A student working on a summer internship in the economic research department of a large corporation studied the relation between sales of a product (Y, in million dollars) and population (X, in million persons) in the firm's 50 marketing districts. The normal error regression model (2.1) was employed. The student first wished to test whether or not a linear association between Y and X existed. The student accessed a simple linear regression program and obtained the following information on the regression coefficients:

Parameter Estimated Value 95 Percent Confidence limits
Intercept 7.43119 -1.18518 16.0476
Slope 0.755048 0.452886 1.05721

a. The student concluded from these results that there is a linear association between Y and X. Is the conclusion warranre? What is the implied level of significance?
b. Someone questioned the negative lower confidence limit for the intercept, pointing out that dollar sales cannot be negative even if the population in a district is zero. Discuss.

Step-by-step explanation:

Hello!

Y: sales of a product in a marketing district (million dollars)

X: population in a marketing district (million persons)

The objective is to test if there is a linear association between the sales of a product and the population of a marketing district.

Parameter; Estimated Value; 95 Percent Confidence limits

Intercept          7.43119                -1.18518        16.0476

Slope             0.755048            0.452886       1.05721

a.

To test if there is or not an association between these two variables, you have to do a hypothesis test for the slope. If the slope is equal to zero, there is no linear association between the two variables, if the slope is different from zero, then there is a linear association between the variables.

So the student's hypotheses are:

H₀: β = 0

H₁: β ≠ 0

The data the student used to conclude is the 95%CI for the slope [0.452886;1.05721]

To be able to decide over an hypothesis test using a confidence interval there are several conditions to be met, one of them is that the confidence level and the significance level should be complementary, this means that if the interval was constructed using 1 - α= 0.95, then the hypothesis test should be conducted with a significance level of α= 0.05.

Considering that the value of the slope stated in the null hypothesis is not included in the given interval, i.e. zero is not included in the CI, then the decision is to reject the null hypothesis.

Then it can be concluded that there is a linear association between the sales of a product and the population in a marketing district.

b.

Although in the context of the variables of study it makes no sense that the estimate of the intercept takes negative numbers, keep in mind that mathematically if possible and correct. This means that obtaining a negative estimate of the intercept does not represent a calculation error or problem for the regression model. In general, when this occurs, a footnote is made indicating has no biological meaning or sense in context.

After all, what you have done with the estimate is a mathematical assertion of the social variables.

I hope it helps!

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).