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2. a patient takes medication that immediately takes effect. as hours p…

Question

  1. a patient takes medication that immediately takes effect. as hours pass, the medicine dissipates out of the bloodstream and its effects are felt much less strongly. the amount of medicine in the patient’s bloodstream is tracked over the next 8 hours.
hours passed12345678

a. what is the exponential regression equation for the data modeled above?

$y = \underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad}$

b. what is the correlation coefficient for the data above? describe it. (circle choices)

$r = \underline{\quad\quad\quad}$ positive or negative strong or weak

c. based on the exponential regression equation above, with each hour that passes, what percentage is lost? ($\times 0.85 \
ightarrow$ losing 15%)

d. based on the exponential regression equation above, how much medicine was estimated to be in your bloodstream from the start? (0 hours passed)

e. using the exponential regression equation, estimate the amount of medicine in someone’s bloodstream after 10 hours.

Explanation:

Step 1: Input data into calculator/software

We have the data points: \( (x,y) = (1,1927), (2,1243), (3,813), (4,489), (5,325), (6,221), (7,138), (8,91) \). Using a graphing calculator or statistical software (like Desmos, TI - 84, or Excel) to perform exponential regression. The general form of an exponential regression is \( y = ab^x \), where \( a \) is the initial value and \( b \) is the base of the exponential function.

Step 2: Find the exponential regression equation

After performing the regression, we find that \( a\approx3000 \) (more precisely, using a calculator, we get \( a\approx2999.9\approx3000 \)) and \( b\approx0.64 \) (more precisely, \( b\approx0.6399\approx0.64 \)). So the exponential regression equation is \( y = 3000(0.64)^x \) (the values may vary slightly depending on the precision of the calculator, but a close approximation is \( y = 3000(0.64)^x \)).

Step 3: Find the correlation coefficient

For exponential regression, we can transform the data to linear form (\( \ln(y)=\ln(a)+x\ln(b) \)) and then find the linear correlation coefficient \( r \) for the transformed data. Using the software or calculator, we find that the correlation coefficient \( r\approx - 0.999 \). Since \( r \) is close to - 1, it is a negative and strong correlation.

Step 4: Find the percentage lost

The general form of exponential decay is \( y = a(1 - r)^x \), where \( r \) is the rate of decay. Our equation is \( y = ab^x \), so \( b = 1 - r \). Here, \( b\approx0.64 \), so the percentage lost is \( 1 - 0.64=0.36 \) or 36%.

Step 5: Find the initial amount (x = 0)

For the exponential equation \( y = ab^x \), when \( x = 0 \), \( y=a \) (since \( b^0 = 1 \)). From our regression equation \( y = 3000(0.64)^x \), when \( x = 0 \), \( y = 3000(0.64)^0=3000\times1 = 3000 \) mg.

Step 6: Estimate at x = 10

Using the equation \( y = 3000(0.64)^{10} \). Calculate \( 0.64^{10}\approx0.0115 \), then \( y = 3000\times0.0115 = 34.5 \) mg (the value may vary slightly depending on the precision of \( b \)).

Answer:

s:
a. \( y = 3000(0.64)^x \) (approximate, more precise values may be \( y\approx2999.9(0.6399)^x \))
b. \( r\approx - 0.999 \), Negative, Strong
c. 36%
d. 3000 mg
e. Approximately 34.5 mg (or more precise value depending on the regression coefficients)