## Tuesday, December 25, 2018

### Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

June 2017, Part III

All Questions in Part III are worth up to 4 credits. Partial credit is possible.

35. Graph y = log2(x + 3) - 5 on the set of axes below. Use an appropriate scale to include both intercepts.

Describe the behavior of the given function as x approaches -3 and as x approaches positive infinity.

If you graph y = log2(x + 3) - 5, you will see that it is not defined for x %lt 3.
The y-intercept is at y = log2(0 + 3) - 5 = -3.42.
Solving 0 = log2(x + 3) - 5 gives you the x-intercept at 29.
You can use the calculator, or work it out:

0 = log2(x + 3) - 5
5 = log2(x + 3)
x + 3 = 25
x + 3 = 32
x = 29

The best scale to use for the x-axis is 3. Use a scale of 1 for the y-axis.
Check the image below. The table of values is provided for you. It wasn't necessary for the exam, but it would be a good idea to label the intercepts.

Second part: As x approaches -3, the function goes to negative infinity. As x approaches infinity, the function approaches infinity.

36. Charlie’s Automotive Dealership is considering implementing a new check-in procedure for customers who are bringing their vehicles for routine maintenance. The dealership will launch the procedure if 50% or more of the customers give the new procedure a favorable rating when compared to the current procedure. The dealership devises a simulation based on the minimal requirement that 50% of the customers prefer the new procedure. Each dot on the graph below represents the proportion of the customers who preferred the new check-in procedure, each of sample size 40, simulated 100 times.

Assume the set of data is approximately normal and the dealership wants to be 95% confident of its results. Determine an interval containing the plausible sample values for which the dealership will launch the new procedure. Round your answer to the nearest hundredth.

Forty customers are selected randomly to undergo the new check-in procedure and the proportion of customers who prefer the new procedure is 32.5%. The dealership decides not to implement the new check-in procedure based on the results of the study. Use statistical evidence to explain this decision.

The mean is given as 0.506, and one standard deviation is 0.078. To be 95% confident in the result requires two standard deviations, or 2 * 0.078 above or below the mean. So the interval would be:

0.506 - 2 * 0.078 < 0.506 < 0.506 + 2 * 0.078
0.506 - 0.156 < 0.506 < 0.506 + 2 * 0.156
0.354 < 0.506 < 0.656
0.35 < 0.506 < 0.66
Be sure to round to the nearest hundredth.

In the second part, since 32.5% is below 35.4%, the amount is outside of the 95% confidence interval.

More Algebra 2 problems.

#### 1 comment:

Don Payne said...

Thanks for all your work on this site.
In the question about Charlie's Automotive Dealership, I don't understand how the calculations leading to the official answer (which you explain clearly) satisfy the scenario. That is, I think the dealer is correct in not implementing the new procedure, but for an incorrect reason.

I interpret "dealership wants to be 95% confident of its results" as meaning that the dealership wants to have 95% confidence that "50% or more of the customers [...] prefer the new procedure". At least, that's what I would want if I were the dealer considering the change.

I'm not an expert in statistics, but it seems to me that:
The 95% confidence interval means that they have 95% confidence that, given that 50% of all customers prefer the new procedure, a random sample of 40 of them will have between 35% and 66% who prefer it. That's not the same thing, is it?

It looks like the dealership will implement the new procedure unless they have 95% confidence that the % of customers who prefer the new procedure is not 50%.

In the third part, when "Forty customers are selected randomly", what if, say, 36% of customers preferred the new procedure? Would they implement? Yes. Would they be 95% confident that 50% of customers prefer the new procedure? No way; more like 5% confidence.

I believe that to have that 95% confidence that 50% of customers will prefer the new procedure, around 65% of the 40 random customers in the sample would have to prefer it, right?

What's worse, what if, say, 67% preferred the new procedure? Would they not implement the new procedure because that's outside the 95% confidence interval?