An image is partitioned into two regions, one white and the other black. A reading taken from a randomly chosen point in the white section will be normally distributed with μ = 4 and σ 2 = 4 , whereas one taken from a randomly chosen point in the black region will have a normally distributed reading with parameters (6, 9). A point is randomly chosen on the image and has a reading of 5. If the fraction of the image that is black is α , for what value of α would the probability of making an error be the same, regardless of whether one concluded that the point was in the black region or in the white region?
An image is partitioned into two regions, one white and the other black. A reading taken from a randomly chosen point in the white section will be normally distributed with μ = 4 and σ 2 = 4 , whereas one taken from a randomly chosen point in the black region will have a normally distributed reading with parameters (6, 9). A point is randomly chosen on the image and has a reading of 5. If the fraction of the image that is black is α , for what value of α would the probability of making an error be the same, regardless of whether one concluded that the point was in the black region or in the white region?
Solution Summary: The author explains that the value of probability of making an error be the same.
An image is partitioned into two regions, one white and the other black. A reading taken from a randomly chosen point in the white section will be normally distributed with
μ
=
4
and
σ
2
=
4
, whereas one taken from a randomly chosen point in the black region will have a normally distributed reading with parameters (6, 9). A point is randomly chosen on the image and has a reading of 5. If the fraction of the image that is black is
α
, for what value of
α
would the probability of making an error be the same, regardless of whether one concluded that the point was in the black region or in the white region?
Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.
The concentration of carbon dioxide (COy) in the atmosphere is increasing rapidly due to our use of fossil fuels. Because plants use COz to fuel photosynthesis, more COz may cause trees and other plants to grow faster. An elaborate apparatus allows researchers to pipe extra CO, to a 30-meter circle of forest. They selected two nearby circles in each of three parts of a pine forest and randomly chose one of each pair to receive extra COz. The response variable is the mean increase in base area for 30 to 40 trees in a circle during a growing season. We measure this in percentage increase per year. Here are one year's data
Pair
1
2
3
Control Plot
9.752
7.263
5.742
Treated Plot
10.587
9.244
8.675
Treated - Control
0.835
1.981
2.933
State the null and alternative hypotheses. Explain clearly why the investigators used a onesided alternative.
Use at procedure to carry out a test and report your conclusion in simple language.
The investigators used the test you just carried out.Any use of the…
The concentration of carbon dioxide (CO2) in the atmosphere is increasing rapidly due to our use of fossil fuels. Because plants use CO2 to fuel photosynthesis, more CO, may cause trees and other plants to
grow faster. An elaborate apparatus allows researchers to pipe extra CO, to a 30-meter circle of forest. They selected two nearby circles in each of three parts of a pine forest and randomly chose one of each
pair to receive extra CO,. The response variable is the mean increase in base area for 30 to 40 trees in a circle during a growing season. We measure this in percent increase per year. Here are one year's
data.
Control plot
- Control
Treated plot
10.547
Pair
Treated
9.732
0.815
7.273
9.244
1.971
5.762
8.675
2.913
(a) State the null and alternative hypotheses. (Where u is the mean difference, treated minus control.)
Но: и
На: и
Explain clearly why the investigators used a one-sided alternative.
The researchers have no preconceived idea what extra CO, will do to the growth rate.…
At a large state university the average number of times that students meet with their academic advisors is μ = 3.6 with a σ = 1.1. The dean of student activities wondered what the relation was between involvement in extracurricular activities and being involved academically. She assumed that meeting with an academic advisor indicated students took their academics seriously. She obtained a random sample of 76 students involved in extracurriculars and found M = 4.8 for the number of times they met with their academic advisors. Did the students involved in extracurriculars have a different number of meetings with their academic advisors compared to the population of students in general? Use all hypothesis steps with p < .05.
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