A researcher with the Ministry of Transportation is commissioned to study the drive times to work (one-way) for U. S. cities. The underlying hypothesis is that average commute times are different across cities. To test the hypothesis, the researcher randomly selects six people from each of the four cities and records their one-way commute times to work. Refer to the below data on one-way commute times (in minutes) to work. Note that the grand mean is 36.625.
Houston Charlotte Tucson Akron
45 25 25 10
65 30 30 15
105 35 19 15
55 10 30 10
85 50 10 5
90 70 35 10
x^-i 74.167 36.667 24.833 10.833
s^2i 524.167 436.667 82.167 14.167
The competing hypotheses about the mean commute times are .
Multiple Choice:
a. H0: μ1 = μ2 = μ3, HA: Not all population means are equal
b. H0: Not all population means are equal, HA: μ1 = μ2 = μ3
c. H0: μ1 = μ2 = μ3 = μ4, HA: Not all population means are equal
d. H0: Not all population means are equal, HA: μ1 = μ2 = μ3 = μ4

c. H0: μ1 = μ2 = μ3 = μ4, HA: Not all population means are equal

Step-by-step explanation:

Hello!

The variable of interest is

Xi: one-way commute time to work in a city of interest.

This variable is measured in 4 cities:

1) Houston ⇒ X₁: One-way commute time to work in Houston

2) Charlotte ⇒ X₂: One-way commute time to work in Charlotte

3) Tucson ⇒ X₃: One-way commute time to work in Tucson

4) Akron ⇒ X₄: One-way commute time to work in Akron.

If the ministry whats to compare the commute time in these 4 cities, he has to compare all population means (μi ∀ i= 1, 2, 3, 4) (And presumably conduct an ANOVA)

So the hypotheses are:

H₀: μ₁ = μ₂ = μ₃ = μ₄

H₁: Not all population means are equal.

The correct answer is c

I hope you have a SUPER day!

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.

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