (GCF) for each pair of numbers. 25, 55 The GCE IS

Answers

Answer:

Step-by-step explanation:

5 l 25,55

l 5,11

Solve the equation:

a+20=11

Answers

Answer:

Step-by-step explanation:

a +20=11

-20-20

a= -9

subtract 20 on each side, you get a=-9

John drives 257 miles and uses 9 gallons of gas. How many miles per gallon did he get?

Answers

Answer:

29 miles were used per gallon

Step-by-step explanation:

257 / 9 = 28.555..

We can round 28.555 to about 29.

So John got 29 miles per gallon.

1. Refer to the equation 2x - 4y = 12(a) Create a table of values for at least 4 points. Show your work on how you found the values for each coordinate pair, and validated the points were on the line
Work for Point 1
Work for Point 2
Work for Points
Work for Point 4
Put your points here
O

Answers

The table of values for points x= 1 to x = 4 relating to the linear expression2x - 4y = 12 is given below :

x ______ y
1 ______-2.5
2 ______-2
3 ______ -1.5
4 ______ - 1

Giventhe equation :

2x - 4y = 12

We could express the equation in slope - intercept form thus :

-4y = 12 - 2x

Divide both sides by - 4

y = - 3 + 0.5x

y = 0.5x - 3

Creating the pair of points :

At ; x = 1

y = 0.5(1) - 3

y = 0.5 - 3 = -2.5

(1, - 2.5)

At point x = 2 ;

y = 0.5(2) - 3

y = 1 - 3 = - 2

(2, - 2)

At point x = 3 ;

y = 0.5(3) - 3

y = 1.5 - 3 = -1.5

(3, -1.5)

At point x = 4 ;

y = 0.5(4) - 3

y = 2 - 3 = -1

(4, - 1)

The graph of the equation is attached below.

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