Answer:
Answer:
a).
The amount spent on school materials for each term of all ST201students
b).
a).
It is not a random sample. This looks like a convenience sampling and there is sampling bias. This sample is not representative of the entire population. Since it is not a random sample it is not appropriate to generalize the results to all students.
b).
The sample size is 80 which is greater than 30. It is large enough to assume normal distribution according to central limit theorem.
c).
mean: $617
z critical value at 95%: 1.96
standard error = σ/sqrt(n) =500/sqrt(80) = 55.9017
lower limit= mean-1.96*se = 617-1.96*55.9017=507.43
upper limit= mean+1.96*se = 617+1.96*55.9017=726.57
d).
The amount spent on school materials for each term for the 80 ST201students is $617. We are 95% confident that amount spent on school materials for each term of all ST201students falls in the interval ($507.43, $726.57).
Step-by-step explanation:
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 50 bulbs was selected, the lifetime of each bulb determined, and the following information obtained:
Average lifetime = 738.44 hours and a standard deviation of lifetimes equal to 38.2 hours.
Should the customer purchase the light bulbs?
(a) Make the decision on a significance level of .05?
(b) Make the decision on a significance level of .01?
Answer:
(a) Customer will not purchase the light bulbs at significance level of 0.05
(b) Customer will purchase the light bulbs at significance level of 0.01 .
Step-by-step explanation:
We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected, and the following information obtained:
Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.
Let Null hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 750 {means that the true average lifetime is same as what is advertised}
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] < 750 {means that the true average lifetime is smaller than what is advertised}
Now, the test statistics is given by;
T.S. = [tex]\frac{Xbar-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, X bar = sample mean = 738.44 hours
s = sample standard deviation = 38.2 hours
n = sample size = 50
So, test statistics = [tex]\frac{738.44-750}{\frac{38.2}{\sqrt{50} } }[/tex] ~ [tex]t_4_9[/tex]
= -2.14
(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.
Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.
(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.
You are staffing a new department, and you have applications on your desk. You will hire 2 of the 5 software engineers who have applied, and 3 of the 7 computer engineers who have applied. What is the total number of possible complete staffs you could hire
Answer:
The total number of possible complete staffs you could hire is 350.
Step-by-step explanation:
The order that the software engineers are is not important. For example, hiring John and Laura as the software engineers is the same as hiring Laura and John. The same applies for the computer engineers. So we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
What is the total number of possible complete staffs you could hire?
2 software engineers from a set of 5
3 computer engineers from a set of 7
So
[tex]T = C_{5,2}*C_{7,3} = \frac{5!}{2!3!}*\frac{7!}{3!4!} = 10*35 = 350[/tex]
The total number of possible complete staffs you could hire is 350.
In ΔABC, b = 68 inches, ∠B=65° and ∠C=93°. Find the length of a, to the nearest inch.
Answer:
28 inches
Step-by-step explanation:
∠A + ∠B + ∠C = 180°
∠A + 65° +93° = 180°
∠A + 158° = 180°
∠A= 180°-158° = 22°
Using Law of Sines
a/sinA= b/sinB
a/sin22= 68 inches/sin65
a/sin22 = 68/0.9063 = 75.03
a = 75.03 x sin 22
a = 75.03 x 0.3746 = 28.106238≈28 inches
An aerospace company has submitted bids on two separate federal government defense contracts. The company president believes that there is a 40% probability of winning the first contract. If they win the first contract, the probability of winning the second is 65%. However, if they lose the first contract, the president thinks that the probability of winning the second contract decreases to 49%.
What is the probability that they win both contracts?
Answer:
26% probability that they win both contracts.
Step-by-step explanation:
These following probabilities are important to solve this question:
0.4 = 40% probability of winning the first contract.
0.65 = 65% probability of winning the second contract if the first contract is won.
What is the probability that they win both contracts?
[tex]P = 0.4*0.65 = 0.26[/tex]
26% probability that they win both contracts.
In 2001, a total of 15,555 homicide deaths occurred among males and 4,753 homicide deaths occurred among females. The estimated 2001 midyear populations for males and females were 139,813,000 and 144,984,000 respectively
a) Calculate the homicide-related death rates for males per 100,000.
b) Calculate the homicide-related death rates for females per 100,000.
c) What type(s) of mortality rates did you calculate in Questions 17and 18?
d) Calculate the ratio of homicide-mortality rates for males compared to females.
e) Interpret the rates you calculated in Question 20 as if you were presenting information to a policymaker.
Answer:
a. 11
b. 3
c. homicide mortality rate
d. 11:3
Step-by-step explanation:
a.) If 15,555 homicide cases were recorded among males of 139,813,000 population, then in every 100,000 males, the number of homicides cases will become:
= [15,555/139,813,000] * 100,000
= 11.13 aproximately 11 homicide cases in every 100,000 males.
b.) If 4,753 homicide cases were recorded among Females of 144,984,000 population, then in every 100,000 Females, the number of homicides cases will become:
= [4753/144,984,000] * 100,000
= 3.28 approximately 3 homicide cases in every 100,000 females
C.) Homicide-mortality rate
d.) ratio of male to female homicide rate = 11 : 3
e.) What those rates means is that in every 100,000
Males in 2001, 11 of them were Victims of homicide and in every 100,000 females, 3 of them are victims of homicide.
The number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays. Assuming the number of accidents on a day follows a Poisson distribution.
What is the probability there are no car accidents on that stretch on Monday?
Answer:
1.83% probability there are no car accidents on that stretch on Monday
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
The number of accidents on a certain section of I-40 averages 4 accidents per weekday independent across weekdays.
This means that [tex]\mu = 4[/tex]
What is the probability there are no car accidents on that stretch on Monday?
This is P(X = 0).
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-4}*(4)^{0}}{(0)!} = 0.0183[/tex]
1.83% probability there are no car accidents on that stretch on Monday
The probability of there being no car accidents on Monday on a certain section of I-40 can be calculated using the Poisson distribution. In this case, with an average of 4 accidents per weekday, the probability is approximately 1.83%.
Explanation:To calculate the probability of there being no car accidents on Monday, we can use the Poisson distribution. In this case, the average number of accidents per weekday is given as 4. The Poisson distribution can be used to calculate the probability of a specific number of events occurring in a given time period.
The formula for calculating the probability of x events occurring in a Poisson distribution is:
P(x) = (e^-λ * λ^x) / x!
Where λ is the average number of events, and x is the number of events we want to calculate the probability for.
In this case, the average number of accidents per weekday is 4, so λ = 4. And we want to calculate the probability of there being no accidents, so x = 0.
Using the formula, we can calculate:
P(0) = (e^-4 * 4^0) / 0! = (e^-4 * 1) / 1 = e^-4, approximately 0.0183Therefore, the probability of there being no car accidents on Monday is approximately 0.0183, or 1.83%.
Positive Test Result Negative Test Result
Hepatitis C 335 10
No Hepatitis C 2 1153
Based on the results of this study, how many false negatives should you expect out of 1000 tests
Answer:
In a sample of 1000 test the expected number of false negatives is 8.6.
Step-by-step explanation:
The table provided is:
Positive Negative TOTAL
Hepatitis C 335 10 345
No Hepatitis C 2 1153 1155
TOTAL 337 1163 1500
A false negative test result implies that, the person's test result for Hepatitis C was negative but actually he/she had Hepatitis C.
Compute the probability of false negative as follows:
[tex]P(Negative|No\ Hepapatits\ C)=\frac{n(Negative\cap No\ Hepapatits\ C)}{n(No\ Hepapatits\ C)} \\=\frac{10}{1163}\\ =0.0086[/tex]
Compute the expected number of false negatives in a sample is n = 1000 tests as follows:
E (False negative) = n × P (Negative|No Hepatitis C)
[tex]=1000\times0.0086\\=8.6[/tex]
Thus, in a sample of 1000 test the expected number of false negatives is 8.6.
Find each difference in the photo below
Answer:
[tex]= - 2 x^{3} + 4x -8[/tex]
Step-by-step explanation:
The first step is to open the parenthesis,
Since there is a negative sign before the second parenthesis, so the sign of all the values in second parenthesis will be changed and the equation will look something like this
[tex]= 2x^{3} + 4x -2 - 4 x^{3} + 6[/tex]
The second step is to re arrange the equation
[tex]= 2x^{3} - 4 x^{3} + 4x - 2 + 6[/tex]
The last and final step is to solve the equation
[tex]= - 2 x^{3} + 4x -8[/tex]
This is our answer
Answer:
The difference is -2x³ + 4x + 4.
Step-by-step explanation:
Subtract the two expression as follows:
[tex](2x^{3}+4x-2)-(4x^{3}-6)=2x^{3}+4x-2-4x^{3}+6\\[/tex]
Combine the like terms together:
[tex]=2x^{3}-4x^{3}+4x-2+6[/tex]
Simplify as follows:
[tex]=-2x^{3}+4x+4[/tex]
Thus, the difference is -2x³ + 4x + 4.
Listed below are the numbers of manatee deaths caused each year by collisions with watercraft. The data are listed in order for each year of the past decade.
(a) Find the range, variance, and standard deviation of the data set.
(b) What important feature of the data is not revealed through the different measures of variation?
80 68 71 72 95 89 97 72 75 81
Answer:
Range = 29
Variance= (X₁- U) ² / N= 973/10 = 97.3
Standard Deviation= √variance= √97.3= 9.864
Step-by-step explanation:
Range = Difference between the highest and lowest value = 97-68= 29
Variance
X₁ X₁-U (X₁- U) ²
80 0 zero
68 -12 144
71 -9 81
72 -8 64
95 15 225
89 9 81
97 17 289
72 -8 64
75 -5 25
81 1 1
∑ 800 ZERO 973
u= ∑X₁ /10=800/10=80
Variance= (X₁- U) ² / N= 973/10 = 97.3
Standard Deviation= √variance= √97.3= 9.864
(b) The important feature of the data is not revealed through the different measures of variation is that the variability of two or more than two sets of data cannot be compared unless a relative measure of dispersion is used .
If the Durbin-Watson statistic has a value close to 0, which assumption is violated? In other words, which assumption is the Durbin-Watson statistic checking to see is violated?
a - Independence of errors
b- Normality of errors
c- Homoscedasticity
d- none of the above
Answer:
Null Hypothesis: No first order autocorrelation
Alternative hypothesis: first order correlation exists
The assumptions to run this test are:
1) Errors are normally distributed with mean 0
2) Errors follows an stationary process
3) Independence condition between the erros
The statistic is defined as:
[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]
And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.
Step-by-step explanation:
The Durbin Watson test is a way to check autocorrelation in residuals for a time seeries or a regression.
We need to remember that the autocorrelation is the similarity of the time series in successive intervals. When we conduct this type of test we are checking if the time series can be modeled with and AR(1) process autoregressive.
The system of hypothesis on this case are:
Null Hypothesis: No first order autocorrelation
Alternative hypothesis: First order correlation exists
The assumptions to run this test are:
1) Errors are normally distributed with mean 0
2) Errors follows an stationary process
3) Independence condition between the errors
The statistic is defined as:
[tex] DW = \frac{\sum_{t=2}^ T (e_t -e_{t-1})^2}{\sum_{t=1}^T e^2_t}[/tex]
And if the value for the DW is near to 0 we can conclude that the assumption of Independence is not satisfied.
This table gives a few (x,y) pairs of a line in the coordinate plane
x (48) (61) (74)
y (-30) (-45) (-60)
what is the x-intercept of the line?
Answer:(22,0)
Step-by-step explanation: you have to find when y equals 0
plz help me answer this question
Answer:
The answer to your question is distance = 3500000 km
Step-by-step explanation:
Data
Scale 1 : 500000
7 cm
Process
1.- To solve this problem use direct proportions or rule of three.
1 : 500 000 :: 7 : x
2.- Multiply the middle numbers and the result divide it by the edge.
x = (7 x 500000) / 1
Simplification
x = 3500000 km
The Whitt Window Company, a company with only three employees, makes two different kinds of hand-crafted windows: a wood-framed and an aluminum-framed window. The company earns $300 profit for each wood-framed window and $150 profit for each aluminum-framed window. Doug makes the wood frames and can make 6 per day. Linda makes the aluminum frames and can make 4 per day. Bob forms and cuts the glass and can make 48 square feet of glass per day. Each wood-framed window uses 6 square feet of glass and each aluminum-framed window uses 8 square feet of glass.
The company wishes to determine how many windows of each type to produce per day to maximize total profit.
(a) Describe the analogy between this problem and the Wyndor Glass Co. problem discussed in Sec. 3.1. Then construct and fill in a table like Table 3.1 for this problem, identifying both the activities and the resources.
(b) Formulate a linear programming model for this problem.(c) Use the graphical method to solve this model.
Answer:
Maximize Z = 6x1 + 3x2
other answers are as follows in the explanation
Step-by-step explanation:
Employee Glass Needed per product(sq feet) Glass available per production
Product
Wood framed glass Aluminium framed glass
doug 6 0 36
linda 0 8 32
Bob 6 8 48
profit $300 $150
per batch
Z = 6x1 + 3x2,
with the constraint
6x1 ≤ 36 8x2 ≤ 32 6x1 + 8x2 ≤ 48
and x1 ≥ 0, x2 ≥ 0
Maximize Z = 6x1 + 3x2
to get the points of the boundary on the graph we say
when 6x1= 36
x1=6
when
8x2= 32
x2=4
to get the line of intersect , we go to
6x1 + 8x2 ≤ 48
so, 6x1 + 8x2 = 48
When X1=0
8x2=48
x2=6
when x2=0
x1=8
the optimal point can be seen on the graph as attached
In this exercise we have to write the maximum function of a company, in this way we find that:
A)[tex]M(Z) = 6x_1+ 3x_2[/tex]
B)[tex]X_1= 8 \ and \ x_2 = 6 \ or \ 0[/tex]
A)So to calculate the maximum equation we have:
[tex]Z = 6x_1 + 3x_2\\6x_1 \leq 36 \\ 8x_2 \leq 32 \\ 6x_1 + 8x_2 \leq 48[/tex]
B) To calculate the limits of the graph we have to do:
[tex]6x_1= 36\\x_1=6\\8x_2= 32\\x_2=4\\6x_1 + 8x_2 = 48\\X_1=0\\8x_2=48\\x_2=6\\x_2=0\\x_1=8[/tex]
See more about graphs at brainly.com/question/14375099
Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not conform to the strength specifications, whereas of 300 forgings selected from process 2, 20 are nonconforming. a) Esetimate the fraction nonconforming for each process. b) Test the hypothesis that the two process have identical fractions nonconforming. Use alpha =0.05. c) Construct a 90% confidence interval on the difference in fraction nonconforming between the two processes.
Answer:
a.
[tex]\bar p_1=0.05\\\bar p_2=0.067[/tex]
b-Check illustration below
c.(-0.0517,0.0177
Step-by-step explanation:
a.let [tex]p_1 \& p_2[/tex] denote processes 1 & 2.
For [tex]p_1[/tex]: T1=10,n1=200
For [tex]p_2[/tex]:T2=20,n2=300
Therefore
[tex]\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067[/tex]
b. To test for hypothesis:-
i.
[tex]H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05[/tex]
ii.For a two sample Proportion test
[tex]Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\[/tex]
iii. for [tex]\frac{\alpha}{2}=(-1.96,+1.96)[/tex] (0.5 alpha IS 0.025),
reject [tex]H_o[/tex] if[tex]|Z|>1.96[/tex]
iv. Do not reject [tex]H_o[/tex]. The noncomforting proportions are not significantly different as calculated below:
[tex]z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}[/tex]
z=-0.78
c.[tex](1-\alpha).100\%[/tex] for the p1-p2 is given as:
[tex](\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645 \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\[/tex]
=(-0.0517,+0.0177)
*CI contains o, which implies that proportions are NOT significantly different.
One percent of all individuals in a certain population are carriers of a particular disease. A diagnostic test for this disease has a 93% detection rate for carriers and a 2% false positive rate. Suppose that an individual is tested. What is the probability that an individual who tests negative does not carry the disease? What is the specificity of the test?
Answer:
(1) The probability that an individual who tests negative does not carry the disease is 0.9709.
(2) The specificity of the test is 98%.
Step-by-step explanation:
Denote the events as follows:
X = a person carries the disease
Y = the test detected the disease.
Given:
[tex]P(X) = 0.01\\P(Y|X)=0.93\\P(Y|X^{c})=0.02[/tex]
The probability of a person not carrying the disease is:
[tex]P(X^{c})=1-P(X)=1-0.01=0.99[/tex]
The probability that the test does not detects the disease when the person is carrying it is:
[tex]P(Y^{c}|X)=1-P(Y|X)=1-0.93=0.07[/tex]
The probability that the test does detects the disease when the person is not carrying it is:
[tex]P(Y^{c}|X^{c})=1-P(Y|X^{c})=1-0.02=0.98[/tex]
(1)
Compute the probability that an individual who tests negative does not carry the disease as follows:
[tex]P(X^{c}|Y^{c})=\frac{P(Y^{c}|X^{c})P(X^{c})}{P(Y^{c}|X^{c})P(X^{c})+P(Y^{c}|X)P(X)} \\=\frac{(0.98\times 0.99)}{(0.98\times 0.99)+(0.07\times 0.01)} \\=0.9709[/tex]
Thus, the probability that an individual who tests negative does not carry the disease is 0.9709.
(2)
By specificity it implies that how accurate the test is.
Compute the probability of negative result when the person is not a carrier as follows:
[tex]P(Y^{c}|X^{c})=1-P(Y|X^{c})=1-0.02=0.98[/tex]
Thus, the specificity of the test is 98%.
) Let y(1) = y0, y 0 (1) = v0. Solve the initial value problem. What is the longest interval on which the initial value problem is certain to have a unique twice differentiable solution?
QUESTION IS INCOMPLETE.
Nevertheless, I will explain how to find, without solving, the longest interval in which an initial value problem is certain to have a unique twice differentiable solution.
Step-by-step explanation:
Consider the Existence and uniqueness theorem:
Let p(t) , q(t) and r(t) be continuous on an interval a ≤ t ≤ b, then the differential equation given by:
y''+ p(t) y' +q(t) y = r(t) ;
y(t_0) = y_0, y'(t_0) = y'_0
has a unique solution defined for all t in the stated interval.
Example:
Consider the differential equation
ty'' + 9y = t
y(1) = y_0,
y'(1) = v_0
ty'' + 9y = t .................................(1)
First, write the differential equation (1) in the form:
y'' + p(t)y' + q(t)y = r(t) ..................(2)
by dividing (1) by t
So
y''+ (9/t)y = 1 ....................................(3)
Comparing (3) with (2)
p(t) = 0
q(t) = 9/t
r(t) = 1
For t = 0, p(t) and r(t) are continuous, but q(t) is undefined.
q(t) is continuous everywhere apart from the point t = 0.
We say (-∞, 0) and (0,∞) are the points where p(t), q(t) and r(t) are continuous.
But t = 1, which is contained in the initial conditions y(1) = y_0 and y'(1) = v_0 is found in (0,∞).
So, we conclude that this interval is the longest interval in which the initial value problem has a unique twice differentiable solution.
what are some questions to ask about a function when sketching its graph?
Let -? and ? denote two distinct objects, neither of which is in R. Define an addition and scalar multiplication on R U {?} U {-?} as you could guess from the notation. Specifically, the sum and product of two real numbers is as usual, and for t ? R define
t? = { -? if t<0 , 0 if t=0, ? if t>0
t(-?) = { ? if t<0, 0 if t=0, -? if t>0
t+? = ?+t=?, t+(-?)=(-?)+t=-?, ?+?=?, (-?)+(-?)=-?, ?+(-?)=0
IsR U {?} U {-?} a vector space over R? Please Explain.
Answer and Step-by-step explanation:
Obviously addition is closed for R including both infinities
As t+infty = infty +t = infty and t+(-infty) =(-infty)+t =(-infty)
and inverse are -infty for infty and infty for -infty
Hence a group under addition
------------------------------------------------------------
But regarding multiplication
we can say 1/infty = 2/infty =0
Hence infinity*0 is not unique.
Also infinity and -infty do not have multiplicative inverse as there is no t such that t*infty = 1
Hence cannot be a vector space.
Suppose a survey of 500 people age 18 to 34 indicated that 32.2% of them live with one or both of their parents. Calculate and interpret a confidence interval estimate for the true proportion of all people age 18 to 34 who live with one or both parents. Use a 94% confidence level. ____________________________________________________________________________________________________________________________ CHAPTER 8: FLOW CHART VIEW OF FORMULAS FOR CONFIDENCE INTERVAL
Answer:
[tex]0.322 - 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.283[/tex]
[tex]0.322 + 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.361[/tex]
The 94% confidence interval would be given by (0.283;0.361)
Step-by-step explanation:
Notation and definitions
[tex]n=500[/tex] random sample taken
[tex]\hat p=0.322[/tex] estimated proportion of people between 18 to 34 who live with their parents
[tex]p[/tex] true population proportion of people between 18 to 34 who live with their parents
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 94% of confidence, our significance level would be given by [tex]\alpha=1-0.94=0.06[/tex] and [tex]\alpha/2 =0.03[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.88, z_{1-\alpha/2}=1.88[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.322 - 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.283[/tex]
[tex]0.322 + 1.88\sqrt{\frac{0.322(1-0.322)}{500}}=0.361[/tex]
The 94% confidence interval would be given by (0.283;0.361)
A study of 420,026 cell phone users found that 136 of them developed cancer of the brain or nervous system. Prior to their study of cell phone use, the rate of such cancer was found to be 0.0251 % for those not using cell phones. Complete parts a and b. a. Use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. (Round to three decimal places as needed). b. Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell phones? Why or why not?
Answer:
a) The 90% confidence interval is:
[tex]0.0003237 \leq \pi \leq 0.0003239[/tex]
b) Yes. Because the proportion of cancer rate for the cell phone users population is bigger than 0.00324%, which is over the rate of non-cell phone users (0.0251%).
Step-by-step explanation:
a) We will construct the 90% confidence interval based on the information given by the sample taken in this study.
The sample proportion is:
[tex]p=\frac{X}{n}=\frac{136}{420,026} =0.000324[/tex]
The standard deviation is estimated as:
[tex]\sigma=\sqrt{\frac{p(1-p)}{n}}= \sqrt{\frac{0.000324*0.999676}{420,026}}=\sqrt{7.7\cdot 10^{-10}} = 0.000028[/tex]
As the sample size is big enough, we use the z-score. For a 90% CI, the value of z is z=1.645.
The margin of error of the CI is:
[tex]E=z\sigma/\sqrt{n}=1.645* 0.000028 /\sqrt{420,026}\\\\E= 0.000046 /648= 0.00000007[/tex]
The 90% CI is:
[tex]p-z\sigma/\sqrt{n}\leq \pi\leq p+z\sigma/\sqrt{n}\\\\0.000324- 0.00000007 \leq\pi\leq 0.000324+ 0.00000007\\\\ 0.0003237 \leq \pi \leq 0.0003239[/tex]
Final Answer:
a. The 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, rounded to three decimal places, is approximately: [0.028%, 0.037%]
b. The known rate for those not using cell phones is 0.0251%.
Explanation:
a. To construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, we need to know the number of successes (in this case, the number of people who developed cancer), the total number of trials (the total number of cell phone users), and use the z-score that corresponds to the desired confidence level (90%).
Given:
Number of cell phone users (n) = 420,026
Number of cell phone users who developed cancer (x) = 136
Confidence level (CL) = 90%
First, we calculate the sample proportion (p):
p = x / n = 136 / 420,026 ≈ 0.000324
For a 90% confidence interval, the z-score associated with the two-tailed confidence level is approximately 1.645 (you would find this value in a z-score table or by using statistical software).
Next, we calculate the standard error (SE) for the proportion:
SE = sqrt((p(1 - p)) / n)
SE = sqrt((0.000324(1 - 0.000324)) / 420,026)
SE ≈ sqrt((0.000324 * 0.999676) / 420,026)
SE ≈ sqrt(0.00000032443 / 420,026)
SE ≈ sqrt(7.72476e-10)
SE ≈ 2.779e-5
Now, we calculate the margin of error (ME):
ME = z * SE
ME = 1.645 * 2.779e-5
ME ≈ 4.570635e-5
Finally, we construct the confidence interval:
Lower bound (LB) = p - ME
LB ≈ 0.000324 - 4.570635e-5
LB ≈ 0.000278
Upper bound (UB) = p + ME
UB ≈ 0.000324 + 4.570635e-5
UB ≈ 0.000370
So the 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, rounded to three decimal places, is approximately:
[0.028%, 0.037%]
b. To determine if cell phone users appear to have a rate of cancer that is different from the rate among those not using cell phones, we compare the confidence interval we just calculated with the known rate for non-cell phone users.
The known rate for those not using cell phones is 0.0251%.
Since the entire 90% confidence interval of [0.028%, 0.037%] for cell phone users is above 0.0251%, this suggests that the rate of cancer among cell phone users might be higher than the rate among non-users. However, this does not provide enough evidence to definitively conclude that cell phone use causes a higher rate of cancer since other factors might be at play and causation cannot be determined from this data alone. It is also important to note that the confidence interval is statistically very close to the known rate for non-users, so the difference might not be practically significant. Additionally, further research with more rigorous control conditions would be necessary to establish a cause-and-effect relationship.
A family is selected at random from the city. Find the probability that the size of the family is between 2 and 5 inclusive. Round approximations to three decimal places.
Final answer:
To find the probability that the size of the family is between 2 and 5 inclusive, calculate the proportion of the given sample that falls within that range. In this case, the probability is approximately 0.792.
Explanation:
To find the probability that the size of the family is between 2 and 5 inclusive, we need to calculate the proportion of the given sample that falls within that range. In the provided sample of college math class, the family sizes are:
545443643355633274522232We can see that there are 19 family sizes falling between 2 and 5, inclusive. Therefore, the probability is:
Probability = Number of favorable outcomes / Total number of outcomes
Probability = 19 / 24 = 0.792
So, the probability that the size of the family is between 2 and 5 inclusive is approximately 0.792.
A market analyst is developing a regression model to predict monthly household expenditures on groceries as a function of family size, household income, and household neighborhood (urban, suburban, and rural). The response variable in this model is _____.
Answer:
Monthly household expenditures on groceries
Step-by-step explanation:
The response variable is the one for which measurements are desired and that depends on other variables.
In this case, family size, household income, and household neighborhood are independent variables, while the response variable is the monthly household expenditures on groceries.
In the game of Pick-A-Ball without replacement, there are 10 colored balls: 3 red, 4 white, and 3 blue. The balls have been placed into a small bucket, and the bucket has been shaken thoroughly. You will be asked to reach into the bucket without looking and select 2 balls. Because the bucket has been shaken thoroughly, you can assume that each individual ball is selected at random with equal likelihood of being chosen. Now, close your eyes! Reach into the bucket, and pick a ball. (Click "Pick-A-Ball!" to simulate reaching into the bucket and drawing your ball.) Pick-A-Ball! What is the probability of selecting the color of ball that you just selected? (Enter your answer in decimal format and round it to two decimal places.) Don’t put your first ball back into the bucket. Now, reach in (again, no peeking!), and pick your second ball. (Click "Pick-A-Ball!" to simulate reaching into the bucket and selecting your next ball.) Pick-A-Ball! What is the probability of selecting the color of ball that you just selected? (Enter your answer in decimal format and round it to two decimal places.)
Final answer:
The probability of selecting the color of the first ball is 3/10. The probability of selecting the same color for the second ball depends on the remaining number of balls of that color divided by the total remaining number of balls.
Explanation:
The probability of selecting the color of the first ball is determined by the number of balls of that color divided by the total number of balls. In this case, there are 3 red balls out of 10, so the probability is 3/10.
After selecting the first ball without replacement, the probability of selecting the same color for the second ball depends on the remaining number of balls of that color divided by the total remaining number of balls. If the first ball was red, there are 2 red balls left out of 9, resulting in a probability of 2/9.
The probability of selecting a ball of the same color as the first ball drawn, without replacement, is as follows:
For red:[tex]\boxed{0.22}[/tex]
For white:[tex]\boxed{0.33}[/tex]
For blue:[tex]\boxed{0.22}[/tex]
Let's think step by step.To solve this problem, we need to calculate the probability of selecting a ball of the same color as the first ball drawn, without replacement. We will consider the two events separately: first, the probability of selecting a ball of the same color as the first ball when there are 10 balls in total, and second, the probability of selecting a ball of the same color as the first ball when there are only 9 balls left in the bucket (since the first ball is not replaced).
First Ball Selection:
When the first ball is picked, there are 10 balls in total, with 3 red, 4 white, and 3 blue balls. The probability of picking a ball of any specific color is the number of balls of that color divided by the total number of balls.
For example, if the first ball drawn is red, the probability of drawing a red ball is:
[tex]\[ P(\text{first red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{10} \][/tex]
Similarly, if the first ball is white, the probability is:
[tex]\[ P(\text{first white}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{4}{10} \][/tex]
And if the first ball is blue, the probability is:
[tex]\[ P(\text{first blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{3}{10} \][/tex]
Second Ball Selection:
After the first ball is drawn and not replaced, there are 9 balls left in the bucket.
If the first ball drawn was red, there are now 2 red balls left out of 9 total balls. The probability of drawing another red ball is:
[tex]\[ P(\text{second red} | \text{first red}) = \frac{\text{Number of red balls left}}{\text{Total number of balls left}} = \frac{2}{9} \][/tex]
If the first ball was white, there are now 3 white balls left out of 9 total balls. The probability of drawing another white ball is:
[tex]\[ P(\text{second white} | \text{first white}) = \frac{\text{Number of white balls left}}{\text{Total number of balls left}} = \frac{3}{9} = \frac{1}{3} \][/tex]
If the first ball was blue, there are now 2 blue balls left out of 9 total balls. The probability of drawing another blue ball is:
[tex]\[ P(\text{second blue} | \text{first blue}) = \frac{\text{Number of blue balls left}}{\text{Total number of balls left}} = \frac{2}{9} \][/tex]
Now, let's calculate the probabilities for each color and round them to two decimal places:
For red:
[tex]\[ P(\text{second red} | \text{first red}) = \frac{2}{9} \approx 0.22 \][/tex]
For white:
[tex]\[ P(\text{second white} | \text{first white}) = \frac{1}{3} \approx 0.33 \][/tex]
For blue:
[tex]\[ P(\text{second blue} | \text{first blue}) = \frac{2}{9} \approx 0.22 \][/tex]
Final
The probability of selecting a ball of the same color as the first ball drawn, without replacement, is as follows:
For red:[tex]\boxed{0.22}[/tex]
For white:[tex]\boxed{0.33}[/tex]
For blue:[tex]\boxed{0.22}[/tex]
These probabilities are rounded to two decimal places.
The answer is: 0.22.
Consider two people being randomly selected. (For simplicity, ignore leap years.)
(a) What is the probability that two people have a birthday on the 9th of any month?
(b) What is the probability that two people have a birthday on the same day of the same month?
Answer:
[tex](a) = \frac{144}{133225} \\\\(b) = \frac{1}{365}[/tex]
Step-by-step explanation:
Part (a) the probability that two people have a birthday on the 9th of any month.
Neglecting leap year, there are 365 days in a year.
There are 12 possible 9th in months that make a year calendar.
If two people have birthday on 9th; P(1st person) and P(2nd person).
[tex]=\frac{12}{365} X\frac{12}{365} = \frac{144}{133225}[/tex]
Part (b) the probability that two people have a birthday on the same day of the same month
P(2 people selected have birthday on the same day of same month) + P(2 people selected not having birthday on same day of same month) = 1
P(2 people selected not having birthday on same day of same month):
[tex]= \frac{365}{365} X \frac{364}{365} =\frac{364}{365}[/tex]
P(2 people selected have birthday on the same day of same month) [tex]= 1-\frac{364}{365} \\\\= \frac{1}{365}[/tex]
Final answer:
The probability that two people have a birthday on the 9th of any month is 1/133,225. The probability that two people have a birthday on the same day of the same month is also 1/133,225.
Explanation:
To calculate the probability that two people have a birthday on the 9th of any month, we need to consider the number of possible outcomes and the number of favorable outcomes. There are 12 months in a year, so the number of possible outcomes is 12. The probability of each person having a birthday on the 9th is 1/365. Therefore, the probability that two people have a birthday on the 9th of any month is (1/365) x (1/365) = 1/133,225.
To calculate the probability that two people have a birthday on the same day of the same month, we need to consider the number of possible outcomes and the number of favorable outcomes. There are 12 months in a year and each month has 30 or 31 days. So the number of possible outcomes is 12 x 31 = 372. The probability of each person having a birthday on a specific day is 1/365. Therefore, the probability that two people have a birthday on the same day of the same month is (1/365) x (1/365) = 1/133,225.
A major television manufacturer has determined that its 44 inch screens have a mean service life that can be modeled by a normal distribution with a mean of 6 years and a standard deviation of one-half year (6 months). What is the probability that the service life of that product is between 5 and 7 years
Answer:
[tex]P(5<X<7)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{5-6}{0.5}<Z<\frac{7-6}{0.5})=P(-2<z<2)[/tex]
And we can find this probability with thie difference:
[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(6,0.5)[/tex]
Where [tex]\mu=6[/tex] and [tex]\sigma=0.5[/tex]
We are interested on this probability
[tex]P(5<X<7)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(5<X<7)=P(\frac{5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{7-\mu}{\sigma})=P(\frac{5-6}{0.5}<Z<\frac{7-6}{0.5})=P(-2<z<2)[/tex]
And we can find this probability with thie difference:
[tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-2<z<2)=P(z<2)-P(z<-2)=0.97725-0.02275=0.9545[/tex]
Final answer:
The probability that a 44 inch television screen will have a service life between 5 and 7 years, given a normal distribution with a mean of 6 years and a standard deviation of 0.5 years, is approximately 95.45%.
Explanation:
To calculate the probability that the service life of a 44 inch television screen is between 5 and 7 years, we use the properties of the normal distribution where the mean (μ) is 6 years and the standard deviation (σ) is 0.5 years. We need to find the z-scores for 5 and 7 years and then use the standard normal distribution table or a calculator to find the probability that the service life falls between these two z-scores.
First, calculate the z-score for 5 years:
z = (X - μ) / σ
z = (5 - 6) / 0.5
z = -2.0
Next, calculate the z-score for 7 years:
z = (7 - 6) / 0.5
z = 2.0
Once we have the z-scores, we look up the corresponding probabilities in the normal distribution table or use a calculator with normal distribution functions. The probability of z being between -2.0 and 2.0 in a standard normal distribution is approximately 0.9545.
Therefore, the probability that a television will last between 5 and 7 years is approximately 95.45%.
If you deposit money today in an account that pays 5.5% annual interest, how long will it take to double your money? Round your answer to two decimal places.
Final answer:
To calculate how long it will take to double an investment with a 5.5% annual interest rate, you can use the Rule of 72, which gives you approximately 13.09 years for doubling the investment.
Explanation:
To determine how long it will take to double your money with an annual interest rate of 5.5%, you can use the Rule of 72. The Rule of 72 is a simple formula to estimate the number of years required to double the invested money at a given annual fixed interest rate. You divide 72 by the annual interest rate.
In this situation, divide 72 by 5.5 to find out how many years it will take to double the investment:
72 / 5.5 = 13.09 years
Thus, it will take approximately 13.09 years to double your money at an annual interest rate of 5.5%, when the interest is compounded annually. This is a rough approximation and the actual number might vary slightly depending on the compounding method used by the account.
Agee Storage issued 35 million shares of its $1 per common stock at $16 per share several years ago. Last year, for the first time, Agee reacquired 1 million shares at $14 per share. If Agee now retires 1 million shares at $19 per share. By what amount will Agee's total paid-in capital decline?
Answer:
$18 Millions
Step-by-step explanation:
Decline in total paid in capital=Total value of reacquisition - Decline in retained earnings
$19-$1=$18 millions
Answer:
$18 000 000
Step-by-step explanation:
The first step is to calculate the par value of shares
$1*35000000=35000000
=$35000000
Issued shares of the the par
amount over par
=$16-$1=$15
Then paid capital in excess of par
$15*1000000
=$525000000
The first reacquire remember there were issued at $16
so for first reacquire
$16-$14=$2
$2*35000000=$70 000000
so Agee total paid-in capital will decline by
$15+$2+$1=$18 per share
What value of z divides the standard normal distribution so that half the area is on one side and half is on the other? Round your answer to two decimal places.
Answer:
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex] the deviation
We know that the z score is given by:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And by properties the value that separate the half area on one side and half is on the other is z=0, since we have this:
[tex] P(Z<0) =0.5[/tex]
[tex] P(Z>0)=0.5[/tex]
So then the correct answer for this case would be z =0.00
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/tex]
Where [tex]\mu[/tex] the mean and [tex]\sigma[/tex] the deviation
We know that the z score is given by:
[tex] z = \frac{X -\mu}{\sigma}[/tex]
And by properties the value that separate the half area on one side and half is on the other is z=0, since we have this:
[tex] P(Z<0) =0.5[/tex]
[tex] P(Z>0)=0.5[/tex]
So then the correct answer for this case would be z =0.00
The z-score that divides the standard normal distribution into two equal halves is 0.00 as the mean of this distribution is also 0. Z-scores are standardized values expressing how many standard deviations a value is above or below the mean and can be determined using a Z-Table of Standard Normal Distribution.
Explanation:The value of z that divides the standard normal distribution so that half the area is on one side and half is on the other is 0.00. The standard normal distribution is a symmetrical distribution where half of the values are less than the mean and half of the values are greater than the mean. In a standard normal distribution, the mean is 0. Hence, the z-score that will divide the distribution into half is also 0.
A z-score is a standardized value, expressing how many standard deviations a value is above or below the mean. These scores are particularly useful when comparing values from different data sets that have different means and standard deviations, such as exam scores from different classes or schools. They help us understand whether a particular score is common, or exceptionally high or low.
To find this value, you can use a Z-Table of Standard Normal Distribution, which shows the cumulative probability of a random variable from a standard normal distribution (with mean 0 and standard deviation 1), as being less than a certain value.
Learn more about Standard Normal Distribution here:https://brainly.com/question/30390016
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A product has a 4 week lead time. The standard deviation of demand for each of the week is given below. What is the standard deviation of demand over the lead time? (Answer to 2 decimal places) Week Standard deviation of demand 1 16 2 15 3 17 4 13
÷Answer:
Standard Deviation = 176.5
Step-by-step explanation:
To calculate the standard deviation, calculate the mean score for the 4 standard deviation scores:
mean, m = Σx ÷ n
where Σx represents summation of each value = 162 + 153 + 317 + 413
= 1045
n = number of samples to be considered = 4
mean, m = 1045 ÷4
= 261.25
To calculate the standard deviation, use the formula below
SD = [tex]\sqrt{\frac{Σ(x-m)}{n} ^{2} }[/tex]
where x = each value from the week lead time
m = mean = 261.25
n = the size = 4
The Standard deviation formula can be simplified further
when x = 162
[tex]\sqrt{\frac{(x1-m)}{n} ^{2} }[/tex] = 49.625
when x = 153
[tex]\sqrt{\frac{(x2-m)}{n} ^{2} }[/tex] = 23.125
when x = 317
[tex]\sqrt{\frac{(x3-m)}{n} ^{2} }[/tex]= 27.875
when x = 413
[tex]\sqrt{\frac{(x4-m)}{n} ^{2} }[/tex]= 75.875
Note that the above 4 equations can be lumped up into one giant equation by applying a big square root function instead of breaking it down
SD = 49.625 + 23.125 + 27.875 + 75.875
SD = 176.5
An article in Medicine and Science in Sports and Exercise "Maximal Leg-Strength Training Improves Cycling Economy in Previously Untrained Men," (2005, Vol. 37 pp. 131–1236) studied cycling performance before and after eight weeks of leg-strength training. Seven previously untrained males performed leg-strength training three days per week for eight weeks (with four sets of five replications at 85% of one repetition maximum). Peak power during incremental cycling increased to a mean of 315 watts with a standard deviation of 16 watts. Construct a 99% two-sided confidence interval for the mean peak power after training. Assume population is approximately normally distributed.
The 99% confidence interval for the mean peak power after training is approximately [tex](292.61, 337.39)[/tex] watts.
Identify the given data:
Mean peak power after training: [tex]\bar{x} = 315[/tex] watts
Standard deviation: [tex]s = 16[/tex] watts
Sample size: [tex]n = 7[/tex]
Confidence level: 99%
Find the critical value:
Since the sample size is small (n < 30) and the population standard deviation is not known, we use the t-distribution. For a 99% confidence interval with [tex](n-1) = 6[/tex] degrees of freedom, the critical value (t-value) can be found from the t-table. Using the t-table, [tex]t_{\frac{\alpha}{2},6} = 3.707[/tex].
Calculate the standard error (SE):
[tex]SE = \frac{s}{\sqrt{n}} = \frac{16}{\sqrt{7}} = \frac{16}{2.6458} \approx 6.04[/tex] watts
Compute the margin of error (ME):
[tex]ME = t_{\frac{\alpha}{2}} \times SE = 3.707 \times 6.04 \approx 22.39[/tex] watts
Construct the confidence interval:
Lower bound: [tex]\bar{x} - ME = 315 - 22.39 \approx 292.61[/tex] watts
Upper bound: [tex]\bar{x} + ME = 315 + 22.39 \approx 337.39[/tex] watts