Mathematics, 13.07.2021 15:10 chrismcb875

Consider the following results for two independent random samples taken from two populations. Sample 1 Sample 2
n1=50 n2=35
x¯1=13.6 x¯2=11.6
σ1=2.2 σ2=3.0

Required:
a. What is the point estimate of the difference between the two population means?
b. Provide a 90% confidence interval for the difference between the two population means.
c. Provide a 95% confidence interval for the difference between the two population means.

Answers: 1

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Answers

Answer from: BARRION1981

a. 2

b. The 90% confidence interval for the difference between the two population means is (1.02, 2.98).

c. The 95% confidence interval for the difference between the two population means is (0.83, 3.17).

Step-by-step explanation:

Before solving this question, we need to understand the central limit theorem and the subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

Sample 1:

$\mu_1 = 13.6, s_1 = \frac{2.2}{\sqrt{50}} = 0.3111$

Sample 2:

$\mu_2 = 11.6, s_2 = \frac{3}{\sqrt{35}} = 0.5071$

Distribution of the difference:

$\mu = \mu_1 - \mu_2 = 13.6 - 11.6 = 2$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.3111^2+0.5071^2} = 0.595$

a. What is the point estimate of the difference between the two population means?

Sample difference, so $\mu = 2$

b. Provide a 90% confidence interval for the difference between the two population means.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of 1 - 0.05 = 0.95 , so Z = 1.645.

The margin of error is:

M = zs = 1.645(0.595) = 0.98

The lower end of the interval is the sample mean subtracted by M. So it is 2 - 0.98 = 1.02

The upper end of the interval is the sample mean added to M. So it is 2 + 0.98 = 2.98

The 90% confidence interval for the difference between the two population means is (1.02, 2.98).

c. Provide a 95% confidence interval for the difference between the two population means.

Following the same logic as b., we have that Z = 1.96 . So

M = zs = 1.96(0.595) = 1.17

The lower end of the interval is the sample mean subtracted by M. So it is 2 - 1.17 = 0.83

The upper end of the interval is the sample mean added to M. So it is 2 + 1.17 = 3.17

The 95% confidence interval for the difference between the two population means is (0.83, 3.17).

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ 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)

=====================================

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

=====================================

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).

Answer from: Quest

b. 2 and 4

step-by-step explanation: