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To study how recognition memory decreases with time, the following experiment was conducted. The subject read a list of 20 words slowly aloud, and later, at different time intervals, was shown a list of 40 words containing the 20 words that he or she had read. The percentage, [tex]$P$[/tex], of words recognized was recorded as a function of [tex]$t$[/tex], the time elapsed in minutes. The table below shows the averages for 5 different subjects. This is modeled by [tex]$P = a \ln t + b$[/tex].

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$t \, (\text{min})$ & 5 & 15 & 30 & 60 & 120 & 240 \\
\hline
$P \, (\%)$ & 73 & 61.7 & 58.3 & 55.7 & 50.3 & 46.7 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
$t \, (\text{min})$ & 480 & 720 & 1440 & 2880 & 5760 & 10080 \\
\hline
$P \, (\%)$ & 40.3 & 38.3 & 29 & 24 & 18.7 & 10.3 \\
\hline
\end{tabular}
\][/tex]

(a) Find [tex]$\ln t$[/tex] for each value of [tex]$t$[/tex] in the table above, and then use linear regression on a calculator to estimate [tex]$a$[/tex] and [tex]$b$[/tex] in the linear fit [tex]$P = a \ln t + b$[/tex]. In the blanks below, enter the corresponding values for [tex]$a$[/tex] and [tex]$b$[/tex]:
[tex]\[ a = \square \][/tex]
[tex]\[ b = \square \][/tex]
(round values to 4 decimal places)

(b) When does this model (in part a) predict that the subjects will recognize no words?

In [tex]$\square$[/tex] DAYS (1440 minutes [tex]$= 1$[/tex] day)

(c) When does this model (in part a) predict that the subjects will recognize all words?

In [tex]$\square$[/tex] SECONDS.

Sagot :

GinnyAnsweridn
Let's walk through the steps to solve the question.

### Part (a):

First, we need to find the natural logarithm ([tex]\(\ln t\)[/tex]) for each value of [tex]\(t\)[/tex] in the table and then perform a linear regression to estimate the parameters [tex]\(a\)[/tex] and [tex]\(b\)[/tex] in the model [tex]\(P = a \ln t + b\)[/tex].

Given [tex]\(t\)[/tex] values (in minutes):
[tex]\[ 5, 15, 30, 60, 120, 240, 480, 720, 1440, 2880, 5760, 10080 \][/tex]

And the corresponding [tex]\(P\)[/tex] values (percentages):
[tex]\[ 73, 61.7, 58.3, 55.7, 50.3, 46.7, 40.3, 38.3, 29, 24, 18.7, 10.3 \][/tex]

Using linear regression, the estimated values for [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are calculated as:
[tex]\[ a = -7.7867 \][/tex]
[tex]\[ b = 86.2831 \][/tex]
These values are rounded to 4 decimal places.

### Part (b):

To predict when the subjects will recognize no words ([tex]\(P = 0\)[/tex]), we solve for [tex]\(t\)[/tex] in the equation [tex]\(0 = a \ln t + b\)[/tex].

[tex]\[ 0 = -7.7867 \ln t + 86.2831 \][/tex]

Rearrange the equation to solve for [tex]\(\ln t\)[/tex]:

[tex]\[ \ln t = \frac{86.2831}{7.7867} \][/tex]

[tex]\[ \ln t \approx 11.083 \][/tex]

Now, exponentiate both sides to solve for [tex]\(t\)[/tex]:

[tex]\[ t = e^{11.083} \approx 64914.4321 \text{ minutes} \][/tex]

Convert minutes to days:
[tex]\[ \text{Days} = \frac{64914.4321}{1440} \approx 45.0797 \][/tex]

Thus:
[tex]\[ t \approx 45.0797 \text{ days} \][/tex]

### Part (c):

To predict when the subjects will recognize all words ([tex]\(P = 100\)[/tex]), we solve for [tex]\(t\)[/tex] in the equation [tex]\(100 = a \ln t + b\)[/tex].

[tex]\[ 100 = -7.7867 \ln t + 86.2831 \][/tex]

Rearrange the equation to solve for [tex]\(\ln t\)[/tex]:

[tex]\[ \ln t = \frac{100 - 86.2831}{-7.7867} \][/tex]

[tex]\[ \ln t \approx -1.764 \][/tex]

Now, exponentiate both sides to solve for [tex]\(t\)[/tex]:

[tex]\[ t = e^{-1.764} \approx 0.1718 \text{ minutes} \][/tex]

Convert minutes to seconds:
[tex]\[ t \times 60 \approx 0.1718 \times 60 \approx 10.3064 \text{ seconds} \][/tex]

Thus:
[tex]\[ t \approx 10.3064 \text{ seconds} \][/tex]

### Summary:

(a)
- [tex]\( a = -7.7867 \)[/tex]
- [tex]\( b = 86.2831 \)[/tex]

(b) Subjects will recognize no words in approximately 45.0797 days.

(c) Subjects will recognize all words in approximately 10.3064 seconds.
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