Question
A developer wants to know whether adding fibers to concrete used in paving driveways will reduce the severity of cracking, because any driveway with severe cracks will have to be repaired by the developer. The developer conducts a completely randomized experiment with 60 new homes that need driveways. Thirty of the driveways will be randomly assigned to receive concrete that contains fibers, and the other 30 driveways will receive concrete that does not contain fibers. After one year, the developer will record the severity of cracks in each driveway on a scale of 0 to 10 , with 0 representing not cracked at all 10 representing severely cracked.
(a) Based on the information provided about the developer’s experiment, identify each of the following.
- Experimental units
- Treatments
- Response variable
(b) Describe an appropriate method the developer could use to randomly assign concrete that contains fibers and concrete that does not contain fibers to the 60 driveways.
Suppose the developer finds that there is a statistically significant reduction in the mean severity of cracks in driveways using the concrete that contains fibers compared to the driveways using concrete that does not contain fibers.
(c) In terms of the developer’s conclusion, what is the benefit of randomly assigning the driveways to either the concrete that contains fibers or the concrete that does not contain fibers?
▶️Answer/Explanation
Ans:
(a) Experimental units: 60 new homes that need drive ways
Treatments: concrete that contains fibers and concrete with no fibers
Response variable: severity of cracks on a scale of 1 to 10
(b) Number the 60 new homes 1-60. Use a random number generator to get 30 unique numbers (ignore repeats) between 1 and 60. These 30 numbers correspond to the homes that get driveways with fibers. The remaining 30 homes get driveways without fibers.
(c) Random assignment creates roughly equivalent groups, so that the only difference between the groups is the treatments. When the developer finds a statistically significant reduction in mean severity of cracks for the driveways with fibers, they can conclude that the difference is due to the treatment and not some other confounding variable. In other words, random assignment allows the developer to conclude the fibers in the driveways caused a reduction in the severity of cracks.
Question
Bath fizzies are mineral tablets that dissolve and create bubbles when added to bathwater. In order to increase sales, the Fizzy Bath Company has produced a new line of bath fizzies that have a cash prize in every bath fizzy. Let the random variable, \(X\), represent the dollar value of the cash prize in a bath fizzy. The probability distribution of \(X\) is shown in the table.
(a) Based on the probability distribution of \(X\), answer the following. Show your work.
(i) Calculate the proportion of bath fizzies that contain \(\$ 1\).
(ii) Calculate the proportion of bath fizzies that contain at least \(\$ 10\).
(b) Based on the probability distribution of \(X\), calculate the probability that a randomly selected bath fizzy contains \(\$ 100\), given that it contains at least \(\$ 10\). Show your work.
(c) Based on the probability distribution of \(X\), calculate and interpret the expected value of the distribution of the cash prize in the bath fizzies. Show your work.
(d) The Fizzy Bath Company would like to sell the bath fizzies in France, where the currency is euros. Suppose the conversion rate for dollars to euros is 1 dollar \(=0.89\) euros. Using your expected value from
part (c), calculate the expected value, in euros, of the distribution of the cash prize in the bath fizzies. Show your work.
▶️Answer/Explanation
Ans:
(a) (i) \(P(x=1)=1-(0.2+0.05+0.05+0.01+0.01)=0.68\)
(ii) \(P(x \geq 10)=0.05+0.05+0.01+0.01=0.12\)
(b) \(P(x=100 \mid x \geq 10)=\frac{0.01}{0.12}=0.083\)
(c)
$
\begin{aligned}
E(x) & =1(0.68)+5(0.2)+10(0.05)+20(0.05)+50(0.01)+100(0.01) \\
& =\$ 4.68
\end{aligned}
$
If many, many bath fizzy were opened, the average amount for the prize is about \(\$ 4.68\) in the long run.
(d) \(\$ 4.68 \times \frac{0.89 \text { euros }}{\$ 1}=4.1652\) euros