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# Tutor profile: Bob W.

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Bob W.
Math, AP Stats Teacher and Learning Support Specialist
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## Questions

### Subject:Trigonometry

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Question:

How can I use trigonometry to measure the height of a tall object?

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Bob W.

### Subject:Statistics

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Question:

What is the difference between a one-sample z-test and a one-sample t-test for significance?

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Bob W.

First, let us consider what they have in common. They both require the sampling distribution (applicable by applying the Central Limit Theorem and a normal probability plot for small samples) to be approximately normal, for data points to be independent, and for the data to be acquired through a random sample. They both are hypothesis tests. A z-test takes into account the error in approximating the sampling distribution. However, it does not take into account the error in estimating the population's standard deviation. Therefore, whenever the population's standard deviation is disclosed (which is uncommon in real life applications), a z-test can be used. When the population's standard deviation is unknown, a t-test takes into account the error for using the sample's standard deviation to approximate the population's standard deviation. At its core, that is the key difference. There is one caveat - when a sample size is large (let's say larger than 25), the z-test and t-test are very similar. Some instructors, courses, or exams will allow for you to use a z-test as a close-enough approximation of the t-test. In this case, the t-test will give you a slightly more conservative and mathematically accurate result.

### Subject:Algebra

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Question:

A brand new sedan costs \$25,000, while another used sedan on Craigslist can be purchased for \$10,000. Suppose each vehicle depreciates at 15% per year, both cars will meet your needs, and you will not keep the cars longer than 10 years. Which is the better investment and why?

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Bob W.

First, we will use algebra to model the value depreciation of both vehicles and discuss the investment implications. To do this, we need to label our variables: New Car Value = N, Used Car Value = U, and time in years = t. Now we can model the value of each car over time using exponential growth / decay (we use this equation because the cars depreciate a fixed percentage each year). The general equation is Value = Initial Value*(1+ rate)^time. Since our rate decreases the value, it has a negative rate of 15% or -.15. Let's substitute our known values to create a depreciation equation for each vehicle: N = 25000(1-.15)^t U = 10000(1-1.5)^t Now we can use that equation to graph the decay of the value of both vehicles. For example, after five years, the new car is worth \$11,092 while the used car is worth \$4,437. At ten years, the new car is worth \$4,921 and the used car is worth \$1,968 Even though the new car is always worth more, after five years the new car has lost \$13,908 in value while the used car only lost \$5,563 in value. As long as the maintenance and insurance costs for the used car are not significantly more (\$8,345 more over five years to be exact), then the used car is a much wiser purchase. Similarly, after ten years the new car has lost \$20,078 in value while the used car only lost \$8,031 in value. As long as the maintenance and insurance costs for the used car are not significantly more (\$12,046 more over five years to be exact), then the used car is a much wiser purchase. As an extension, if the \$15,000 saved by buying a used car was invested in a mutual fund earning 9% growth for 30 years, we can use the same equation to find the future value. V=15000(1+.09)^30 = \$199,015.17. In addition to the depreciation, there is a significant opportunity cost that could be up to roughly \$200,000 after 30 years.

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