The Food and Drug Administration (FDA) is a U.S. government agency that regulates (you guessed it) food and drugs for consumer safety. One thing the FDA regulates is the allowable insect parts in various foods. You may be surprised to know that much of the processed food we eat contains insect parts. An example is flour. When wheat is ground into flour, insects that were in the wheat are ground up as well. The mean number of insect parts allowed in 100 grams (about 3 ounces) of wheat flour is 75. If the FDA finds more than this number, they conduct further tests to determine if the flour is too contaminated by insect parts to be fit for human consumption. The null hypothesis is that the mean number of insect parts per 100 grams is 75. The alternative hypothesis is that the mean number of insect parts per 100 grams is greater than 75. Is the following a Type I error or a Type II error or neither? The test fails to show that the mean number of insect parts is greater than 75 per 100 grams when it is. Group of answer choices Type I error Type II error
Answers
Answer:
Type II error
Step-by-step explanation:
Let's remember the definition of Type I error and Type II error:
A type I error is the rejection of a true null hypothesis, this means that we would get a "false positive" with this error.
A type II error is the non rejection of a not true null hypothesis, this error would give us a "false negative".
In this problem the mean number of insect parts per 100 grams is 75. However, the test fails to show that this number is greater than 75 when it is, this means that the test is not detecting these insect parts and therefore is giving a "false negative"
Thus, this is a Type II error.
Final answer:
This situation is an example of a Type II error. This occurs when the test fails to reject the null hypothesis when it should be rejected. In this context, it means the test was unable to detect the actual average content of insect parts is higher than the allowable limit.
Explanation:
The situation presented constitutes a Type II error in the field of hypothesis testing in statistics. A Type II error occurs when the tester fails to reject the null hypothesis when it should be rejected. In this context, it means the test failed to show that the mean number of insect parts per 100 grams is greater than 75, when in fact, it is. This can potentially mean allowing more contaminated flour into the market because the test did not pick up on the true mean being higher than 75 insect parts per 100 grams.
Learn more about Type II Error here:
#SPJ11
Related Questions
The following data give the prices of seven textbooks randomly selected from a university bookstore. $93 $173 $107 $125 $56 $163 $144 a. Find the mean for these data. Calculate the deviations of the data values from the mean. Is the sum of these deviations zero? Mean = $ 123 Deviation from the mean for $173 = $ 0 Sum of these deviations = $ -433 b. Calculate the range, variance, and standard deviation. [Round your answers to 2 decimal places.] Range = $ Variance = Standard deviation = $ Click if you would like to Show Work for this question: Open Show Work LINK TO TEXT
Is the perpendicular bisector of . What is the length of ? A.4B.6C.12D.7
A crew of highway workers paved 12 mile in 14 hour. If they work at the same rate, how much will they pave in one hour?
Solve the equation −11=−3m+7
B. They are complementary angles
C. They are supplementary angles.
D. They are vertical angles.
Answers
Answer:
A. They are adjacent angles
Step-by-step explanation:
Angles 1 and 2 are next to each other, that makes them adjacent angles
A They are adjacent angles (next to each other and share a side) True
B. They are complementary angles (add to 90 degrees) False- Angle 2 is 90 degrees)
C. They are supplementary angles. (add to 180) False, they do not form a straight line
D. They are vertical angles. (opposite angles) False, they are not made by intersecting lines
Answer:
The answer is adjacent angles:)
Step-by-step explanation:
Which floor do you live on?
Answers
Answers
Given Information:
Annual interest rate = r = 12%
Principal amount = P = $1000
Number of years = t = 3
Required Information
Accumulated amount = A = ?
Answer:
Annual compounding = A = $1404.93
Semi-annuall compounding = A = $1418.52
Quarterly compounding = A = $1425.76
Monthly compounding = A = $1432.30
Daily compounding = A = $1433.14
Step-by-step explanation:
The accumulated amounts in terms of compound interest is given by
Where P is the initial amount invested and A is the accumulated amount.
For annual compounding:
i = 0.12
N = 3
For semiannually compounding:
i = 0.12/2 = 0.06
N = 2*3 = 6
For quarerterly compounding:
i = 0.12/4 = 0.03
N = 4*3 = 12
For monthly compounding:
i = 0.12/30 = 0.004
N = 30*3 = 90
For daily compounding:
i = 0.12/365 = 0.0003287
N = 365*3 = 1095
Answers
Answer:
22 in
Step-by-step explanation:
It travels 35/60 ths of the complete circumference of a circle with r =6 in
diameter = 12 circumference = pi * d
35/ 60 * pi * 12 = ~22 inches
Answers
Answer:
-3 and 4
Step-by-step explanation:
We can start naming the first number x
Hence, the second number would be 10+2x
Set up an equation.
10+2x+x=1
Combine like terms
10+3x=1
Subtract 10 from both sides
3x=-9
Divide both sides by 3.
x=-3
The first number is he first is -3.
Plug that into the expression for the second number.
10+2x
10+2(-3)
10-6
4
The two numbers are -3 and 4.
Answers
Answer: The expected number of spades that you will draw is 0.751 spades
Step-by-step explanation:
The expected value can be calculated as:
∑xₙ*pₙ
Where xₙ is the n-th event, and pₙ is the probability of that event.
First, let's count the possible events and calculate the probability for each one.
x₀ = drawing 0 spades.
Out of 52 cards, we have only 13 spades, then 52 - 13 = 39 are not spades.
Then the probability of not drawing a spade in the first draw is:
p1 = 39/52
In the second draw we will have a card less than before in the deck (so we have 38 cards that are not spades, and 51 cards in total), then the probability of not drawing a spade is:
p2 = 38/51
And with the same reasoning, in the third draw the probability is:
p3 = 37/50
The joint probability for this event will be:
p₀ = p1*p2*p3 = (39/52)*(38/51)*(37/50) = 0.413
Second event:
x₁ = drawing one spade.
Let's suppose that in the first draw we get the spade, the probability will be:
p1 = 13/52
In the second draw, we get no spade, then the probability is:
p2 = 39/51
in the third draw we also get no spade, the probability is:
p3 = 38/50
And we also have the case where the spade is drawn in the second draw, and in the third draw, then we have 3 permutations, this means that the probability of drawing only one spade is:
p₁ = 3*p1*p2*p3 = 3*(13/52)(39/51)*(38/50) = 0.436
third event:
x₂ = drawing two spades:
Let's assume that in the first draw we do not get a spade, then the probabilities are:
p1 = 39/52
p2 = 13/51
p3 = 12/50
And same as before, we will have 3 permutations, because we could not draw a spade in the second draw, or in the third, then the probability for this case is:
p₂ = 3*p1*p2*p3 = 3*( 39/52)*(13/51)*(12/50) = 0.138
And the last event:
x₃ = drawing 3 spades.
The probabilities will be:
p1 = 13/52
p2 = 12/51
p3 = 11/50
And there are no permutations here, so the joint probability is:
p₃ = p1*p2*p3 = (13/52)*(12/51)*(11/50) = 0.013
Now we can calculate the expected value:
EV = 0*0.413 + 1*0.436 + 2*0.138 + 3*0.013 = 0.751
The expected number of spades that you will draw is 0.751 spades
The expected number of spades drawn when drawing three cards without replacement from a standard deck is approximately 0.75 spades.
To calculate this, we can use the concept of conditional probability. Initially, there are 13 spades out of 52 cards in the deck, giving us a 13/52 chance of drawing a spade on the first card.
If the first card drawn is a spade, there are now 12 spades left out of 51 cards, so the probability of drawing a spade on the second card is 12/51.
If the first two cards are spades, there are 11 spades left out of 50 cards for the third draw, with a probability of 11/50.
Now, we multiply these probabilities together and sum up the possible scenarios (0, 1, 2, or 3 spades drawn) to get the expected value: (0 * (39/52 * 38/51 * 37/50)) + (1 * (13/52 * 39/51 * 38/50 + 39/52 * 12/51 * 38/50 + 39/52 * 38/51 * 11/50)) + (2 * (13/52 * 12/51 * 39/50 + 13/52 * 39/51 * 11/50 + 39/52 * 12/51 * 11/50)) + (3 * (13/52 * 12/51 * 11/50)) ≈ 0.75 spades.
So, the expected number of spades drawn when selecting three cards without replacement from a standard deck is approximately 0.75.
This means, on average, you can expect to draw about 3/4 of a spade.
for such more questions on number
#SPJ3