Answer:
(a) 0.85
(b) 0.7225
Step-by-step explanation:
(a) The point estimate for the proportion of all such components that are not defective is given by the number of non-defective units in the sample divided by the sample size:
[tex]p=\frac{80-12}{80}\\p=0.85[/tex]
The proportion is 0.85.
(b) Assuming that the sample is large enough to accurately provide a point estimate for the whole population, this can be treated as a binomial model with probability of success (non-defective part) p = 0.85. Since both components must be non-defective for the system to work, the probability of two successes in two trials is:
[tex]P(x=2) = 0.85^2\\P(x=2) = 0.7225[/tex]
An estimate of 0.7225 for the proportion of all such systems that will function properly.
Find the common difference for the sequence shown.
56, 49, 42, 35, ...
-7
-12
7
Answer:
-7
Step-by-step explanation:
Given : sequence 56, 49, 42, 35, ...
To find : Find the common difference.
Solution : We have given that 56, 49, 42, 35, ...
By the common difference of an arithmetic sequence is:
d = a_{2} -a_{1}a2−a1
Where, a_{1}a1 is the first term and a_{1}a1 is the second term.
Then common difference = 49 -56
= -7 .
As a result of complaints by the staff about noise, the coffee and recreation area for student interns at OHaganBooks will now be in a 512 square foot rectangular area in the headquarter's basement against the southern wall. (The specified area was arrived at in complex negotiations between the student intern representative and management.) The construction of the partition will cost $16 per foot for the north wall and $4 per foot for the east and west walls. What are the dimensions of the cheapest recreation area that can be made?
Answer:
The dimensions are 32 ft by 16 ft
Step-by-step explanation:
Area of the coffee and recreation room=512 square foot
LB=512
L=512/B
Perimeter of the Room = Perimeter of north wall+Perimeter of east wall+ Perimeter of west wall =L+2B (West and East are opposite)
Cost of the Perimeter=16L+4(2B)
[tex]C(B)=16(\frac{512}{B})+8B\\C(B)=\frac{8192+8B^2}{B}[/tex]
To minimize cost, first, we take the derivative of C(B)
Using quotient rule
[tex]C^{'}(B)=\frac{8B^2-8192}{B^2}[/tex]
Setting the derivative equal to zero
[tex]8B^2-8192=0\\8B^2=8192\\B^2=1024\\B=32 ft[/tex]
[tex]L=\frac{512}{B}=\frac{512}{32}=16ft[/tex]
The dimension of the cheapest recreation area will be 32 ft by 16 ft
Shiffon Electronics manufactures music player. Its costing system uses two cost categories, direct materials and conversion costs. Each product must pass through the Assembly Department, the Programming department, and the Testing Department. Direct materials are added at the beginning of the production process. Conversion costs are allocated evenly throughout production. Shiffon Electronics uses weighted-average costing. The following information is available for the month of March 2017 for the Assembly department. Work in process, beginning inventory 340 units Conversion costs (25% complete) Units started during March 860 units Work in process, ending inventory: 140 units Conversion costs (60% complete) The cost details for the month of March are as follows: Work in process, beginning inventory: Direct materials $349,000 Conversion costs $360,000 Direct materials costs added during March $703,500 Conversion costs added during March $1,130,000 What amount of conversion costs is assigned to the ending Work-in-Process account for March
Answer:
Possible options:
A.$182,343
B.$122,792
C.$109,406
D.$270,629
Answer C
Step-by-step explanation:
Under this method the percentage completion done in the Beginning Inventory is ignored while calculating the equivalent completed units during the current period. Beginning Inventory Units are treated as fresh units introduced for production.
Cost Accounted for:
- The cost of opening work-in-progress and cost of the current period are aggregated and the aggregate cost is divided by output in terms of completed equivalent units.
- The units of Beginning Inventory of WIP and their cost are taken in full under this method.
Solve the equation.
e + 1.2 =2
e=
Answer:
e= 0.8
Step-by-step explanation:
Get e by itself, so subtrct 1.2 from 2 to get 0.8
The solution to the equation [tex]\(e + 1.2 = 2\) is \(e = 0.8\).[/tex]
To solve the equation \(e + 1.2 = 2\), we need to isolate the variable \(e\) on one side of the equation. Here's how you can do it step by step:
1. **Subtract 1.2 from both sides:**
\[e + 1.2 - 1.2 = 2 - 1.2\]
\[e = 0.8\]
In the context of real numbers, this means that when you substitute \(e = 0.8\) back into the original equation, it holds true:
\[0.8 + 1.2 = 2\]
This equation verifies that \(e = 0.8\) is indeed the solution. In a broader sense, solving equations like this one is a fundamental concept in algebra and mathematics. It allows us to find unknown values in various real-life situations, making it a crucial skill in fields such as physics, engineering, economics, and many others.
Understanding how to manipulate equations helps us model and solve complex problems, making mathematics an essential tool in problem-solving and critical thinking.
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Solve 15y − 1 = 11/2y + 2
Answer:
y = 6/19 = 0.316
Step-by-step explanation:
:)
Answer:
y=6/19
Step-by-step explanation:
15y − 1 = 11/2y + 2
15y-11/2y=3
30/2y-11/2y=3
19/2y=3
19y=6
y=6/19
Someone please help
????????
Answer:
use the distance formula,
by using it, prove the adjacent sides of the quadrilateral DEFG
hence, DEFG would be a rhombus
what is 4/7 divded by 10/21
Answer:
6/5
Step-by-step explanation:
(4/7) divided by (10/21) = (4/7) * (21/10) = (3*4)/10= 6/5
Find the value of y+5 given that -3y-3=6
Simplify
Brian has reduced his cholesterol level by 13% after his last check up. If his original level was 200, what is his cholesterol level now?
Daniel believes that people perform better in the barbell curl, on average, if they are encouraged by a coach. He recruited 29 subjects to participate in an experiment and randomly assigned them into two groups. Daniel gave one group verbal encouragement during the exercise and was quiet the exercise for the other group. He recorded the total number of barbell curls each subject was able to complete before setting the bar down.
(a) Explain the purpose of random assignment in this experiment.
(b) Compare these distributions,
(c) State the hypotheses Daniel should use to test his belief about receiving encouragement during exercise. Make sure to define any parameters you use.
(d) Identify the significance test Daniel should use to analyze the results of his experiment and show that the conditions for this test are met.
(e) The P-value for Daniel's test is 0.107. What conclusion should Daniel make at the a=0.05 significance level?
Answer:
a) to remove bias in the study
b) The number of barbell curls performed by group who was encourged was higher than the group who was not encouraged
c) H0: mean of barbell curls for group encouraged= mean of barbell curls for group not encouraged
Ha: mean of barbell curls for group encouraged≠ mean of barbell curls for group not encouraged
d) two sample t-test.
Requirement has been met as there are two separate groups whose means have to be compared. Median of box plot is equal to mean. Standard Deviation is equal to range divided by four. Sample size will have to be assumed
e) Encouragement affects the number of barbell curl performed by an individua
Step-by-step explanation:
a) random assignemnt removes bias in any study.
b) box plot is attached
c) null hypthesis and alternate hypothesis would compare means of the two distributions
d) As we are comparing two groups, we will use two sample t-test.
The median of box plot is equal to mean for both groups.
The rnge/4 is equal to standard deviation.
mean of encouraged group= 30
mean of not encoouraged group=19
SD of encouraged group= (75-5)/4= 15
SD of not encouraged group= (60-0)/4= 15
Assume number of inidivudals in encouraged group =15
assume number of individuals in not encouraged group = 14
e) test statistic= (mean1-mean2)/√(SD1²/(N1-1)+ SD2²/(N2-1))
= (30-19)/(√(15²/14+ 15²/13)
= 1.9734
Since test statistic is greater than p-value, null hypothesis is rejected.
Encouragement affects the number of barbell curl performed by an individual
The administration at Pierce College conducted a survey to determine the proportion of students who ride a bike to campus. Of the 125 students surveyed 6 ride a bike to campus. Which of the following is a reason the administration should not calculate a confidence interval to estimate the proportion of all students who ride a bike to campus? Check all that apply.
A. The sample needs to be random but we don’t know if it is.
B. The actual count of bike riders is too small.
C. The actual count of those who do not ride a bike to campus is too small.
D. n(p-hat) is not greater than 10.
E. n(1 minus p-hat) is not greater than 10.
Answer:
The correct option is (D).
Step-by-step explanation:
To construct the (1 - α)% confidence interval for population proportion the distribution of proportions must be approximated by the normal distribution.
A Normal approximation to binomial can be applied to approximate the distribution of proportion p, if the following conditions are satisfied:
[tex]n\hat p \geq 10[/tex][tex]n ( 1 - \hat p) \geq 10[/tex]In this case p is defined as the proportions of students who ride a bike to campus.
A sample of n = 125 students are selected. Of these 125 students X = 6 ride a bike to campus.
Compute the sample proportion as follows:
[tex]\hat p=\frac{X}{n}=\frac{6}{125}=0.048[/tex]
Check whether the conditions of Normal approximation are satisfied:
[tex]n\hat p =125\times 0.048=6<10\\n(1-\hat p) =125\times (1-0.048)=119>10[/tex]
Since [tex]n\hat p <10[/tex], the Normal approximation to Binomial cannot be applied.
Thus, the confidence interval cannot be used to estimate the proportion of all students who ride a bike to campus.
Thus, the correct option is (D).
Simplify 18 - 2[X + (x - 5)].
8- 4x
28 - 4x
28 - 2x
Answer:
28 - 4x
Step-by-step explanation:
[tex]18 -2 [x + (x - 5)] \\ = 18 - 2 [x + x - 5] \\ = 18 - 2 [2x - 5] \\ = 18 -4x + 10 \\ = 28 - 4x \\ [/tex]
what is a area of a triangle with a heigt of 5 inches and a base of 10inches
Answer: 25 in²
Step-by-step explanation: To find the area of a triangle, start with the formula for the area of a triangle which is shown below.
[tex]Area =[/tex] [tex]\frac{1}{2} bh[/tex]
In this problem, we're given that the base is 10 inches
and the height is 5 inches.
Now, plugging into the formula, we have [tex](\frac{1}{2})(10in.)(5 in.)[/tex].
Now, it doesn't matter which order we multiply.
So we can begin by multiplying (1/2) (10 in.) to get 5 inches.
Now, (5 in.) (5 in.) is 25 in².
So the area of the triangle is 25 in².
mean of 153 176 249 150 620 242
Answer: 265
Step-by-step explanation:
Add up all the numbers and divide this value by the total number of terms,in this case 6 terms.
(153+ 176+ 249+ 150+ 620+ 242)/6 =265
Please help!!!
If the area of a rectangle is 42m^2, find value x
Answer:
Give me more fees back on your equation
Step-by-step explanation:
basketball players score 19 times in one game he scored a total of 33.2 of each point to point shot in one of each free-throw how many to point shots did he make how many free throws
The answer is 17 free throws because I said so and that’s the correct answer
Answer:
my answer was gunna be his but that person answered first so ill just sound like i copied that person
Step-by-step explanation:
evaluate the expression when y=-6
y squared+7y+4
Answer:
In the equation of a straight line (when the equation is written as "y = mx + b"), the slope is the number "m" that is multiplied on the x, and "b" is the y-intercept (that is, the point where the line crosses the vertical y-axis). This useful form of the line equation is sensibly named the "slope-intercept form".
Step-by-step explanation:
This is just something to keep in mind
Answer:
-2
Step-by-step explanation:
Well, lt me hope the expression is complete amd correct.
Let's go on with the information you submitted.
y² + 7y + 4
When y= -6.
Let's substitute the value, y= -6 in the expression. We'll have to be careful with the signs.
y² + 7y + 4
-6² + 7(-6) + 4
36 - 42 + 4
40 - 42 =
-2
In a study of 346 comma 145 cell phone users, it was found that 26 developed cancer of the brain or nervous system. Assuming that cell phones have no effect, there is a 0.000113 probability of a person developing cancer of the brain or nervous system. We therefore expect about 40 cases of such cancer in a group of 346 comma 145 people. Estimate the probability of 26 or fewer cases of such cancer in a group of 346 comma 145 people. What do these results suggest about media reports that cell phones cause cancer of the brain or nervous system?
Answer:
a) P ( X ≤ 26 ) = 0.0134
b) The significance of the cell phone causing brain cancer is 1.34 or 0.0134
Step-by-step explanation:
Solution:-
- From the entire population of cell phone users os size N = 346,145 media reported n = 26 people who developed cancer of brain or nervous system.
- The probability of a person developing cancer of the brain or nervous system, assuming that cell phones have no effect, is p = 0.000113.
- Denote a random variable X: The number people who developed cancer of brain or nervous system are normally distributed.
- The expected or mean number of people who developed cancer of brain or nervous system are u = 40.
- The standard deviation ( s ) for the distribution would be:
[tex]s = \sqrt{u*( 1 - p ) }\\\\s = \sqrt{40*( 1 - 0.000113 ) }\\\\s = 6.3242\\[/tex]
- The random variate X follows normal distribution with parameters:
X ~ Norm ( 40 , 6.3242^2 )
- The probability of X ≤ 26 cases from the total population of N = 346,145 can be determined by evaluating the Z-score standard value of the test statistics:
[tex]P ( X \leq 26 ) = P ( Z \leq [ \frac{X - u }{s} ) \\\\P ( Z \leq [ \frac{ 26 - 40 }{6.3242} ) = P ( Z \leq -2.21371 )[/tex]
- Using standard normal table compute the probability on the left side of Z-score value -2.21371. Hence,
P ( X ≤ 26 ) = 0.0134
- The probability of the less than 26 number of cases who developed cancer of the brain or nervous system is 0.0134 as per media reports.
- So 1.34% of population is a case brain cancer due to cell phones according to media reports.
Which is significant less than the expected number of cases ( 40 ) that occur regardless of cell phone effect.
Answer: The significance of the cell phone causing brain cancer is 1.34 or 0.0134.
describe the products you get if you multiply 8 by factors less than 1. describe the products you get if you multiply 8 by factors greater than 1. give some examples that justify your answer
Answer:
When you multiply 8 by factors less than 1, your product is a number less than 8. An example of this would be when you multiply 8 by 1/2. Your product is 4. Another example is when you multiply 8 by 1/4. Your product is 2. When you multiply 8 by factors that are greater than 1, your product is a number greater than 8. One example of this is when you multiply 8 by 5. Your answer is 40. Another example is when you multiply the number 8 by 2. Your product is 16.
Step-by-step explanation:
Final answer:
When you multiply 8 by factors less than 1, the products will be smaller than 8. When you multiply 8 by factors greater than 1, the products will be larger than 8.
Explanation:
When you multiply 8 by factors less than 1, the products will be smaller than 8. For example, if you multiply 8 by 0.5, you get a product of 4. The further the factor is from 1, the smaller the product will be. Some other examples include multiplying 8 by 0.25 to get a product of 2, or multiplying 8 by 0.1 to get a product of 0.8.
On the other hand, when you multiply 8 by factors greater than 1, the products will be larger than 8. For example, if you multiply 8 by 2, you get a product of 16. The further the factor is from 1, the larger the product will be. Some other examples include multiplying 8 by 3 to get a product of 24, or multiplying 8 by 10 to get a product of 80.
Suppose that a study of elementary school students reports that the mean age at which children begin reading is 5.8 years with a standard deviation of 0.7 years. (2 pts) Step 1. If a sampling distribution is created using samples of the ages at which 61 children begin reading, what would be the mean of the sampling distribution of sample means
Answer:
5.8 years
Step-by-step explanation:
The sampling distribution of the sample means has a mean that is equal to mean of the population from which the sample has been drawn.
study of elementary school students reports that the mean age at which children begin reading is 5.8 years with a standard deviation of 0.7 years.
This means the population mean is
[tex] \mu = 5.8 \: years[/tex]
and the population standard deviation is
[tex] \sigma = 0.7 \: years[/tex]
If a sampling distribution is created using samples of the ages at which 61 children begin reading, the mean of the sampling distribution of the sample mean will be
[tex]5.8 \: years[/tex]
The mean of the sampling distribution of sample means can be calculated by dividing the mean of the population by the square root of the sample size.
Explanation:To find the mean of the sampling distribution of sample means, we need to divide the mean of the population by the square root of the sample size. In this case, the mean of the population is 5.8 years and the sample size is 61 children. So, the mean of the sampling distribution of sample means can be calculated as:
Mean of sampling distribution of sample means = 5.8 years / sqrt(61) ≈ 0.745 years
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Which expressions represent 1 less than the product of 3 and 4?
Answer:3+4-1
Step-by-step explanation:
how many 1/2 are in 7
14
All you have to do is multiply the 1/2 by the 7.
Answer:
14
Step-by-step explanation:
there are 14 1/2 in 7
HELP ASAP PLEASE!!!
Given the figure below, which of the following points name a line segment, a line, or a ray?
a. point B and point C
b. point A and point C
c. point D and point A
d. point E and point A
U may pick more than one.
The time until recharge for a battery in a laptop computer under common conditions is normally distributed with mean of 265 minutes and a standard deviation of 50 minutes. a) What is the probability that a battery lasts more than four hours? Enter your answer in accordance to the item a) of the question statement (Round the answer to 3 decimal places.) b) What are the quartiles (the 25% and 75% values) of battery life? 25% value = Enter your answer; 25% value = _ minutes minutes (Round the answer to the nearest integer.) 75% value = Enter your answer; 75% value = _ minutes minutes (Round the answer to the nearest integer.) c) What value of life in minutes is exceeded with 95% probability? Enter your answer in accordance to the item c) of the question statement (Round the answer to the nearest integer.)
Answer:
a) 0.691 = 69.1% probability that a battery lasts more than four hours
b) 25% value = 231
75% value = 299
c) 183 minutes
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 265, \sigma = 50[/tex]
a) What is the probability that a battery lasts more than four hours?
4 hours = 4*60 = 240 minutes
This is 1 subtracted by the pvalue of Z when X = 240. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{240 - 265}{50}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.309
1 - 0.309 = 0.691
0.691 = 69.1% probability that a battery lasts more than four hours
b) What are the quartiles (the 25% and 75% values) of battery life?
25th percentile:
X when Z has a pvalue of 0.25. So X when Z = -0.675
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{X - 265}{50}[/tex]
[tex]X - 265 = -0.675*50[/tex]
[tex]X = 231[/tex]
75th percentile:
X when Z has a pvalue of 0.75. So X when Z = 0.675
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 265}{50}[/tex]
[tex]X - 265 = 0.675*50[/tex]
[tex]X = 299[/tex]
25% value = 231
75% value = 299
c) What value of life in minutes is exceeded with 95% probability?
The 100-95 = 5th percentile, which is the value of X when Z has a pvalue of 0.05. So X when Z = -1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 265}{50}[/tex]
[tex]X - 265 = -1.645*50[/tex]
[tex]X = 183[/tex]
Consider the following sets of sample data:
A:
$36,900
, $19,400, $22,200, $21,900, $35,300, $20,500, $35,400, $24,000, $37,700, $35,300, $38,300, $29,600, $26,000, $38,400
B:
2.1
, 5.0, 3.5, 3.7, 2.5, 2.1, 3.7, 4.6, 2.7, 4.1, 1.7
For each of the above sets of sample data, calculate the coefficient of variation, CV. Round to one decimal place.
Answer:
The coefficient of variation for A is 24.6%.
The coefficient of variation for B is 33.7%.
Step-by-step explanation:
The coefficient of variation (CV) is well defined as the ratio of the standard deviation to the mean. It exhibits the degree of variation in association to the mean of the population.
The formula to compute the coefficient of variation is,
[tex]CV=\frac{SD}{Mean}\times 100\%[/tex]
Consider the data set A.
Compute the mean of the data set A as follows:
[tex]Mean_{A}=\frac{1}{n}\sum X[/tex]
[tex]=\frac{1}{14}\times [36900+19400+...+26000+38400]\\=30064.2857[/tex]
Compute the standard deviation of the data set A as follows:
[tex]SD_{A}= \sqrt{ \frac{ \sum{\left(x_i - Mean_{A}\right)^2 }}{n-1} }[/tex]
[tex]= \sqrt{ \frac{ 712852142.8571 }{ 14 - 1} } \\\approx 7405.051[/tex]
Compute the coefficient of variation for A as follows:
[tex]CV=\frac{SD_{A}}{Mean_{A}}\times 100\%[/tex]
[tex]=\frac{7405.051}{30064.2857}\times 100\%\\=24.6\%[/tex]
The coefficient of variation for A is 24.6%.
Consider the data set B.
Compute the mean of the data set B as follows:
[tex]Mean_{B}=\frac{1}{n}\sum X[/tex]
[tex]=\frac{1}{11}\times [2.1+5.0+...+4.1+1.7]\\=3.2455[/tex]
Compute the standard deviation of the data set B as follows:
[tex]SD_{B}= \sqrt{ \frac{ \sum{\left(x_i - Mean_{B}\right)^2 }}{n-1} }[/tex]
[tex]= \sqrt{ \frac{ 11.9873 }{ 11 - 1} } \\\approx 1.0949[/tex]
Compute the coefficient of variation for B as follows:
[tex]CV=\frac{SD_{B}}{Mean_{B}}\times 100\%[/tex]
[tex]=\frac{1.0949}{3.2455}\times 100\%\\=33.7\%[/tex]
The coefficient of variation for B is 33.7%.
What is pi? Explain it in your own words.
Answer:
pi is the ratio if the circumference over the diameter
Step-by-step explanation:
What is the value of 2 in 255.6
Answer:
200?
Step-by-step explanation:
its kinda simple but the 2 is in the hundreds place therefore it is 200.
hope it helped ;)
The mean potassium content of a popular sports drink is listed as 138 mg in a 32-oz bottle. Analysis of 40 bottles indicates a sample mean of 136.9 mg. (a) State the hypotheses for a two-tailed test of the claimed potassium content.
Answer:
The value for p <0.1 i.e. 0.02088 that means ,we had success in test on potassium content
Step-by-step explanation:
Given:
Means: 138 mg
sample size =40
New mean=136.9 mg
To find : Hypothesis on test.
Solution:
We know that ,
Z=(X-mean)/standard deviation)/{(sqrt(sample size)}
Consider the standard deviation of 3.00 mg
Z=(136.9-138)/(3/sqrt(40))
=-1.1/0.4743
Z=-2.3192
two tailed test
=2{(1-p(<z))}
P value for Z=2.3192 is 0.98956
=2{1-0.98956}
=0.02088
The value for p <0.1
we had success in test on potassium content.
The hypothesis test for the sports drink's potassium content involves a null hypothesis (H0) that the mean is equal to the claimed 138 mg, and an alternative hypothesis (Ha) that it is not equal.
Explanation:The question involves conducting a hypothesis test for a sports drink's potassium content.
Establishing hypotheses for a two-tailed test involves setting a null hypothesis (H0) that the mean potassium content is equal to the claimed amount, and an alternative hypothesis (Ha) that the mean potassium content is not equal to the claimed amount.
Specifically:
H0: μ = 138 mg (The mean potassium content is 138 mg as listed.)Ha: μ ≠ 138 mg (The mean potassium content is not 138 mg as listed.)This is a basic procedure in statistical testing where the null hypothesis often represents a statement of 'no effect' or 'no difference', and the alternative hypothesis contradicts this assumption.
Moose Drool Makes Grass More Appetizing Different species can interact in interesting ways. One type of grass produces the toxin ergovaline at levels about 1.0 part per million in order to keep grazing animals away. However, a recent study27 has found that the saliva from a moose counteracts these toxins and makes the grass more appetizing (for the moose). Scientists estimate that, after treatment with moose drool, mean level of the toxin ergovaline (in ppm) on the grass is 0.183. The standard error for this estimate is 0.016.
Give notation for the quantity being estimated, and define any parameters used.
Give notation for the quantity that gives the best estimate, and give its value.
Give a 95% confidence interval for the quantity being estimated. Interpret the interval in context.
3.68 Bisphenol A in Your Soup Cans Bisphenol A (BPA) is in the lining of most canned goods, and recent studies have shown a positive association between BPA exposure and behavior and health problems. How much does canned soup consumption increase urinary BPA concentration? That was the question addressed in a recent study34 in which consumption of canned soup over five days was associated with a more than 1000% increase in urinary BPA. In the study, 75 participants ate either canned soup or fresh soup for lunch for five days. On the fifth day, urinary BPA levels were measured. After a two-day break, the participants switched groups and repeated the process. The difference in BPA levels between the two treatments was measured for each participant. The study reports that a 95% confidence interval for the difference in means (canned minus fresh) is 19.6 to 25.5 μg/L.
Is this a randomized comparative experiment or a matched pairs experiment? Why might this type of experiment have been used?
What parameter are we estimating?
Interpret the confidence interval in terms of BPA concentrations.
If the study had included 500 participants instead of 75, would you expect the confidence interval to be wider or narrower?
3.70 Effect of Overeating for One Month: Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later, a Swedish study shows.35 Researchers recruited 18 healthy and normal-weight people with an average age of 26. For a four-week period, participants increased calorie intake by 70% (mostly by eating fast food) and limited daily activity to a maximum of 5000 steps per day (considered sedentary). Not surprisingly, weight and body fat of the participants went up significantly during the study and then decreased after the study ended. Participants are believed to have returned to the diet and lifestyle they had before the experiment. However, two and a half years after the experiment, the mean weight gain for participants was 6.8 lbs with a standard error of 1.2 lbs. A control group that did not binge had no change in weight.
What is the relevant parameter?
How could we find the actual exact value of the parameter?
Give a 95% confidence interval for the parameter and interpret it.
Give the margin of error and interpret it.
In the first study, moose drool reduces the toxin levels on grass. In the second study, canned soup consumption increases urinary BPA concentrations. In the third study, overeating for four weeks leads to long-term weight gain.
Explanation:Question 1:
The notation for the quantity being estimated is the mean level of the toxin ergovaline on the grass (in ppm). Scientists estimate the mean level after treatment with moose drool to be 0.183 ppm.
The quantity that gives the best estimate is the mean level of the toxin ergovaline on the grass after treatment with moose drool (0.183 ppm).
A 95% confidence interval for the quantity being estimated is (0.183 - 1.96 * 0.016, 0.183 + 1.96 * 0.016), which is approximately (0.152, 0.214) ppm. This means there is a 95% chance that the true mean level of the toxin ergovaline on the grass after treatment with moose drool is between 0.152 and 0.214 ppm.
Question 2:
This is a matched pairs experiment because each participant is subjected to both treatments (canned soup and fresh soup). It was used to eliminate individual differences and control for variables that might affect the urinary BPA concentrations.
The parameter being estimated is the difference in means (canned minus fresh) of urinary BPA concentrations.
The 95% confidence interval for the difference in means is (19.6, 25.5) μg/L. This means that we are 95% confident that the true difference in mean urinary BPA concentrations between canned soup and fresh soup is between 19.6 and 25.5 μg/L.
If the study had included 500 participants instead of 75, we would expect the confidence interval to be narrower because a larger sample size reduces the standard error, leading to a more precise estimate.
Question 3:
The relevant parameter is the mean weight gain for participants two and a half years after the experiment.
The actual exact value of the parameter cannot be found unless we measure the weight gain for all individuals in the population.
A 95% confidence interval for the parameter is (6.8 - 1.96 * 1.2, 6.8 + 1.96 * 1.2), which is approximately (4.44, 9.16) lbs. This means there is a 95% chance that the true mean weight gain for participants two and a half years after the experiment is between 4.44 and 9.16 lbs.
The margin of error is 1.96 * 1.2 lbs, which is approximately 2.35 lbs. This means that the estimated mean weight gain may vary by up to 2.35 lbs from the true mean weight gain.
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The compressive strength of concrete is normally distributed with mu = 2500 psi and sigma = 50 psi. A random sample of n = 8 specimens is collected. What is the standard error of the sample mean?
Round your final answer to three decimal places (e.g. 12.345).
The standard error of the sample mean is __ psi.
Answer:
The standard error of the sample mean is _17.677_ psi.
Step-by-step explanation:
Explanation:-
A random sample of n = 8 specimens is collected.
Given sample size is n = 8
Given mean of the population 'μ' = 2500 psi
standard deviation 'σ' = 50 psi
Let x⁻ is the mean of the observed sample
Standard error of the sample mean = [tex]\frac{S.D}{\sqrt{n} }[/tex] ...(i)
Given Population of standard error (S.D) 'σ' = 50 psi
Now substitute all values in (i)
[tex]S.E = \frac{50}{\sqrt{8} } =17.677[/tex]
Conclusion:-
The standard error of the sample mean is _17.677psi.
The standard error of the sample mean for a sample of 8 concrete specimens is approximately 17.68 psi
To calculate the standard error of the sample mean, we use the formula:
[tex]\[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} \][/tex]
Where:
- [tex]\sigma[/tex] is the standard deviation of the population.
- n is the sample size.
Given:
- [tex]\sigma[/tex] = 50 psi
- n = 8
Substituting these values into the formula:
[tex]\[ \text{Standard Error} = \frac{50}{\sqrt{8}} \][/tex]
[tex]\[ \text{Standard Error} = \frac{50}{2.828} \][/tex]
[tex]\[ \text{Standard Error} \approx 17.68 \][/tex]
So, the standard error of the sample mean is approximately 17.68 psi.