Prepare a prototype residual plot for each of the following cases: (1) error variance decreases with X; (2) true regression func
tion is U shaped, but a linear regression function is fitted.
1 answer:
You might be interested in
Answer:
a) 57.35%
b) 99.99%
c) 68.27%
Step-by-step explanation:
When we have a random variable X that is normally distributed with mean and standard deviation
, then
The probability that the random variable has a value less than a, P(X < a) = P(X ≤ a) is the area under the normal curve with mean and standard deviation
to the left of a.
The probability that the random variable has a value greater than b, P(X > b) = P(X ≥ b) is the area under the normal curve with mean and standard deviation
to the right of b.
The probability that the random variable has a value between a and b, P(a < X < b) = P(a ≤ X ≤ b) = P(a < X ≤ b)= P(a ≤ X < b) is the area under the normal curve with mean and standard deviation
between a and b.
In this case, the random variable is the collagen amount found in the extract of the plant. The mean is 63 g/ml and the standard deviation is 5.4 g/ml
(a) What is the probability that the amount of collagen is greater than 62 grams per mililiter?
As we have seen, we need to find the area under the normal curve with mean 63 and standard deviation 5.4 to the right of 62 (see picture).
You can find this value easily with a calculator or a spreadsheet. If you prefer the old-style, then you have to standardize the values and look up in a table.
<em>If you have access to Excel or OpenOffice Calc, you can find this value by introducing the formula: </em>
<em>1- NORMDIST(62,63,5.4,1) in Excel </em>
<em>1 - NORMDIST(62;63;5.4;1) in OpenOffice Calc </em>
<em>and we will get a value of 0.5735 or 57.35% </em>
(b) What is the probability that the amount of collagen is less than 90 grams per mililiter?
Now we want the area to the left of 90
<em>NORMDIST(90,63,5.4,1) in Excel </em>
<em>NORMDIST(90;63;5.4;1) in OpenOffice Calc </em>
You will get a value of 0.9999 or 99.99%
(c) What percentage of compounds formed from the extract of this plant fall within 1 standard deviations of the mean?
You can use either the rule that 68.27% of the data falls between and
or compute area between 63 - 5.4 and 63 + 5.4, that is to say, the area between 57.6 and 68.4
<em>In Excel </em>
<em>NORMDIST(68.4,63,5.4,1) - NORMDIST(57.6,63,5.4,1) </em>
<em>In OpenOffice Calc </em>
<em>NORMDIST(68.4;63;5.4;1) - NORMDIST(57.6;63;5.4;1) </em>
In any case we get a value of 0.6827 or 68.27%
