Answer:
99.8% confidence:
[0.6833, 0.7656]
99% confidence:
0.6902 < p < 0.7588
Step-by-step explanation:
Lets call p the probability that a U.S adult pay its credit catrd in full each month. Lets call Y the random variable that counts the total of persons that paid their credit card in full from a random sample of 1118 adults. Y is a random variable with distribution Y ≈ Bi(1118,p) . The expected value of Y is μ = 1118*p and its variance is σ² = 1118*p(1-p). The proportion of adults that paid their credit card is a random variable, lets call it X, obtained from Y by dividing by 1118. The expected value is p and the variance is p(1-p)/1118. The Central Limit Theorem states that X can be approximated by a random variable with Normal Distribution, with mean μ = p, and standard deviation σ = √(p(1-p)/1118).
The standarization of X, W, is a random variable with distribution (approximately) N(0,1) obtained from X by substracting μ and dividing by σ. Thus
[tex] W = \frac{X - \mu}{\sigma} = \frac{X - p}{\sqrt{\frac{p(1-p)}{1118}}} [/tex]
If we want a 99.8% confidence interval, then we can find a value Z such that P(-Z < W < Z) = 0.998. If we do so, then P(W < Z) = 0.999, therefore, Ф(Z) = 0.999, were Ф is the cummuative distribution function of the standard normal distribution. The values of Ф can be found on the attached file. We can find that Ф(3.08) = 0.999, thus, Z = 3.08.
We have that P(-3.08 < W < 3.08) = 0.998, in other words
[tex] P(-3.08 < \frac{X-p}{\sqrt{\frac{p(1-p)}{1118}}} < 3.08) = 0.998 [/tex]
[tex] P(-3.08 * \sqrt{\frac{p(1-p)}{1118}} < X-p < 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]
[tex] P(-X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < -p < -X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]
Taking out the - sign from the -p and reversing the inequalities, we finally obtain
[tex] P(X -3.08 * \sqrt{\frac{p(1-p)}{1118}} < p < X + 3.08 * \sqrt{\frac{p(1-p)}{1118}}) = 0.998 [/tex]
As a conclusion, replacing p by its approximation X, a 99.8% confidence interval for p is
[tex] [X -3.08 * \sqrt{\frac{X(1-X)}{1118}}\, , \, X + 3.08 * \sqrt{\frac{X(1-X)}{1118}}] [/tex]
replacing X with the proportion of the sample, 810/1118 = 0.7245, we have that our confidence interval is
[tex] [0.7245 -3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}\, , \, 0.7245 + 3.08 * \sqrt{\frac{0.7245(1-0.7245)}{1118}}] [/tex]
By solving the fraction and multypling by 3.08, we have
[0.7245 - 0.0411 < p < 0.7245 + 0.0411]
FInally, the 99.8 % confidence interval for p is
[0.6833, 0.7656]
If p is 0.99 (99% confidence), then we would want Z such that Ф(Z) = 0.995, by looking at the table we have that Z is 2.57, therefore the 99% interval for p is
[tex] [X -2.57 * \sqrt{\frac{X(1-X)}{1118}}\, , \, X + 2.57 * \sqrt{\frac{X(1-X)}{1118}}] [/tex]
and, by replacing X by 0.7245 we have that the 99% confidence interval is
[0.6902, 0.7588]
thus, 0.6902 < p < 0.7588
I hope i could help you!
Final answer:
To construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full, we use the sample proportion, the z-value for the confidence level, and the sample size to calculate the margin of error and produce the interval.
Explanation:
To construct a 99.8% confidence interval for the proportion of U.S. adults who pay their credit cards in full each month from a survey result where 810 out of 1118 respondents paid their credit cards in full, we will use the formula for the confidence interval of a proportion: CI = p ± Z*sqrt((p(1-p))/n). Here, 'p' is the sample proportion, 'Z' is the z-value corresponding to the desired confidence level, and 'n' is the sample size.
The sample proportion (p) is 810/1118. To find the appropriate 'Z' value for a 99.8% confidence interval, we can refer to a standard normal distribution table or use a calculator to find that Z is approximately 3.08.
Step-by-step:
Find the sample proportion: p = 810/1118
Determine the 'Z' value for 99.8% confidence level: Z ≈ 3.08
Calculate the standard error (SE): SE = sqrt(p(1-p)/n)
Calculate the margin of error (ME): ME = Z * SE
Construct the confidence interval: (p - ME, p + ME)
After calculations, we can state that the confidence interval for the proportion of U.S. adults who pay their credit cards in full each month is between two values with three decimal places precision (the exact numbers would need to be calculated).
Suppose the force acting on a column that helps to support a building is a normally distributed random variable X with mean value 15.0 (kips) and standard deviation 1.25 (kips). Calculate the following probabilities: (1) (2 points) P(X ≥ 17.5). (2) (2 points) P(14 ≤ X ≤ 18).
Answer:
(1) 0.4207
(2) 0.7799
Step-by-step explanation:
Given,
Mean value,
[tex]\mu = 15.0[/tex]
Standard deviation,
[tex]\sigma = 1.25[/tex]
(1) P(X ≥ 17.5) = 1 - P( X ≤ 17.5)
[tex]= 1- P(\frac{x-\mu}{\sigma} \leq \frac{17.5-\mu}{\sigma})[/tex]
[tex]=1-P(z\leq \frac{17.5 - 15}{1.25})[/tex]
[tex]=1-P(z\leq \frac{2.5}{1.25})[/tex]
[tex]=1-P(z\leq 2)[/tex]
[tex]=1- 0.5793[/tex] ( By using z-score table )
= 0.4207
(2) P(14 ≤ X ≤ 18) = P(X ≤ 18) - P(X ≤ 14)
[tex]=P(z\leq \frac{18 - 15}{1.25}) - P(z\leq \frac{14 - 15}{1.25})[/tex]
[tex]=P(z\leq \frac{3}{1.25}) - P(z\leq -\frac{1}{1.25})[/tex]
[tex]=P(z\leq 2.4) - P(z\leq -0.8)[/tex]
= 0.9918 - 0.2119
= 0.7799
The probabiliity that the force is greater than 17.5 kips is 2.28% while the probabiliity that the force is between 14 and 18 kips is 20.37%
What is z score?
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
z = (raw score - mean) / standard deviation
Mean value 15.0 (kips) and standard deviation 1.25 (kips).
1) For 17.5:
z = (17.5 - 15) / 1.25 = 2
P(x > 17.4) = 1 - P(z < 2) = 1 - 0.9772 = 0.0228
2) For 14:
z = (14 - 15)/1.25 = 0.8
For 18:
z = (18 - 15)/1.25 = 2.4
P(14 < x < 18) = P(z < 2.4) - P(z < 0.8) = 0.9918 - 0.7881 = 0.2037
The probabiliity that the force is greater than 17.5 kips is 2.28% while the probabiliity that the force is between 14 and 18 kips is 20.37%
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Toys for boys and for girls are donated to a benefit event by two groups. The Lakeville Do-Gooders donated 5 toys for boys and 7 for girls. The Southside Champions Club donated 6 toys for boys and 5 for girls. The master of ceremonies pulls the first toy out of a bin and it’s a toy for a boy. Find the probability that it was donated by the Do-Gooders.
Answer:
0.4545 or 45%
Step-by-step explanation:
Do-Gooders toys for boys = 5
Southside toys for boys = 6
Given that the first toy pulled out of the bin is a toy for boys, to find the probability that this toy was donated by the Do-Gooders, the number of toys for girls donate shouldn't be considered since there's a zero-chance that the toy in question is for girls. Therefore:
[tex]Total\ sample \ space = 5+6 =11\\P(Do-Gooders)=\frac{5}{11} =0.4545[/tex]
There's a 0.4545 or 45% probability that it was donated by the Do-Gooders.
Calamity Jane goes to the bank to make a withdrawal, and is equally likely to find 0 or 1 customers ahead of her. The service time of the customer ahead, if present, is exponentially distributed with parameter λ. What is the CDF of Jane’s waiting time?
Answer:
[tex]F_{T}(t)=\frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})[/tex]
Step-by-step explanation:
The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
[tex]P(X=x)=\lambda e^{-\lambda x}[/tex]
We are interested on the cumulative distribution function (CDF) of Jane’s waiting time. Let T the random variable that represent the Jane’s waiting time, and t possible value for the random variable T.
For [tex]t\geq 0[/tex] the CDF for the waiting time is given by:
[tex]F_{T}(t) =P(T\leq t)[/tex]
For this case we can use the total rule of probability and the conidtional probability. We have two options, find 0 or 1 customer ahead of Jane, and for each option we have one possibility in order to find the CDF, like this:
[tex]F_{T}(t)=P(T\leq t|0 customers)P(0 customers)+ P(T\leq t|1 customers)P(1 customers)[/tex]
On this case the probability that we need to wait if we have o customers ahead is [tex]P(T\leq t|0 customers)=1[/tex] since if we don't have a customer ahead, so on this case Jane will not wait. And assuming that the probability of find 0 or 1 customers ahead of Jane is equal we have [tex]P(0 customers)=P(1 customers)=\frac{1}{2}[/tex]. And replacing we have:
[tex]F_{T}(t)=\frac{1}{2} +\frac{1}{2}\int_{0}^{t} \lambda e^{-\lambda \tau} d\tau = \frac{1}{2}+\frac{1}{2}[-e^{-\lambda t}+e^{-\lambda 0}]= \frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})[/tex]
Jane's waiting time scenario can be split into two cases, when there are no customers ahead of her, and when there is one customer ahead of her. For these two cases, the Cumulative Distribution Function (CDF) of Jane's waiting time can be calculated as [tex]0.5 + 0.5 \times (1 - e^{-\lambda w})[/tex]. This answer is derived using the law of total probability and the definition of the CDF for the exponential distribution.
Explanation:The question asks for the cumulative distribution function (CDF) of Jane’s waiting time at the bank. This is a probability question related to the exponential distribution and cumulative distribution function. The situation can be represented using two sub-situations: one, when there is no customer ahead (happens with 0.5 probability), Jane has no waiting time, and two, when there is a customer (happens with 0.5 probability) and Jane has to wait for an exponentially distributed time with parameter λ.
The cumulative distribution function (CDF) of an exponential random variable is given by [tex]F(x) = 1 - e^{-\lambda x}[/tex] when x ≥ 0. Using the law of total probability, the cumulative distribution function (CDF) of Jane's waiting time W can be obtained as [tex]P(W \leq w) = P(0 \text{ customer}) \cdot P(W \leq w | 0 \text{ customer}) + P(1 \text{ customer}) \cdot P(W \leq w | 1 \text{ customer}) = 0.5 \cdot 1 + 0.5 \cdot (1 - e^{-\lambda w})[/tex]when w ≥ 0. Or in other words, Jane's waiting time has a CDF of [tex]0.5 + 0.5 \times (1 - e^{-\lambda w})[/tex]
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Consider the differential equation ay′′ + by′ + cy = 0, where b2 − 4ac < 0 and the characteristic equation has complex roots λ ± iμ. Substitute the functions u(t) = eλt cosμt and v(t) = eλt sinμt for y in the differential equation and thereby confirm that they are solutions.
Answer:
Please see attachment
Step-by-step explanation:
The ASQ (American Society for Quality) regularly conducts a salary survey of its membership, primarily quality management professionals. Based on the most recently published mean and standard deviation, a quality control specialist calculated thez-score associated with his own salary and found it was minus2.50. What is his salary?A. 2.5 standard deviations below the average salary.B. 2 and a half times less than the average salary.C. 2.5 standard deviations above the average salary.D. 2 and a half times more than the average salary.
Answer:
A. 2.5 standard deviations below the average salary.
Step-by-step explanation:
Hello!
Usually, when you calculate a probability of an X₀ value from a normally distributed variable you need to standardize it to reach the proper value. In this case, you already have the Z-value and need to reverse the standardization to obtain the X₀ value.
Z= X₀ - μ
δ
δ*Z = X₀ - μ
X₀= (δ*Z) + μ
replace Z= -2.50
X₀= (δ*-2.50) + μ
X₀= μ - (δ*2.50)
You can replace the Z value any time. I've just choose to do it at the end because it's more confortable for me.
I hope it helps!
Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 104 and standard deviation 5
a. What is the probability that chloride concentration equals 105? b. What is the probability that chloride concentration is less than 105?
b. What is the probability that chloride concentration differs from the mean by more than 1 standard deviation?
Answer:
a) 0
b) 0.579 is the probability that chloride concentration is less than 105.
c) 0.32 is the probability that chloride concentration differs from the mean by more than 1 standard deviation.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 104
Standard Deviation, σ = 5
We are given that the distribution of blood chloride concentration is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) Since normal distribution s a continuous distribution, the probability for a particular value is zero. Therefore,
[tex]P(x =105) = 0[/tex]
b) P(chloride concentration is less than 105)
[tex]P(x < 105) = P(z < \displaystyle\frac{105-104}{5}) = P(z < 0.2)[/tex]
Calculating the value from the standard normal table we have,
[tex]P(z<0.2) = 0.579 = 57.9\%\\P( x < 105) = 57.9\%[/tex]
c) P(chloride concentration differs from the mean by more than 1 standard deviation)
Since it is a normal distribution, the Empirical rule shows that 68% falls within the first standard deviation, 95% within the first two standard deviations, and 99.7% within the first three standard deviations.
= 1 - P(chloride concentration within the mean by 1 standard deviation)
= 1 - 0.68 = 0.32
Final answer:
The probability of a single value is 0 in a normal distribution. The probability that the chloride concentration is less than 105 mmol/L can be found using the z-score and CDF of the normal distribution. The probability of the concentration being more than one standard deviation from the mean involves calculating the z-scores for one standard deviation away from the mean and using standard normal distribution probabilities.
Explanation:
The question involves the application of probability theory and statistics, specifically the properties of the normal distribution. Here are the answers to the student's questions:
(a) In a normal distribution, the probability of a continuous random variable taking on any single value is technically zero because there are an infinite number of potential values it could take. Therefore, the probability that the chloride concentration equals exactly 105 mmol/L is 0.
(b) To find the probability that the chloride concentration is less than 105 mmol/L, we use the cumulative distribution function (CDF) for a normal distribution. The z-score is calculated by subtracting the mean from the observed value and dividing by the standard deviation: z = (105 - 104) / 5 = 0.20. Using standard normal distribution tables or a calculator, we find the probability corresponding to a z-score of 0.20.
(b) The probability that the chloride concentration differs from the mean by more than 1 standard deviation involves finding the probability of being outside the interval [99, 109], since one standard deviation from the mean is 104 ± 5. We calculate the z-scores for 99 and 109, which are -1 and +1, respectively. Using the normal distribution, we find the probabilities associated with these z-scores and then subtract them from 1 to get the probability of being more than one standard deviation away from the mean.
Amachineworksforanexponentiallydistributedtimewithrateμandthenfails. A repair crew checks the machine at times distributed according to a Poisson process with rate λ; if the machine is found to have failed then it is immediately replaced. Find the expected time between replacements of machines.
Answer:[tex]\frac{1}{\mu }+\frac{1}{\lambda }[/tex]
Step-by-step explanation:
If a machine is replaced at some time t, then the expected time until next failure is [tex]\frac{1}{\mu }[/tex]
and the time between the checks is exponentially distributed with rate \lambda, the expected time until next failure is [tex]\frac{1}{\lambda }[/tex]
Because of memory less property of the exponential The answer is
[tex]\frac{1}{\mu }+\frac{1}{\lambda }[/tex]
Which statements are true about the solid, which has a length of
6 inches, a width of 3 inches, and a height of 2 inches?
Check all that apply.
The perimeter of one of its faces is 8 inches.
The perimeter of one of its faces is 10 inches.
The area of one of its faces is 6 inches.
The area of one of its faces is 12 in.2.
The volume of the solid is 36 in.3.
Answer:
Step-by-step explanation:
The solid is a cuboid with length of 6 inches, a width of 3 inches, and a height of 2 inches.
The volume is l× w×h = 6×3×2 = 36inches^3
Therefore, the following statements are true about the Solid
1) The volume of the solid is 36 in.3.
2) The perimeter of one of its faces is 10 inches.(2width + 2 height = 2×2 + 2×3 = 10 inches)
3) The area of one of its faces is 6 inches^2. (width × height = 3×2 = 6 inches)
4) The area of one of its faces is 12 in.2.(length × height = 6×2 = 12 inches^2)
Answer : The correct options are:
The perimeter of one of its faces is 10 inches.
The area of one of its faces is 6 inch².
The area of one of its faces is 12 inch².
The volume of the solid is 36 inches³.
Step-by-step explanation :
Formula used for perimeter of one face (rectangle) is:
Perimeter = 2(l+b)
Perimeter = 2(b+h)
Perimeter = 2(l+h)
Formula used for area of one face (rectangle) is:
Area = l × b
Area = b × h
Area = l × h
Formula used for volume of solid (cuboid) is:
volume of solid = l × b × h
Given:
l = length = 6 inches
b = width = 3 inches
h = height = 2 inches
Now we have to calculate the perimeter of one face.
Perimeter = 2(l+b)
Perimeter = 2(6+3) = 18 inch
Perimeter = 2(b+h)
Perimeter = 2(3+2) = 10 inch
Perimeter = 2(l+h)
Perimeter = 2(6+2) = 16 inch
Now we have to calculate the area of one face.
Area = l × b
Area = 6 × 3 = 18 inch²
Area = b × h
Area = 3 × 2= 6 inch²
Area = l × h
Area = 6 × 2 = 12 inch²
Now we have to calculate the volume of solid.
volume of solid = l × b × h
volume of solid = 6 × 3 × 2
volume of solid = 36 inches³
The volume of solid is, 36 inches³
The probability distribution for damage claims paid by the Newton Automobile Insurance Company on collision insurance follows. Payment ($) Probability 0 0.85 500 0.04 1,000 0.04 3,000 0.03 5,000 0.02 8,000 0.01 10,000 0.01
(a) Use the expected collision payment to determine the collision insurance premium that would enable the company to break even. If required, round your answers to two decimal places. If your answer is zero, enter "0". x f(x) xf(x) 0 500 1000 3000 5000 8000 10000 Total
(b) The insurance company charges an annual rate of $520 for the collision coverage. What is the expected value of the collision policy for a policyholder? (Hint: It is the expected payments from the company minus the cost of coverage.) If required, enter negative value as negative number. $ Why does the policyholder purchase a collision policy with this expected value? The input in the box below will not be graded, but may be reviewed and considered by your instructor.
Answer:
a) $430
b) -$90
Step-by-step explanation:
Given a random variable X with possible values
[tex]\large x_1,x_2,...x_n[/tex]
with respective probabilities of occurrence
[tex]\large P(x_1),P(x_2),...P(x_n)[/tex]
then the expected value E(X) of X is
[tex]\large x_1*P(x_1)+x_2*P(x_2)+...+x_n*P(x_n)[/tex]
a)
The expected collision payment would then be
0*0.85 + 500*0.04 + 1000*0.04 + 3000*0.03 + 5000*0.02 + 8000*0.01 + 10000*0.01 = $430
So the insurance premium that would enable the company to break even is $430
b)
The expected value of the collision policy for a policyholder is the expected payments from the company minus the cost of coverage:
$430 - $520 = -$90
Why does the policyholder purchase a collision policy with this expected value?
Because the policyholder does not know what the probability of having an accident is in her particular case.
Besides, it is better to have the policy and not need it than to need it and not have it.
The probability computed shows that the collision insurance premium that would enable the company to break even is $430.
How to calculate the probability?The collision insurance premium that would enable the company to break even will be calculated thus:
= (0 × 0.85) + (500 × 0.04) + (1000 × 0.04) + (3000 × 0.03) + (5000 × 0.02) + (8000 × 0.01 + (10000 × 0.01)
= $430
The expected value of the collision policy for a policyholder will be:
= $430 - 520
= -$90
In conclusion, the policyholder purchase a collision policy with this expected because he doesn't know the probability of having an accident.
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A researcher at a major hospital wishes to estimate the proportion of the adult population of the United States that has high blood pressure. How large must the sample be in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 5%?
The researcher would need a sample size of 271 individuals to be 90% confident that the sample proportion will not differ from the true proportion of adults
How to find the sample size ?
To determine the sample size needed to estimate the proportion of the adult population of the United States with high blood pressure with a 90% confidence level and a margin of error (precision) of 5%, you can use the formula for sample size for estimating proportions. The formula is:
[tex]\[n = \frac{Z^2 \cdot p(1-p)}{E^2}\][/tex]
E is the margin of error as a proportion (in this case, 5% or 0.05).
Plug in the values:
[tex]\(n = \frac{(1.645)^2 \cdot 0.5(1-0.5)}{(0.05)^2}\)[/tex]
[tex]\(n = \frac{2.706025 \cdot 0.25}{0.0025}\)[/tex]
[tex]\(n = \frac{0.67650625}{0.0025}\)[/tex]
n = 270.6025
= 271 people
Evaluate f(-12) given f(x) = -21x - 12
Answer:
The answer for the given function is f( -12 ) = 240.
Step-by-step explanation:
Given:
[tex]f(x)=-21x-12[/tex]
To Find :
F ( -12 ) = ?
Solution:
Function:
A function is like a machine that gives an output for a given input. A function has an independent variable which is called the input of the function. The output for a given input is called the dependent variable.For Example
f ( x ) = x + 3
x = independent variable
f ( x ) = Output for a given input
Here we have
[tex]f(x)=-21x-12\\Put\ x=-21\\\therefore f(-12) = -21\times -12 - 12\\\therefore f(-12) = 252 - 12\\\therefore f(-12) = 240[/tex]
Therefore f(-12) = 240
Answer for f(-12) = 240 for function f(x) = -21x -12
Step-by-step explanation:
Theory: f(x) is the" Function " and defined as a relation between the set of input values and a respective set of output values. set of input values is term as " Domain " and a respective set of output values is term as " Range "
In given question, f(x) = -21x -12
Note: Question didn't mention the domain so that, we can take it as Real numbers.
To find value of f( -12 )
Replacing value of x with (-12) in f(x)
we get,
f(-12) = -21(-12) -12
f(-12) = 240.
A study was conducted to determine the proportion of American teenagers between 13 and 17 who smoke. Previous surveys showed that 15% percent of all teenagers smoke. A Gallup survey interviewed a nationally representative sample of 785 teenagers aged 13 to 17. Seventy-one teenagers in the survey acknowledged having smoked at least once in the past week. Does the study provide adequate evidence to conclude that the percentage of teenagers who smoke is now different than 15%
1. Are the requirements for both a confidence intervale and a hypothesis test met?
2. Determine the appropriate 95% confidence intervale for the true proportion of teenage smokers. Input your answers for the margin of error, lower bound and upper bound.
Final answer:
The requirements for constructing a 95% confidence interval and performing a hypothesis test are met. The 95% confidence interval for the true proportion of teenage smokers is approximately 6.99% to 11.1%, indicating that the proportion may have changed from the previous 15%.
Explanation:
To ascertain if there is sufficient evidence to conclude that the percentage of teenagers who smoke has changed from 15%, we should conduct both a confidence interval analysis and a hypothesis test.
Requirements for Confidence Interval and Hypothesis Testing
Both require a random sample and a sufficiently large sample size to warrant using a normal approximation for the distribution of sample proportions. For a 95% confidence interval and hypothesis testing regarding proportions, the sample should satisfy the conditions np≥15 and n(1-p)≥15 where n is the sample size and p is the hypothesized proportion.
In this case, with a sample size of 785 and the previous proportion of 0.15, the conditions 785(0.15)≥ 15 and 785(0.85)≥ 15 are both met, satisfying the criteria to proceed with the confidence interval and hypothesis test.
95% Confidence Interval for True Proportion
To calculate the 95% confidence interval for the proportion of teenage smokers, use the formula:
Proportion = x/n
Margin of Error (ME) = z*(√[p(1-p)/n])
Lower Bound = Proportion - ME
Upper Bound = Proportion + ME
Where:
x is the number of successes in the sample (71 teenagers acknowledged smoking)
n is the sample size (785)
z is the z-score corresponding to the desired confidence level (approximately 1.96 for 95%)
Substituting in the values we get:
Proportion = 71/785 = 0.0904
ME = 1.96*√[0.0904(1-0.0904)/785] ≈ 0.0205
Lower Bound = 0.0904 - 0.0205 ≈ 0.0699
Upper Bound = 0.0904 + 0.0205 ≈ 0.111
Based on this confidence interval, we have evidence that the proportion of smokers may differ from the previous proportion of 15% as it does not include 0.15 in the interval.
At Maple Grove Middle School, the record for the 100 m dash is 12.45 seconds. The
table show the length of time it took for the four fastest students to complete the 100-m
dash this year relative to the school record.
Based on the results in the table, which statement is true?
Student
Pablo
Time(s)
-0.12
Cindy
0.34
Eddie
-0.15
Sam
0.21
A. Pablo and Eddie both broke the school record this year.
B. The order of the runners from fastest to slowest was Cindy, Sam, Pablo, and Eddie.
C. Sam ran the 100-m dash faster than Cindy.
D. Eddie was the fastest runner in the race.
Answer:
Step-by-step explanation:
the record for the 100 m dash is 12.45 seconds. Since the length of times in the table is relative to the record time, 12.45 seconds,
Pablo's time - 12.45 = -0.12
Pablo's time = -0.12 + 12.45 = 12.33 seconds.
Cindy's time - 12.45 = 0.34
Cindy's time = 0.34 + 12.45 = 12.79 seconds
Eddie's time - 12.45 = -0.15
Eddie's time = -0.15 + 12.45 = 12.3 seconds.
Sam's time - 12.45 = 0.21
Sam's time = 0.21 + 12.45 = 12.66 seconds.
The following statements are true
A. Pablo and Eddie both broke the school record this year.
C. Sam ran the 100-m dash faster than Cindy.
D. Eddie was the fastest runner in the race.
Answer:
A. Pablo and Eddie both broke the school record this year.
Step-by-step explanation:
Current record is 12.45seconds
Pablo record is -0.12 seconds which is 12.45 – 0.12 = 12.33 seconds
Eddie’s record is -0.15 seconds which is 12.45 – 0.15 = 12.30 seconds.
There are 15 balls in an urn. 7 of the balls are red, and 8 of the balls are blue. Drawing a ball with replacement means that you take a ball out, note its color, and then return it. Drawing withoutreplacement means that you take a ball out and do not put it back.
(a) What is the probability that if you draw one ball, it will be blue?(b) If you draw two balls without replacement, what is the probability that you draw a red and then a blue?(c) If you draw two balls with replacement, what is the probability that you draw a red and then a blue?(d) If you draw three balls without replacement, what is the probability that you draw at least two red balls?(e) If you draw three balls with replacement, what is the probability that you draw at least two red balls?(f) We play a game where I draw a ball first without replacement, and then you draw one. I draw a blue ball, and then you draw a red ball. Are these events independent?
Answer:
Dependent
Step-by-step explanation:
Given that there are 15 balls in an urn. 7 of the balls are red, and 8 of the balls are blue
when we do with replacement each time probability for red or blue remains the same.
a) the probability that if you draw one ball, it will be blue=[tex]\frac{8}{15}[/tex]
(b) If you draw two balls without replacement, the probability that you draw a red and then a blue
=[tex]\frac{7}{15} *\frac{8}{14} \\=\frac{4}{15}[/tex]
(c) If you draw two balls with replacement, the probability that you draw a red and then a blue
=[tex]\frac{7}{15} *\frac{8}{15} \\=\frac{56}{225}[/tex]
(d) If you draw three balls without replacement, the probability that you draw at least two red balls
=P(2 red, 1 blue)+P(3 blue)
=[tex]\frac{7C2*8C1}{15C3} +\frac{7C3}{15C3} \\=\frac{203}{455} =\frac{29}{65}[/tex]
(e) If you draw three balls with replacement, the probability that you draw at least two red balls
= P(X≥2) where x is binomial with n=3 and p = 7/15
= 0.4506
(f) We play a game where I draw a ball first without replacement, and then you draw one. I draw a blue ball, and then you draw a red ball.
No when first ball is drawn without replacement the next probability for blue and red would be changed. So dependent.
The probability of various scenarios involving drawing balls from an urn is calculated, including with and without replacement. The events of drawing a blue ball and a red ball after each other are examined in terms of their independence.
Explanation:In order to answer the given question, we need to calculate the probability for each scenario:
(a) The probability of drawing a blue ball with replacement is
8/15
.
(b) The probability of drawing a red and then a blue ball without replacement is
(7/15) * (8/14)
.
(c) The probability of drawing a red and then a blue ball with replacement is
(7/15) * (8/15)
.
(d) The probability of drawing at least two red balls without replacement can be calculated by finding the probability of drawing two red balls and the probability of drawing three red balls, and then adding them together:
(7/15) * (6/14) + (7/15) * (6/14) * (5/13)
.
(e) The probability of drawing at least two red balls with replacement can be calculated by finding the probability of drawing two red balls and the probability of drawing three red balls, and then adding them together:
(7/15) * (7/15) + (7/15) * (7/15) * (7/15)
(f) No, the events of you drawing a blue ball and the student drawing a red ball after that are
not independent
because drawing the blue ball affects the probability of drawing a red ball for the student.
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A publisher reports that 49% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 200 found that 42% of the readers owned a personal computer. Determine the P-value of the test statistic.
Answer:
0.0239
Step-by-step explanation:
A publisher reports that 49% of their readers own a personal computer.
Claim : . A marketing executive wants to test the claim that the percentage is actually different from the reported percentage.
[tex]H_0:p = 0.49\\H_a:p\neq 0.49[/tex]
A random sample of 200 found that 42% of the readers owned a personal computer.
No. of people owned a personal computer = [tex]42\% \times 200[/tex]
= [tex]\frac{42}{100} \times 200[/tex]
= [tex]84[/tex]
[tex]x = 84\\n = 200[/tex]
We will use one sample proportion test
[tex]\widehat{p}=\frac{x}{n}[/tex]
[tex]\widehat{p}=\frac{84}{200}[/tex]
[tex]\widehat{p}=0.42[/tex]
Formula of test statistic =[tex]\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
=[tex]-1.98[/tex]
Now refer the p value from the z table
p value =0.0239
Hence The p value of test statistic is 0.0239
A random sample of 85 group leaders, supervisors, and similar personnel at General Motors revealed that, on average, they spent 6.5 years in a particular job before being promoted. The standard deviation of the sample was 1.7 years. Construct a 95% confidence interval.
Final answer:
To construct the 95% confidence interval for the average time spent in a particular job at General Motors, we calculated the margin of error and then added and subtracted it from the sample mean, resulting in a confidence interval of approximately (6.1331, 6.8669).
Explanation:
To construct a 95% confidence interval for the average time spent in a particular job at General Motors, we utilize the sample mean, standard deviation, and the sample size to compute the margin of error and the confidence interval. Given that the sample mean is 6.5 years and the standard deviation is 1.7 years with a sample size of 85, we can calculate the confidence interval as follows:
Firstly, determine the critical value (z*) for a 95% confidence level. Since the population standard deviation is unknown and the sample size is less than 30, we use the t-distribution. For a 95% confidence level and 84 degrees of freedom (n-1), the critical value (using a t-table or software) is approximately 1.989.Compute the standard error of the mean by dividing the standard deviation by the square root of the sample size. Standard error (SE) = 1.7 / sqrt(85) ≈ 0.1845.Calculate the margin of error (MOE) by multiplying the critical value with the standard error. MOE = 1.989 * 0.1845 ≈ 0.3669.Finally, construct the confidence interval by adding and subtracting the margin of error from the sample mean. Lower limit = 6.5 - 0.3669 ≈ 6.1331 and upper limit = 6.5 + 0.3669 ≈ 6.8669. Therefore, the 95% confidence interval is (6.1331, 6.8669).This confidence interval suggests that we are 95% confident that the true average time spent in the job before promotion at General Motors lies between approximately 6.1 and 6.9 years.
A group of friends has gotten very competitive with their board game nights. They have found that overall, they each have won an average of 18 games, with a population standard deviation of 6 games. If a sample of only 2 friends is selected at random from the group, select the expected mean and the standard deviation of the sampling distribution from the options below. Remember to round to the nearest whole number.
Answer: [tex]\mu_x=18\text{ hours}[/tex]
[tex]\sigma_x=4\text{ hours}[/tex]
Step-by-step explanation:
We know that mean and standard deviation of sampling distribution is given by :-
[tex]\mu_x=\mu[/tex]
[tex]\sigma_x=\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\mu[/tex] = population mean
[tex]\sigma[/tex] =Population standard deviation.
n= sample size .
In the given situation, we have
[tex]\mu=18\text{ hours}[/tex]
[tex]\sigma=6\text{ hours}[/tex]
n= 2
Then, the expected mean and the standard deviation of the sampling distribution will be :_
[tex]\mu_x=\mu=18\text{ hours}[/tex]
[tex]\sigma_x=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{2}}=4.24264068712\approx4[/tex] [Rounded to the nearest whole number]
Hence, the the expected mean and the standard deviation of the sampling distribution :
[tex]\mu_x=18\text{ hours}[/tex]
[tex]\sigma_x=4\text{ hours}[/tex]
Marie is saving money for home repairs. To date, she has saved $1,329. She needs at least $1,569 for the repairs. She plans to set aside $40 per week to add to her current savings. If this situation is modeled by the inequality below, how many more weeks, x, does she need to continue saving in order to have enough money for the repairs? $1,329 + $40x > $1,569
Answer:
x > 6
Step-by-step explanation:
Subtract $1329 from both sides, then divide by $40.
$1,329 + $40x > $1,569
$40x > $240
x > 6
Marie needs to save for more than 6 more weeks.
_____
Comment on the answer
The given inequality is written using the > symbol. Marie would have exactly enough to cover the projected cost after 6 weeks, but it is appropriate from a finance point of view for her to save more than the required amount. It will take more than 6 weeks for her to do that.
The two intervals (113.4, 114.6) and (113.1, 114.9) are confidence intervals for μ = mean resonance frequency (in hertz) for all tennis rackets of a certain type. The two intervals were calculated using the same sample data. (a) What is the value of the sample mean (in hertz) resonance frequency?
Answer:
114
Step-by-step explanation:
Given that two intervals are confidence intervals for μ = mean resonance frequency (in hertz) for all tennis rackets of a certain type.
They are
i) [tex](113.4, 114.6)[/tex] and
ii) [tex](113.1, 114.9)[/tex]
We know that confidence interval has centre as the mean value
Hence we find the average of lower and upper bounds to find out the sample mean.
Sample mean in i) [tex]\frac{113.4+114.6}{2} =114[/tex]
Sample mean in ii) [tex]\frac{113.1+114.9}{2} =114[/tex]
Thus we find that value of sample mean =114
The value of the sample mean resonance frequency is 114 hertz.
Explanation:The value of the sample mean (in hertz) resonance frequency can be found by taking the average of the lower and upper bounds of the confidence intervals. For the first interval (113.4, 114.6), the sample mean is (113.4 + 114.6) / 2 = 114 hertz. For the second interval (113.1, 114.9), the sample mean is (113.1 + 114.9) / 2 = 114 hertz. So the value of the sample mean resonance frequency is 114 hertz.
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Suppose a sociologist is interested in testing if the average age at puberty of girls living is Thailand is different from the average age of puberty of girls living in Japan. To test her theory, she selects a random sample of 50 young Thai women and 50 young Japanese women and determines that the average age at puberty of the women in Japan is 1.7 years younger than the average age at puberty of the Thai women. Which of the following would change if instead the average age at puberty for Japanese women is 0.8 years younger?a) The p-value would increase.b) The p-value would decrease.c) There would be no difference because Japanese women still went through puberty at a younger age.d) The sample standard deviation would be larger if the difference were less.e) Exactly 2 of the above are correct.
Answer:
Option b is right
Step-by-step explanation:
Given that a sociologist is interested in testing if the average age at puberty of girls living is Thailand is different from the average age of puberty of girls living in Japan. To test her theory, she selects a random sample of 50 young Thai women and 50 young Japanese women and determines that the average age at puberty of the women in Japan is 1.7 years younger than the average age at puberty of the Thai women
If average of Japanese women is decreased by 0.8 years, we would get mean difference = [tex]1.7+0.8=2.5[/tex]
Since mean difference increases test statistic would increase thus reducing p value, i.e. chances of rejecting H0 higher.
The sample std devition would not change because of this.
The weights of bags of peas are normally distributed with a mean of 12.08 ounces and a standard deviation of 1.03 ounces. Bags in the upper 5% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?
A. 13.88 ouncesB. 12.18 ouncesC. 10.28 ouncesD. 12.03 ounces.
Answer: A bag can weigh 13.77 ounces and not need to be repacked.
Step-by-step explanation:
Since we have given that
Mean = 12.08 ounces
Standard deviation = 1.03 ounces
Bags in the upper 5% are too heavy and must be repackaged.
Using the standard normal table.
z = 1.645
So, [tex]X=\mu+z\sigma\\\\X=12.08+1.645\times 1.03\\\\X=13.77\ ounces[/tex]
Hence, A bag can weigh 13.77 ounces and not need to be repacked.
A surveying team set up some equipment 104104 ft from the base of a tree in order to sight the top of the tree. From ground level, they measure the angle of elevation to be 27.5∘27.5∘. If the calculated height needs to be accurate to within 1.51.5%, what is the allowed error in the angle measurement? (Assume the 104104 ft measurement is 100% accurate.)
In order to estimate the allowed angle error when measuring the height of a tree, you should first use the given distance and the angle of elevation to calculate the tree's height. Then calculate the potential height range considering the 1.5% error, and figure out the corresponding angles with these heights. The angle error can be estimated by the difference between these two angles.
Explanation:To solve this problem, one can use the principles of trigonometry. Specifically, the tangent function in trigonometry (tan θ = height / distance) can be used to estimate the height of the tree. Here, the distance from the tree (104 feet) and the angle of elevation (27.5°) are given.
We must first calculate the tree height using tan θ = height / distance. After that, we find that the tree is approximately 58.4 ft tall.
In this case, a 1.5% error in the height measurement would be 0.015 * 58.4 ft = 0.876 feet. Using the equation θ = arctan (height / distance), we can figure out the angle error. Enter the heights 58.4 + 0.876 ft and 58.4 - 0.876 into the equation and use the distance 104104 ft to calculate the corresponding angles. The allowed error for the angle follows as the difference of these two angles.
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To ensure the tree height is within 1.5% accuracy, we calculated that the allowable error in the angle measurement is ±0.22 degrees. This was determined by converting the tolerable height range back to angles using inverse tangent. The primary method used is trigonometry.
To determine the allowed error in the angle measurement for a tree with a height accurate to within 1.5%, we follow these steps:
Let's denote the height of the tree as h, the measured distance from the tree as d = 104 ft, and the angle of elevation as θ = 27.5∘.Using the tangent function: tan(θ) = h / d, we find the height h using trigonometry: h = d × tan(θ)Given that d = 104 ft and tan(27.5°) ≈ 0.5197, we can calculate: h ≈ 104 × 0.5197 ≈ 54.05 ft
To find the acceptable range of h within 1.5% accuracy: h ± 0.015h:So, 54.05 ft ± 0.015 × 54.05 ft ≈ 54.05 ft ± 0.81 ft
Tree Height thus ranges from approximately 53.24 ft to 54.86 ft.
Next, we need to determine the allowed error in the angle θ:We convert the bounds of height back to angles using inverse tangent:
For minimum height: tan⁻¹(53.24/104) ≈ 27.28°For maximum height: tan⁻¹(54.86/104) ≈ 27.72°The allowed error in the angle measurement is the difference from the original angle:
± |27.5° - 27.28°| ≈ ±0.22° and ± |27.72° - 27.5°| ≈ ±0.22°
Therefore, the allowable error in the angle measurement to ensure a height accuracy within 1.5% is ±0.22 degrees.
People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) people per hour, where t is measured in hours. The functions f and g are nonnegative and differentiable for all times t. Which of the following inequalities indicates that the rate of change of the number of people in the building is increasing at time t? o f (t) > 0 f (t)-9(t) > 0 o f (t)>0 of'(t)-g'(t) > 0
Answer:
The correct option is D) [tex]f'(t)-g'(t) > 0[/tex]
Step-by-step explanation:
Consider the provided information.
People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) people per hour,
The change of number of people in building is:
[tex]h(x)=f(t)-g(t)[/tex]
Where f(t) is people entering in building and g(t) is exiting from the building.
It is given that "The functions f and g are non negative and differentiable for all times t."
We need to find the the rate of change of the number of people in the building.
Differentiate the above function with respect to time:
[tex]h'(x)=\frac{d}{dt}[f(t)-g(t)][/tex]
[tex]h'(x)=f'(t)-g'(t)[/tex]
It is given that the rate of change of the number of people in the building is increasing at time t.
That means [tex]h'(x)>0[/tex]
Therefore, [tex]f'(t)-g'(t)>0[/tex]
Hence, the correct option is D) [tex]f'(t)-g'(t) > 0[/tex]
The rate of change of the number of people in the building is increasing at time t and with the help of this statement the correct option is D).
Given :
People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) people per hour, where t is measured in hours. The functions f and g are nonnegative and differentiable for all times t.The change of the number of people in the building is given by:
[tex]h(x) = f(t) - g(t)[/tex]
To determine the inequality, differentiate the above equation with respect to time.
[tex]h'(x)=f'(t)-g'(t)[/tex]
Now, it is given that the rate of change of the number of people in the building is increasing at time t. That means:
[tex]h'(x)>0[/tex]
[tex]f'(t)-g'(t)>0[/tex]
Therefore, the correct option is D).
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A sample of a given size is used to construct a 95% confidence interval for the population mean with a known population standard deviation. If a bigger sample had been used instead, then the 95% confidence interval would have been ________ and the probability of making an error would have been ________.
Answer:
narrower, reduced
Step-by-step explanation:
Given that a sample of a given size is used to construct a 95% confidence interval for the population mean with a known population standard deviation
we have confidence interval is margin of error on either side of mean.
Margin of error= Critical value * std error
Std error in turn is inversely proportinal to square root of n
For same confidence level, if sample size is increased we get a less margin of error and hence a narrower confidence interval.
when a confidence interval is narrowed, probability of making a type I error would be reduced.
Hence we get
If a bigger sample had been used instead, then the 95% confidence interval would have been __narrower______ and the probability of making an error would have been ___reduced_____.
A larger sample size would result in a narrower 95% confidence interval and a decreased probability of making an error. This is due to less variability and increased precision with increased sample size.
Explanation:If a larger sample size had been used to construct a 95% confidence interval for the population mean, the interval would have been narrower. This is because larger sample sizes typically result in less variability. In fact, as the sample size increases, the estimate of the population mean becomes more accurate, and thus the confidence interval around that estimate becomes tighter, containing less of the probability distribution.
The probability of making an error, or in other words, the margin of error, decreases with an increase in sample size. This reduction in the level of uncertainty is due to reduced variability seen with larger samples. By using a larger sample, the chances of the confidence interval not containing the population mean, thus making an error, decrease.
To give an example, if you initially have a 95% confidence interval of (67.02, 68.98) from a smaller sample and you then generate a bigger sample, you could get a confidence interval like (67.18, 68.82). This interval is narrower, representing the reduced variability and increased precision from the larger sample size.
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According to a survey, 67% of murders committed last year were cleared by arrest or exceptional means. Fifty murders committed last year are randomly selected, and the number cleared by arrest or exceptional means is recorded.
a) Find the probability that exactly 41 murders were cleared.
b)The probability that between 36 and 38 of the murders, inclusive were cleared is?
c) Would it be unusual if fewer than 20 of the murders were cleared? why or why not?
Answer:
a) P(X=41)= 0.0086 = 0.86%
b) P(36≤X≤38)== 0.1215 = 12.15%
c) would be unusual that fewer that 20 murders were cleared because P(X≤20) = 0.000081 =0.0081%
Step-by-step explanation:
if the random variable X= number of cleared murders , then X follows a binomial distribution:
P(X=x)= n!/[(n-x)!*x!]*p^x*(1-p)^(n-x)
where P(X=x)= probability of x cleared murders, n= number of murders selected=50 , p= probability for a murder to be cleared (67%)
therefore
a) P(X=41)=50!/[(50-41)!*41!]*0.67^41*0.33^(50-41)= 0.0086
b) P(36≤X≤38)= P(X≤38) - P(X≤36) = 0.9371- 0.8156 = 0.1215
c) P(X≤20) = 0.000081
therefore would be unusual that fewer that 20 murders were cleared
Consider a triangle ABC for which ∠A=100∘,a=34,b=13. If such a triangle can not exist, then write NONE in each answer box. If there could be more than one such triangle, then enter dimensions for the one with the smallest value for side c. Finally, if there is a unique triangle ABC, then enter its dimensions.
B is _______ degrees;
∠C is_________ degrees;
c= _________.
Answer:
B is 22.12 degrees; ∠C is 57.88°; c=29.24
Step-by-step explanation:
So, first, it's important to draw a diagram of the triangle the problem is talking about (see attached picture).
Once the triangle has been drawn, we can visualize it better and determine what to do. So first, we are going to find what the value of angle B is by using law of sines:
[tex]\frac{sin B}{b}=\frac{sin A}{a}[/tex]
which can be solved for angle B:
[tex]sin B=b\frac{sin A}{a}[/tex]
[tex] B= sin^{-1}(b\frac{sin A}{a})[/tex]
and substitute the values we already know:
[tex] B= sin^{-1}(13\frac{sin 100 ^{o}}{34})[/tex]
which yields:
B=22.12°
Once we know what the angle of B is, we can now find the value of angle C by using the fact that the sum of the angles of any triangle is equal to 180°. So:
A+B+C=180°
When solving for C we get:
C=180°-A-B
C=180°-22.12°-|00°=57.88°
So once we know what angle C is, we can go ahead and find the length of side c by using the law of sines again:
[tex]\frac{c}{sin C}=\frac{a}{sin A}[/tex]
and solve for c:
[tex]c=sin C \frac{a}{sin A}[/tex]
so we can now substitute for the values we already know:
[tex]c=sin(57.88^{o})\frac{34}{sin(100^{o})}[/tex]
which yields:
c=29.24
Which function is not linear?
Answer: y=x^2+2
Step-by-step explanation:
y=x^2+2 (answer choice 3) is parabolic and not linear
that is the answer
A CPU manufacturer is interested in studying the relationship between clock speed and the operating temperature that results at that clock speed for a particular CPU model. Let x be the clock speed in MHz and let Y be the temperature in ^{\circ}C . The following data was collected:i xi yi1 350 31.42 360 35.63 370 41.84 380 51.05 390 56.86 400 62.87 410 67.4a) Find the equation of the regression line.b) Estimate the temperature for clock speed x = 430 MHz.c) Find the 95% confidence interval for \beta .d) Compute the coefficient of determination R^{2} ?. Is this a high quality fit?
Final answer:
To find the equation of the regression line, calculate the slope and y-intercept using the given formulas with the given data points.
Explanation:
To find the equation of the regression line, we need to calculate the slope and the y-intercept. The slope, b, can be found using the formula:
b = (nΣ(xy) - ΣxΣy) / (nΣ(x^2) - (Σx)^2)
where n is the number of data points, Σ represents summation, x and y are the given data points. The y-intercept, a, can be calculated using the formula:
a = (Σy - bΣx) / n
Substituting the values from the given data into these formulas will give you the equation of the regression line.
The amount of cola in a 355 ml bottle from a certain company is a random variable with a mean of 355 ml and a standard deviation of 2 ml. For a sample of size 32, perform the following calculations. a. Find an approximate probability that the sample mean is less than 354.8 ml. b. Suppose the amount of cola is distributed as N(355, 4). Find an approximate probability that 10 of the bottles in the sample contain less than 354.8 ml of cola.
The probability of the sample mean being less than 354.8 ml can be calculated using the z-score and the standard normal distribution. For the second part, if the standard deviation is indeed 4, it's necessary to calculate the probability for a single bottle first, then apply the binomial distribution to find the probability for 10 out of 32 bottles.
Explanation:The calculation of the probability that the sample mean is less than 354.8 ml when the population mean is 355 ml and the standard deviation is 2 ml for a sample size of 32 uses the standard normal distribution. Applying the Central Limit Theorem, we can find the z-score and use it to calculate the probability.
For part b, assuming the normal distribution N(355, 4), to find the probability that 10 out of 32 bottles contain less than 354.8 ml, we would use the binomial distribution with the probability calculated in part a. However, the second part of question b appears to have a typo with the standard deviation. If the distribution is N(355, 4), we would need to first find the probability of a single bottle having less than 354.8 ml, then use the binomial formula to find the probability of 10 bottles having less than that amount.
A new species of sea crab has been discovered, and an experiment conducted to determine whether or not the animal can regulate its temperature. That the animal can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability. Ten of these sea crabs were exposed to ambient temperatures of 24 degrees Celsius. Their body temperatures were measured with the results below:24.33, 24.61, 24.67, 24.64, 24.42, 24.97, 25.23, 24.73, 24.90, 24.44For purposes of this example, assume that it is reasonable to regard these 10 crabs as a random sample from the population of all crabs of this species.a)- Calculate a point estimate of the population mean.b)-Construct and interpret a 99% confidence interval for m.c)-Does it appear from these data that the crabs are able to regulate their body temperature? Provide statistical justification for your response.
Answer:
a) [tex]\hat \mu =\bar X=24.694[/tex]
b) The 99% confidence interval would be given by (24.409;24.979)
c) Since the 99% confidence interval not contains the exposed to ambient temperatures of 24 degrees Celsius, and since the animals "that can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability", on this case we can conclude that the sea crabs are able to regulate their body temperature.
Step-by-step explanation:
1) Previous concepts
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".
Data: 24.33, 24.61, 24.67, 24.64, 24.42, 24.97, 25.23, 24.73, 24.90, 24.44
[tex]\bar X[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n=10 represent the sample size
Part a
A point of estimate for the population mean is the sample mean, given by this formula:
[tex]\bar X= \frac{\sum_{i=1}^n x_i}{n}[/tex]
If we apply this formula we got that [tex]\hat \mu =\bar X=24.694[/tex]
Part b
In order to calculate the confidence interval first we need to calculate the sample deviation given by this formula:
[tex]s=\frac{\sum_{i=1}^n (x_i -\bar x)}{n-1}[/tex]
If we use this formula we got that [tex]s=0.277[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=10-1=9[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that [tex]t_{\alpha/2}=3.25[/tex]
Now we have everything in order to replace into formula (1):
[tex]24.694-3.25\frac{0.277}{\sqrt{10}}=24.409[/tex]
[tex]24.694+3.25\frac{0.277}{\sqrt{10}}=24.979[/tex]
So on this case the 99% confidence interval would be given by (24.409;24.979)
Part c
Since the 99% confidence interval not contains the exposed to ambient temperatures of 24 degrees Celsius, and since the animals "that can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability", on this case we can conclude that the sea crabs are able to regulate their body temperature.