Acertain test preparation course is designed to 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 6 students' scores on the exam after completing the course: 19, 15, 9, 17, 7, 16. using these data, construct a 95% confidence interval for the average net change in a students's score after completing the course. assume the population is approx. normal. step 1. calculate the sample mean for the given sample data (round answer to 1 decimal place)step 2. calculate the sample standard deviation for the given sample data ( round answer to 1 decimal place)step 3. find the critical value that should be used in constructing the confidence interval. ( round answer to 3 decimal placed)step 4. construct the 95% confidence interval (round answer to 1 decimal place)

1)

2)

3)

4) The 95% confidence interval would be given by (8.8;18.8)  

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".

We can calculate the sample mean and sample deviation with the following formulas:

Step 1. calculate the sample mean for the given sample data (round answer to 1 decimal place)

Step 2. Calculate the sample standard deviation for the given sample data ( round answer to 1 decimal place)

represent the sample mean for the sample  

population mean (variable of interest)

s represent the sample standard deviation

n=6 represent the sample size  

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

  (1)

Step 3. Find the critical value that should be used in constructing the confidence interval. ( round answer to 3 decimal places)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.95 or 95%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,5)".And we see that

Step 4. construct the 95% confidence interval (round answer to 1 decimal place)

Now we have everything in order to replace into formula (1):

So on this case the 95% confidence interval would be given by (8.8;18.8)    

answer: ur v dumb

step-by-step explanation:

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).