Creating a 95% confidence interval involves capturing 95% of the probability deviation in a normal distribution. The formula for doing so is (x ± (standard deviation/√n)x critical value). By adjusting the sample size or confidence level accordingly, different intervals can be created.
Explanation:The question inquires about the construction of a
confidence interval
for the true average stray-load loss for a certain type of induction motor. When constructing a 95% confidence interval, we are looking to capture the central 95 percent of the probability on a normal distribution, excluding 2.5 percent on each tail of the distribution.
For point one, with n = 25, x = 58.3, and o = 3.0, use the following formula to construct a 95% Confidence Interval: (x ± (standard deviation/√n)x critical value), resulting in (58.3 ± (3/√25)x 1.96). The same method applies to the rest of the points, only changing the sample size, n, or confidence level.
For the final query, the required sample size, n, can be found by rearranging the formula for confidence intervals and solving for n with the half-width of the interval set at 0.5.
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(1). 95% CI (n=25): Standard deviation of 3.0, sample mean of 58.3, yielding a 95% confidence interval of (57.124, 59.476).
(2). 95% CI (n=100): With the same standard deviation and mean, a larger sample size leads to a narrower 95% confidence interval of (57.712, 58.888).
(3). 99% CI (n=100): Expanding the confidence level to 99% increases the margin of error, resulting in a wider interval of (57.5272, 59.0728).
(4). Sample Size (99% CI, ME=0.5): To achieve a half-width of 0.5 for a 99% confidence interval, approximately 956 samples are required.
Given the problem requirements, we will calculate the confidence intervals based on the provided data:
Standard deviation, σ = 3.0Sample mean, x = 58.3Using these values, we’ll calculate the 95% and 99% confidence intervals (CIs) for the population mean, μ:
1. Constructing a 95% CI when n = 25
Calculate the standard error (SE):SE = σ/√n
= 3.0/√25
= 3.0/5
= 0.6
Find the z-value for 95% CI: z = 1.96Margin of error (ME): ME = z * SE= 1.96 * 0.6
= 1.176
Confidence Interval: 58.3 ± 1.176 = (57.124, 59.476)2. Constructing a 95% CI when n = 100
Calculate the standard error (SE):SE = σ/√n
= 3.0/√100
= 3.0/10
= 0.3
Find the z-value for 95% CI: z = 1.96Margin of error (ME): ME = z * SE = 1.96 * 0.3 = 0.588Confidence Interval: 58.3 ± 0.588 = (57.712, 58.888)3. Constructing a 99% CI when n = 100
Calculate the standard error (SE): SE = σ/√n = 3.0/√100 = 3.0/10 = 0.3Find the z-value for 99% CI: z = 2.576Margin of error (ME): ME = z * SE = 2.576 * 0.3 = 0.7728Confidence Interval: 58.3 ± 0.7728 = (57.5272, 59.0728)4. Determining the sample size n for a half-width of 0.5 in the 99% CI
Margin of error (ME): 0.5Find the z-value for 99% CI: z = 2.576Using the formula for margin of error, ME = z * (σ/√n):Solve for n:Therefore, n ≈ 956 to achieve a half-width of 0.5 for a 99% confidence interval.
as a lifeguard, sara earns a base pay of $80 per day. if her day involves swim instruction, sara earns an additional 9t dollars, where t represents the number of hours she gives instruction and $9 is her hourly rate. The total amount Sara earns in a day can be expressed as 80 + 9t.
(a) What does 9t represent in this context?
(b) What are the terms and the coefficients in the expression 80 + 9t
(c) Rewrite the expression for 7 hours of swim instruction.
(d) how much will sara earn in all?
Answer:
Step-by-step explanation:
Sara's base pay is $80
(a) The 9t represents the additional amount of money that she earns if she gives instruction for t hours. 9 stands for $9 per hour
b)The term in the expression is t which stands for the number of hours that she instructs. The coefficient is 9 which stands for the hourly rate
c) The expression for 7 hours would be
We will substitute t into 80 + 9t. It becomes
80 + 9×7
d) in all, she will earn
80 + 9×7 = $143
A random sample of 49 text books purchased at a local bookstore showed an average price of $122 with a population standard deviation of $15. Let u (new) be the true mean cost of a text book sold by this store. Construct a confidence interval with a 90% degree of confidence. Clearly label the following:
a. Point estimate
b. Critical value,
c. Margin of error
d. Confidence interval
e. Interpretation (confidence statement).
Answer:
point estimate is $122critical value for the 90% confidence level (1.645)margin of error is $3.52590% confidence interval is $122±3.525there is 90% probability that true population average price of text books is in the range $122±$3.525Step-by-step explanation:
Confidence Interval can be calculated using P±ME where
P is the point estimate for the mean cost of a text book ( $122 )ME is the margin of error from the meanAnd margin of error (ME) can be calculated using the formula
ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where
z is the critical value for the 90% confidence level (1.645)s is the population standard deviation ($15) N is the sample size (49)Margin of error, ME=[tex]\frac{1.645*15}{\sqrt{49} }[/tex] = 3.525
Then 90% confidence interval is $122±3.525
To interpret this, there is 90% probability that true population average price of text books is in the range $122±$3.525
Let R+ denote the set of positive real numbers. Let f : R × R+ → R be given by f(x, y) = x/y. (a) Is f an injective function? Prove your answer. (b) Is f a surjective function? Again, prove your answer. (c) Is f a bijection? Prove your answer.
Let [tex]f:\Bbb R\times\Bbb R^+\to\Bbb R[/tex].
a. [tex]f[/tex] is injective if [tex]f(x_1,y_1)=f(x_2,y_2)[/tex] for any two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], then the two points must be identical with [tex]x_1=x_2[/tex] and [tex]y_1=y_2[/tex].
[tex]f[/tex] is not injective because we can pick two points for which [tex]f[/tex] gives the same value:
[tex]f(2,1)=\dfrac21=2[/tex]
[tex]f(4,2)=\dfrac42=2[/tex]
b. [tex]f[/tex] is surjective if there exists [tex](x,y)[/tex] for which every point in the codomain [tex]\Bbb R[/tex] is obtained by [tex]f(x,y)[/tex].
This should be somewhat obvious if you consider 3 different cases:
[tex]f(x,y)<0[/tex] and can take on any negative real number for any choice of [tex]x<0[/tex][tex]f(x,y)=0[/tex] if and only if [tex]x=0[/tex][tex]f(x,y)>0[/tex] for any choice of [tex]x>0[/tex]So [tex]f[/tex] is surjective.
c. [tex]f[/tex] is bijective if it is both injective and surjective. The conclusions above show [tex]f[/tex] is not bijective.
A boat tour company finds that if the price p , charged for an one-hour harbor tour, is $20, the average number of passengers per week, x , is 300. When the price is reduced to $18, the average number of passengers per week increases to 360.The demand and supply curves of a certain brand of running shoes are given by
p=0(x)=1 1 2-0.04x, and p-S(x)-0.06x+42, where p is the price in dollars and x is the quantity sold.
(a) Assuming that the demand curve p D(x) is linear, find its formula: P-D(x)
(b) Assuming that the equilibrium price is $22 per tour, the equilibrium demand is E
(c) The consumers' surplus at that demand is $ . (Use a fraction for the coefficient.)
Answer
The answer and procedures of the exercise are attached in a microsoft word document.
Explanation
Please consider the data provided by the exercise. If you have any question please write me back. All the exercises are solved in a single sheet with the formulas indications.
Answer:
a) p(x) = (-1/30)*x + 30
b) xe = 240
c) The consumers' surplus at that demand is $ 960.00
Step-by-step explanation:
Given info
p₁ = 20
x₁ = 300
p₂ = 18
x₂ = 360
a) We can apply the following equation
p - p₁ = m*(x - x₁)
where
m is the slope, which can be obtained as follows
m = (p₂-p₁) / (x₂-x₁)
⇒ m = (18-20) / (360-300) = -2/60 = -1/30
then
p-20 = (-1/30)*(x-300)
⇒ 30*p - 600 = -x + 300
⇒ p(x) = (-1/30)*x + 30
b) If p(x) = 22
⇒ 22 = (-1/30)*x + 30
⇒ xe = 240 passengers
c) If x = 0 ⇒ p(0) = (-1/30)*(0) + 30 = 30
then
we can get h as follows
h = p(0) - p(240) = 30 - 22 = 8
if b = x(equlibrium) = 240
we apply
A = b*h/2 ⇒ A = 240*8/2 = 960
The consumers' surplus at that demand is $ 960.00
In 2008 the Pew Research Center interviewed a random sample of 5,566 registered voters and found that 36% were Democrats, 27% Republicans, and 37% Independents.
Let’s assume that this was the actual distribution for political party affiliation for all registered voters in 2008.
We can use a chi-square goodness-of-fit test to answer which of one the following research questions?
a. Is the proportion of Democrats larger than the proportion of Republicans this year?
b. Is the percentage of registered voters that are Independent higher now than the 37% reported in 2008?
c. Is the distribution of political party affiliation for all registered voters in 2012 the same as stated for 2008?
Final answer:
The research question that can be answered using a chi-square goodness-of-fit test is option c: Is the distribution of political party affiliation for all registered voters in 2012 the same as stated for 2008?
Explanation:
The research question that can be answered using a chi-square goodness-of-fit test in this scenario is option c: Is the distribution of political party affiliation for all registered voters in 2012 the same as stated for 2008?
In a chi-square goodness-of-fit test, we compare the observed frequencies (in this case, the distribution of political party affiliation in 2008) with the expected frequencies (the distribution we are testing, which is the distribution of political party affiliation in 2012). If the observed frequencies differ significantly from the expected frequencies, we can conclude that there is a significant difference between the two distributions.
In this case, the observed distribution is 36% Democrats, 27% Republicans, and 37% Independents in 2008. We want to test if this distribution is the same as the distribution in 2012. By performing a chi-square goodness-of-fit test, we can determine if there is a significant difference between the two distributions.
Jason had $32. He spent all the money buying four CDs for x dollars each and two magazines for y dollars each. If Jason had bought five CDs and two magazines, he would have run short by $4. The following system of equations models this scenario:
4x + 2y = 32
5x + 2y = 36
Use the system of equations to solve for x and y.
(4, 8)
(5, 6)
(8, 4)
(6, 5)
Answer:
(4,8)
Step-by-step explanation:
We have the two equations:
[tex]$ 4x + 2y = 32 \hspace{15mm} .....(1) $[/tex] and
[tex]$ 5x + 2y = 36 \hspace{15mm} .....(2) $[/tex]
Subtracting (1) from (2):
⇒ 5x + 2y -4x - 2y = 36 - 32
⇒ x = 4
Substituting the value of x in (1), we get:
4(4) + 2y = 32
⇒ 2y = 16
⇒ y = 8
We write the solution in ordered pair as (x,y) = (4,8).
Answer:
The answer is the ordered pairs (4,8) because we used the system equations to solve for x and y.
Step-by-step explanation:
A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let’s assume the rocket rises vertically and its speed is 600 ft/s when it has risen 3000 ft.
(a) How fast is the distance from the television camera to the rocket changing at that moment?
(b) If the television camera is always kept aimed at the rocket, how fast is the camera’s angle of elevation changing at that same moment?
Answer:
a) the time speed for distance from the television camera to the rocket is changing at rate of 360 ft/sec
b) camera’s angle of elevation changing at the rate of 0.096 radians/ sec
Step-by-step explanation:
details and step by step explanation is in the images below
image I is for Solution of a)
image 2 is for Solution of b)
The graph of the function f(x) = –(x + 3)(x – 1) is shown below. On a coordinate plane, a parabola opens down. It goes through (negative 3, 0), has a vertex at (negative 1, 4), and goes through (1, 0).
What is true about the domain and range of the function?
A. The domain is all real numbers less than or equal to 4, and the range is all real numbers such that –3 ≤ x ≤ 1.
B. The domain is all real numbers such that –3 ≤ x ≤ 1, and the range is all real numbers less than or equal to 4.
C. The domain is all real numbers, and the range is all real numbers less than or equal to 4.
D. The domain is all real numbers less than or equal to 4, and the range is all real numbers.
Answer:
The Domain is "All Real numbers", and the Range is "All real number less than or equal to 4" which coincides with option C in your problem.
Step-by-step explanation:
Recall that the Domain of a function is the set of all x-values for which there is a y-value obtained by the rule defined by the function.
In this case, whatever real number you enter for "x" in your functional expression, you will find another real value "y". That means that the actual Domain of your function is the full Real number line (all Real numbers).
Recall as well that the Range of a function is the set of y-values that are being "called" (or generated) as you use all the values for x in the Domain. In our case, it is great that they give the graph of the function (see attached image), so you can visualize that the function's graph is that of a "parabola" with branches opening downwards. You can see as well that the parabola presents a maximum value when x = -1, that means that any other value of x you use cannot give you as result a y-value that goes above that maximum.
If you evaluate that maximum value of the vertical coordinate by replacing "x" with "-1" in the actual function, you get:
[tex]f(x)=-(x+3)*(x-1)\\f(-1)=-(-1+3)*(-1-1)\\f(-1)=-(2)*(-2)\\f(-1)=4[/tex]
That means that the maximum y-value one can get from this function is "4". That is, the actual Range of the function can be any number that is smaller or equal to 4.
Bottom line: The Domain is "All Real numbers", and the Range is "All real number less than or equal to 4".
Answer:
its C
Step-by-step explanation:
edge 2020
A farmer needs to enclose three sides of a garden with a fence (the fourth side is a cliff wall). The farmer has 43 feet of fence and wants the garden to have an area of 228 sq-feet. What should the dimensions of the garden be? (For the purpose of this problem, the width will be the smaller dimension(needing two sides); the length will be the longer dimension (needing one side). Additionally? The length should be as long as possible.)
Answer:
9.5 ft wide by 24 ft long
Step-by-step explanation:
A graphing calculator shows there are two solutions to the system of equations ...
2x + y = 43 . . . . . . fence length when x= width
xy = 228 . . . . . . . area
The solutions are ...
(width, length) = (9.5, 24) or (12, 19)
Since we want the length as long as possible, the choice of dimensions is ...
length = 24 feet; width = 9.5 feet.
Data on the pH of rain in Ingham County, Michigan, are as follows: 5.47 5.37 5.38 4.63 5.37 3.74 3.71 4.96 4.64 5.11 5.65 5.39 4.16 5.62 4.57 4.64 5.48 4.57 4.57 4.51 4.86 4.56 4.61 4.32 3.98 5.70 4.15 3.98 5.65 3.10 5.04 4.62 4.51 4.34 4.16 4.64 5.12 3.71 4.64 What proportion of the samples has pH below 5.0?
Answer:
[tex]\hat p =\frac{26}{39}=0.667[/tex]
Step-by-step explanation:
Some important concepts
A proportion refers to "the fraction of the total that possesses a certain attribute"
The data det ordered from the smallest to the largest value is:
3.1 ,3.71 ,3.71 ,3.74 ,3.98 ,3.98 ,4.15 ,4.16 ,4.16 ,4.32 ,4.34 ,4.51 ,4.51 ,4.56 ,4.57 ,4.57 ,4.57 ,4.61 ,4.62 ,4.63 ,4.64 ,4.64 ,4.64 ,4.64 ,4.86 ,4.96 ,5.04 ,5.11 ,5.12 ,5.37 ,5.37 ,5.38 ,5.39 ,5.47 ,5.48 ,5.62, 5.65 ,5.65 ,5.7
If we are interesed on the sample values below 5.0 we need to count how many values are below this number. If we do this we got that 26 numbers are below 5.0 and the total of numbers are 39.
so the proportion on this case would be"
[tex]\hat p =\frac{26}{39}=0.667[/tex]
To determine the proportion of rain samples with a pH below 5, count the number of data points less than 5.0 and divide it by the total number of data points. The pH value of the environment can vary greatly due to factors such as the presence of pollution.
Explanation:This question is about finding out the proportion of rain samples that have a pH below 5.0 in Ingham County, Michigan. To do this, you count the number of data points less than 5.0 and divide it by the total number of data points. The pH of the environment (rain, in this case) can vary greatly due to factors such as pollution, affecting the acidity or alkalinity of the rain. Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO₂ which forms carbonic acid. Anything below 7.0 is acidic and above 7.0 is alkaline.
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A newsletter publisher believes that over 64%64% of their readers own a Rolls Royce. For marketing purposes, a potential advertiser wants to confirm this claim. After performing a test at the 0.010.01 level of significance, the advertiser decides to reject the null hypothesis. What is the conclusion regarding the publisher's claim?
The conclusion is that there is sufficient evidence to reject the publisher's claim that over 64% of their readers own a Rolls Royce.
When conducting a hypothesis test, there are two competing hypotheses: the null hypothesis (H0) and the alternative hypothesis (H1). In this scenario:
Null hypothesis (H0): The proportion of readers who own a Rolls Royce is equal to or less than 64%.
Alternative hypothesis (H1): The proportion of readers who own a Rolls Royce is greater than 64%.
The significance level, denoted by α, represents the probability of rejecting the null hypothesis when it is actually true. In this case, α = 0.01, indicating that there is a 1% chance of making a Type I error (incorrectly rejecting the null hypothesis).
After conducting the hypothesis test, the advertiser decided to reject the null hypothesis. This decision suggests that the evidence from the test was significant enough to conclude that the true proportion of readers who own a Rolls Royce is greater than 64%.
In summary, based on the test results at the 0.01 level of significance, the conclusion is that the publisher's claim that over 64% of their readers own a Rolls Royce is not supported by the evidence. The data suggest that the proportion may be higher than 64%.
Computer output from a regression analysis is provided. Coefficients: Estimate Std. Error t value p-value (Intercept) 7.2960 14.5444 0.502 0.62200 X 1.6370 0.5453 3.002 0.00765 We want to do the hypothesis test to see if the slope in the population is different from zero? That is, do the hypothesis test to see if we have a statistically significant linear relationship. What is your decision on the hypothesis test and why? Use a level of significance of .05.
Answer:
Step-by-step explanation:
Hello!
You need to test the hypothesis that the slope of the regression is cero.
I've run in the statistic software the given data for Y and X and estimated the regression line:
Yi= 7.82 -1.60Xi
Where
a= 7.82
b=-1.60
Sb= 3.38
The hypothesis is:
H₀: β = 0
H₁: β ≠ 0
α: 0.05
This is a two-tailed test, the null hypothesis states that the slope of the regression is cero, this means that if the null hypothesis is true, there is no linear regression between Y and X.
The statistic for this test is a Student-t
t= b - β ~t[tex]_{n-2}[/tex]
Sb
The critical values are:
Left: [tex]t_{n-2; \alpha /2} = t_{2; 0.025} = -4.303[/tex]
Right: [tex]t_{n-2; \alpha /2} = t_{2; 0.975} = 4.303[/tex]
t= -1.60 - 0 = -0.47
3.38
the p-value is also two-tailed, you can calculate it by hand:
P(t ≤ -0.47) + (1 - P(t ≤ 0.47) = 0.3423 + (1 - 0.6603) =0.6820
With the level of significance of 5%, the decision is to not reject the null hypothesis. This means that the slope of the regression is equal to cero, i.e. there is no linear regression between the two variables.
I hope this helps!
The hypothesis test, utilizing a t-statistic of 3.002 and a p-value of 0.00765, rejects the null hypothesis, demonstrating a statistically significant linear relationship. The slope coefficient (β₁ = 1.637) suggests a moderately strong relationship.
Hypothesis Test for a Linear Relationship
In this hypothesis test, we are trying to determine whether there is a statistically significant linear relationship between the independent and dependent variables. The null hypothesis (H₀) states that there is no linear relationship (β₁ = 0), while the alternative hypothesis (H₁) states that there is a linear relationship (β₁ ≠ 0).
The test statistic used in this case is the t-statistic, which is calculated as the ratio of the estimated slope (β₁) to its standard error. The p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed test statistic, assuming that the null hypothesis is true.
In this particular example, the t-statistic is 3.002 and the p-value is 0.00765. The level of significance, which is the threshold for rejecting the null hypothesis, is typically set at 0.05.
Since the p-value (0.00765) is less than the level of significance (0.05), we reject the null hypothesis. This means that we have statistically significant evidence to conclude that there is a linear relationship between the independent and dependent variables. In other words, the slope of the regression line is not equal to zero, indicating that there is a change in the dependent variable as the independent variable changes.
Interpretation of Results
The rejection of the null hypothesis in this case indicates that there is a statistically significant linear relationship between the independent and dependent variables. This means that we can be confident that the observed relationship is not due to chance. The magnitude of the t-statistic (3.002) suggests that the relationship is moderately strong.
The interpretation of the slope coefficient (β₁) is that for every one-unit increase in the independent variable, the dependent variable is expected to increase by 1.637 units, on average. The standard error (0.5453) indicates the variability of the slope estimate.
In conclusion, the hypothesis test provides strong evidence to support the existence of a statistically significant linear relationship between the independent and dependent variables. The magnitude of the slope coefficient indicates that the relationship is moderately strong.
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The speed limit posted on a local highway is 75 mph. Is the average speed on that stretch of highway significantly more than 75mph? Forty vehicle’s speeds were recorded by speed detection devices. What would be the correct alternative hypothesis?
Answer: [tex]H_a: \mu >75[/tex]
Step-by-step explanation:
Alternative hypothesis [tex](H_a)[/tex]: It is a statement which always indicates that there is a significant difference between the groups being tested.
It always contains by < , > or ≠ sign .
Given claim : Is the average speed on that stretch of highway significantly more than 75mph?
Here objective is whether the average speed on that stretch of highway significantly more than 75mph.
Let [tex]\mu[/tex] be the population mean speed on that stretch of highway significantly more than 75mph.
Then, [tex]H_a: \mu >75[/tex]
Hence , the correct alternative hypothesis would be : [tex]H_a: \mu >75[/tex]
The correct alternative hypothesis for this question is that the average speed on the highway is significantly greater than 75mph.
Explanation:The correct alternative hypothesis for this question would be that the average speed on that stretch of highway is significantly greater than 75mph.
To determine if the average speed is significantly more than 75mph, a statistical test can be conducted. This test would compare the speeds of the 40 recorded vehicles to the 75mph speed limit and calculate the average speed. If the average speed is significantly higher than 75mph, it would support the alternative hypothesis.
For example, if the average speed of the 40 recorded vehicles is found to be 80mph with a small p-value, it would indicate that the average speed on that stretch of highway is significantly greater than 75mph.
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I need help with 3 please!
Answer:
Step-by-step explanation:
PQ=24
PS=19
PR=42
TQ=10
QR=19
SR=24
PT=21
SQ=20
m∠QRS=180-m∠PQR=180-106=74°
m∠PQS=m∠QSR=49°
m∠RPS=m∠PRQ=m∠QRS-m∠PRS=74-35=39°
m∠PSQ=m∠RQS=m∠PQR-m∠PQS=106-49=57°
m∠PQR=106°
m∠ QSR=49°
M∠PRS=35°
According to a study conducted by an organization, the proportion of Americans who were afraid to fly in 2006 was 0.10. A random sample of 1 comma 400 Americans results in 154 indicating that they are afraid to fly. Explain why this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased.
Final answer:
The slightly higher proportion of Americans who indicated they are afraid to fly in a random sample is not necessarily evidence of an increased fear of flying overall. This variation could be due to sample variability, and without statistical analysis like hypothesis testing or confidence interval construction, one cannot conclude a genuine increase in fear of flying among Americans.
Explanation:
A study conducted by an organization found that the proportion of Americans who were afraid to fly in 2006 was 0.10. When a random sample of 1,400 Americans results in 154 individuals indicating that they are afraid to fly, it might initially seem like evidence that the proportion of Americans afraid to fly has increased. However, this is not necessarily indicative of an overall trend. To understand why we need to consider statistical variability and the concept of a confidence interval.
Statistical fluctuations in samples can lead to results that differ from the actual population proportion. In this case, the sample proportion of individuals afraid to fly is roughly 0.11 (154/1400), only slightly higher than the reported 0.10 from 2006. Without conducting a hypothesis test or constructing a confidence interval around the sample proportion, it's not possible to definitively say whether this difference is statistically significant or just due to random chance.
Furthermore, sample size plays a crucial role in the reliability of estimates. Although a sample size of 1,400 may seem large, the inherent randomness in sample selection could still lead to results differing from the true population proportion. In summary, a slight increase in the sample proportion of individuals afraid to fly, compared to a previous study, is not conclusive evidence of a trend without further statistical analysis to support such a claim.
The data from a sample of 1,400 Americans revealing 154 individuals who are afraid to fly does not necessarily indicate an increase in the proportion of Americans with this fear compared to 2006. Statistically significant evidence through a hypothesis test is required to make such a conclusion since sample data can fluctuate due to random chance.
Explanation:A random sample of 1,400 Americans resulted in 154 indicating that they are afraid to fly, which might suggest an increase from the 2006 study that showed a 0.10 proportion of Americans had this fear. However, this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased due to potential sampling error or the natural fluctuation inherent in sample data. To determine if there is a statistically significant increase, one would need to perform a hypothesis test comparing the sample proportion to the known proportion of 0.10.
Variability can occur in survey results, and a single sample might not represent the entire population accurately. Also, the difference observed might be a result of random chance. Therefore, it's essential to conduct statistical tests to infer whether the observed change is meaningful and not due to sampling variability.
In a sample of 83 walking canes, the average length was found to be 34.9in. with a standard deviation of 1.5. Give a point estimate for the population standard deviation of the length of the walking canes. Round your answer to two decimal places, if necessary.
Answer:
The point estimate for the population standard deviation of the length of the walking canes is 0.16.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem
The point estimate of the standard deviation is the standard deviation of the sample.
In a sample of 83 walking canes, the average length was found to be 34.9in. with a standard deviation of 1.5. By the Central Limit Theorem, we have that:
[tex]s = \frac{1.5}{\sqrt{83}} = 0.16[/tex]
The point estimate for the population standard deviation of the length of the walking canes is 0.16.
Mattel Corporation produces a remote-controlled car that requires three AA batteries. The mean life of these batteries in this product is 34 hours. The distribution of the battery lives closely follows the normal probability distribution with a standard deviation of 5.5 hours. As a part of their testing program Sony tests samples of 25 batteries.
What can you say about the shape of the distribution of sample mean?
What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal place.)
What proportion of the samples will have a mean useful life of more than 36 hours? (Round your answer to 4 decimal place.)
What proportion of the sample will have a mean useful life greater than 33.5 hours? (Round your answer to 4 decimal place.)
What proportion of the sample will have a mean useful life between 33.5 and 36 hours? (Round your answer to 4 decimal place.)
Answer:
0.0406, 0.8284,0.7887
Step-by-step explanation:
Given that Mattel Corporation produces a remote-controlled car that requires three AA batteries
X is N(34, 5.5)
Hence sample size of 25 would follow a t distribution with df = 24
This is because sample size <30
t distribution with df 24 would be bell shaped symmetrical about the mean and unimodal.
Std error of sample mean = std dev /sqrt n=[tex]\frac{5.5}{5} \\=1.1[/tex]
Prob (X>36) = [tex]P(t>\frac{36-34}{1.1} ) = P(t>1.82)\\= 0.04063[/tex]
i.e nearly 4.1% of the sample would have a mean useful life of more than 36 hours
X>33.5 implies [tex]t>-0.45[/tex]
=0.82837
=0.8284 proportion will have a mean useful life greater than 33.5 hours
Proportion between 33.5 and 36 hours
= [tex]0.3284+0.4593=0.7887[/tex]
The shape of the distribution of the sample mean is approximately normal due to the Central Limit Theorem. The standard error of the distribution of the sample mean can be calculated as 1.1 hours. Proportions of the sample mean falling above and between specific time intervals can be determined using z-scores and the Z-table.
Explanation:The shape of the distribution of the sample mean, in this case, is approximately normal. This is because the distribution of the battery lives closely follows the normal probability distribution. When taking a sample of 25 batteries, the Central Limit Theorem states that the distribution of the sample mean approaches a normal distribution regardless of the shape of the original distribution.
The standard error of the distribution of the sample mean can be calculated using the formula: standard deviation / square root of sample size. In this case, the standard deviation is 5.5 hours and the sample size is 25. Therefore, the standard error is 5.5 / √25 = 1.1 hours.
To determine the proportion of the samples that will have a mean useful life of more than 36 hours, we need to calculate the z-score first. The formula for z-score is: (sample mean - population mean) / standard error. Plugging in the given values, we get (36 - 34) / 1.1 = 1.82. By looking up the z-score in the Z-table, we find the corresponding proportion is approximately 0.0344.
Similarly, to find the proportion of the sample that will have a mean useful life greater than 33.5 hours, we calculate the z-score: (33.5 - 34) / 1.1 = -0.45. By looking up the z-score in the Z-table, we find the corresponding proportion is approximately 0.3264.
To determine the proportion of the sample that will have a mean useful life between 33.5 and 36 hours, we subtract the proportion of the sample that will have a mean useful life greater than 36 hours from the proportion of the sample that will have a mean useful life greater than 33.5 hours. Therefore, 0.3264 - 0.0344 = 0.2920.
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If 1/3 of the school is 6th graders and 1/2 are girls what fraction are 6th grade girls
Answer: 1/6
Step-by-step explanation:
Daniel is a front-desk manager at Refington Hotel, a mid-market hotel. During Daniel's shift last week, he received 29 customer complaints. A total of 428 guests had stayed in the hotel that week. Given this information, find out the errors per million opportunities.
Answer:
67,757 errors per million opportunities
Step-by-step explanation:
Assuming that each customer can only make a single complaint (1 error opportunity per customer), the number of errors per million opportunities (EPMO) is given by:
[tex]EPMO = \frac{complaints}{guests}*1,000,000 \\EPMO = \frac{29}{428}*1,000,000 \\EPMO = 67,757[/tex]
Refington hotel should expect 67,757 complaints per million guests.
According to the Texas Department of State Health Services, there are approximately 17.7 million adults (age 18+) living in Texas, of which 1.45 million have diabetes. In 2010, 366,921 adult Texans were diagnosed with diabetes for the first time.
What is the incidence of diabetes in Texas? Express your answer in units of per 1000 persons per year
Answer:
20.73 ≈ 21 persons per year
Step-by-step explanation:
Data provided in the question:
Total population of Texas = 17.7 million
People having diabetes = 1.45 million
Number of new cases of diabetes diagnosed for the first time = 366,921
Now,
the incidence of diabetes in Texas
= [ Number of new cases diagnosed ] ÷ [ Total population ]
= 366,921 ÷ 17.7 million
= 366,921 ÷ 17,700,000
= 0.02073
thus,
the incidence of diabetes in Texas per 1000 persons
= 0.02073 × 1,000
= 20.73 ≈ 21 persons per year
Identify the asymptotes of each function. 1/(x-3) - 6
[tex]\bf \cfrac{1}{x-3}-6\implies \stackrel{\textit{using the LCD of x-3}}{\cfrac{1-(x-3)6}{x-3}}\implies \cfrac{1-6x+18}{x-3}\implies \cfrac{-6x+19}{x-3}[/tex]
for the vertical asymptote, we simply zero out the denominator and solve for "x"
x - 3 = 0
x = 3 <---- that's the only vertical asymptote
for the horizontal asymptote, well, let's notice the degrees of the numerator and denominator, in this case, the degree of the numerator is 1, and the degree of the denominator is 1, thus when that occurs, the horizontal asymptote occurs at the fraction from the leading terms' coefficients.
[tex]\bf \cfrac{-6x+19}{1x-3}\implies \stackrel{\textit{horizontal asymptote}}{\cfrac{-6}{1}\implies -6 = y}[/tex]
Find the vertices and foci of the hyperbola with equation quantity x plus 4 squared divided by 9 minus the quantity of y plus 3 squared divided by 16 = 1.
Answer:
vertices: (-7, -3), (-1, -3)foci: (-9, -3), (1, -3)Step-by-step explanation:
For a hyperbola of the form ...
[tex]\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1[/tex]
The vertices are located at (h±a, k), and the foci are located at (h±c, k), where ...
[tex]c=\sqrt{a^2+b^2}[/tex]
Here, we have (h, k) = (-4, -3), a=3, b=4, and c=√(9+16) = 5.
So, the points of interest are ...
vertices: (-4±3, -3) . . . . shown red on the graphfoci: (-4±5, -3) . . . . . . . . shown green on the graphAnswer:
previous was correct
Step-by-step explanation:
please help
1 through 5
Answer:
Step-by-step explanation:
Suppose that a magnet high school includes grades 11 and 12, with half of the students in each grade. 60% of the senior class and 10% of the junior class are taking calculus. Suppose a calculus student is randomly selected to accompany the math teachers to a conference. What is the probability that the student is a junior? (Enter your answer as a fraction.)
Answer: Our required probability is [tex]\dfrac{1}{7}[/tex]
Step-by-step explanation:
Since we have given that
P(Junior ) = [tex]\dfrac{1}{2}[/tex]
P(Senior) = [tex]\dfrac{1}{2}[/tex]
Let the given event be 'C' taking calculus.
P(C|J) = 10% = 0.10
P(C|S) = 60% = 0.60
We need to find the probability that the student is a junior.
So, our required probability is given by
[tex]P(J|C)=\dfrac{P(J).P(C|J)}{P(S).P(C|S)+P(J).P(C|J)}\\\\P(J|C)=\dfrac{0.5\times 0.1}{0.5\times 0.1+0.5\times 0.6}\\\\P(J|C)=\dfrac{0.05}{0.05+0.3}\\\\P(J|C)=\dfrac{0.05}{0.35}\\\\P(J|C)=\dfrac{5}{35}\\\\P(J|C)=\dfrac{1}{7}[/tex]
Hence, our required probability is [tex]\dfrac{1}{7}[/tex]
A small town has 2100 inhabitants. At 8 AM, 80 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor? (Do not round k in your calculation. Round the final answer to one decimal place.) hours after the beginning
Answer:
2.7 PM
Step-by-step explanation:
Since, the rate of spread is proportional to the product of fraction y of people who have heard the rumour and the fraction who have not heard,
[tex]\frac{dy}{dt}=ky(1-y)[/tex]
[tex]\frac{dy}{y(1-y)}=kdt[/tex]
[tex](\frac{1}{y}+\frac{1}{1-y})dy = kdt[/tex]
Integrating both sides,
[tex]\int (\frac{1}{y}+\frac{1}{1-y})dy = \int kdt[/tex]
[tex]\ln y - \ln (1-y) = kt + C[/tex]
[tex]\ln (\frac{y}{1-y}) = kt + C[/tex]
[tex]\frac{y}{1-y}=e^{kt + C}[/tex]
[tex]y = e^{kt+C} - y e^{kt+C}[/tex]
[tex]y(1+e^{kt+C}) = e^{kt+C}[/tex]
[tex]y = \frac{e^{kt+C}}{1+e^{kt+C}}[/tex]
If t = 0, ( at 8 AM ), y = [tex]\frac{80}{2100}[/tex]
[tex]\frac{80}{2100}= \frac{e^{0+C}}{1+e^{0+C}}[/tex]
[tex]\frac{4}{105}=\frac{e^C}{1+e^C}[/tex]
[tex]4 + 4e^C = 105e^C[/tex]
[tex]4 = 101e^C[/tex]
[tex]\implies e^C=\frac{4}{101}[/tex]
Now, at noon, i.e t = 4, y = [tex]\frac{1}{2}[/tex]
[tex]\frac{1}{2}=\frac{e^{4k}.\frac{4}{101}}{1+e^{4k}.\frac{4}{101}}[/tex]
[tex]\frac{1}{2}=\frac{4e^{4k}}{101+4e^{4k}}[/tex]
[tex]101 + 4e^{4k}=8 e^{4k}[/tex]
[tex]101 = 4e^{4k}[/tex]
[tex]\frac{101}{4}=e^{4k}[/tex]
[tex](\frac{101}{4})^\frac{1}{4} = e^k[/tex]
If [tex]y = \frac{90}{100}=\frac{9}{10}[/tex]
[tex]\frac{9}{10}= \frac{(\frac{101}{4})^\frac{t}{4}\times \frac{4}{101}}{1+(\frac{101}{4})^\frac{t}{4}\times \frac{4}{101}}[/tex]
Using graphing calculator,
t ≈ 6.722,
Hence, after 6.722 hours since 8 AM, i.e. on 2.7 PM ( approx ) the 90% of the population have heard the rumour.
The mean life span of a brand name tire is 50,000 miles. Assume that the life spans of the tires are normally distributed, and the population standard deviation is 800 miles.
a. If you select one tire, what is the probability that its life span is less than 48,500 miles?
b. If you select 100 tires, what is the probability that their mean life span is more than 50,200 miles?
Answer:
a) [tex]P(X <48500)=0.0304[/tex]
b) [tex]P(\bar X>50200)=1-0.994=0.0062[/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".
Let X the random variable that represent the mean life span of a brand name tire, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu=50000,\sigma=800)[/tex]
Part a
We want this probability:
[tex]P(X<48500)[/tex]
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(X <48500)=P(Z<\frac{48500-50000}{800})=P(Z<-1.875)=0.0304[/tex]
Part b
Let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(50000,\frac{800}{\sqrt{100}})[/tex]
We want this probability:
[tex]P(\bar X>50200)=1-P(\bar X<50200)[/tex]
The best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X >50200)=1-P(Z<\frac{50200-50000}{\frac{800}{\sqrt{100}}})=1-P(Z<2.5)[/tex]
[tex]P(\bar X>50200)=1-0.994=0.0062[/tex]
Final answer:
To determine the probability that a tire's lifespan is less than a certain value, you calculate its Z-score and look up the probability in a standard normal distribution table. For sample means, use the standard error to calculate Z-scores. In hypothesis testing, compare the p-value to α to decide on the claim.
Explanation:
Calculating Probabilities for Normally Distributed Tires' Lifespan
To calculate the probability that a single tire has a lifespan of less than 48,500 miles, we can use the Z-score formula. The formula for the Z-score is (X - μ) / σ, where X is the value we are looking at (48,500 miles), μ is the mean (50,000 miles), and σ is the population standard deviation (800 miles). Calculating the Z-score gives us (48,500 - 50,000) / 800 = -1.875. We can then look up this Z-score on a standard normal distribution table or use a calculator to find the corresponding probability.
For part b, looking at the probability for the mean of 100 tires being more than 50,200 miles, we would use the standard error of the mean, σ/√ n, where n is the sample size (100 in this case). We calculate the Z-score with this new standard deviation and find the probability corresponding to that Z-score using the same method.
It is essential to remember that for larger samples, the standard deviation of the mean decreases, and the calculations have to reflect this change.
Hypothesis Testing for Tires' Lifespan
Using a hypothesis test, we can determine if there is enough evidence to support or reject a claim about the population mean. In a hypothesis test, we compare the p-value to the level of significance (α = 0.05). If the p-value is lower than α, we reject the null hypothesis. In the example given with an alpha of 0.05 and a p-value of 0.0103, we would reject the null hypothesis, concluding that the average lifespan of the tires is likely less than the claimed 50,000 miles.
The accounting department analyzes the variance of the weekly unit costs reported by two production departments. A sample of 16 cost reports for each of the two departments shows cost variances of 2.3 and 5.6, respectively. Is this sample sufficient to conclude that the two production departments differ in terms of unit cost variance? Use a = .10.
Calculate the value of the test statistic (to 2 decimals).
Answer:
Test Statistic is 2.34
Step-by-step explanation:
File Attached. kindly view it.
According to the United Nations, in the year 2002, the population of the world was 6.1 billion people and was growing at an annual rate of about 1.5%. If this pattern were to continue, then every year, the population would be 1.015 times the population of the previous year. Thus, if P(t) is the world population (in billions) t years after the base year 2002.
(a) What was the population in 2004?
(b) What will the population be in 2010?
Answer:
(a) 6.2843 billion
(b) 6.8716 billion
Step-by-step explanation:
Since there is a constant growth rate of 1.5% per year, and the population in 2002 was 6.1 billion people. The general equation for the total world population, in billions, after 2002 is:
[tex]P(t) = 6.1*(1+0.015)^t[/tex]
With t being the time, in years, after 2002.
a) What was the population in 2004?
[tex]t=2004-2002=2\\P(2) = 6.1*(1.015)^2\\P(2) = 6.2843 \ billion[/tex]
b) What was the population in 2010?
[tex]t=2010-2002=8\\P(8) = 6.1*(1.015)^8\\P(8) = 6.8716 \ billion[/tex]
For population parameter μ = true average resonance frequency for all tennis rackets of a certain type, each of these is a confidence interval. (114.4, 115.6) and (114.1, 115.9) What is the value of the sample mean resonance frequency?
a. 114.5
b. 115.8
c. 114.1
d. 115.0
Answer:
X[bar]= 115
Step-by-step explanation:
Hello!
Every Confidence interval to estimate the population mean are constructed following the structure:
"Estimator" ± margin of error"
Wich means that the intervals are centered around the sample mean. To know the value of the sample mean you have to make the following calculation:
[tex]X[bar]= \frac{Upper bond + Low bond}{2}[/tex]
[tex]X[bar]= \frac{115.6+114.4}{2}[/tex] = 115
Since both intervals were calculated with the information of the same sample, you can choose either to calculate the sample mean.
I hope it helps!
A manager must select three coders from her group to write three different software projects. There are 7 junior and 3 senior coders in her group. The first project can be written by any of the coders. The second project must be written by a senior person and the third project must be written by a junior person. How many ways are there for her to assign the three coders to the projects if no person can be assigned to more than one project?
Answer:
The total number of ways are 168.
Step-by-step explanation:
Consider the provided information.
There are 7 junior and 3 senior coders in her group.
The first project can be written by any of the coders. The second project must be written by a senior person and the third project must be written by a junior person.
For second project we have 3 choices and for third project we have 7 choices.
Now there are 2 possible case:
Case I: If first and second coder is senior, then the total number of ways are:
[tex]3\times 2\times 7=42[/tex]
Case II: If first and third coder is junior, then the total number of ways are:
[tex]7\times 3\times 6=126[/tex]
Hence, the total number of ways are: 42+126=168
The manager has 180 different ways to assign the three coders to the three different projects based on the given conditions.
Explanation:The question is about the number of ways to assign three coders to three different projects, respecting certain conditions. This is a combinatorics problem. The first coder can be selected from any coder in the group (10 coders) so we have ten options. The second coder must be a senior so we have 1×3 options (as we have already selected one coder for the first project and we can't assign them to more than one project). Similarly, the third coder must be a junior from the remaining 9 coders (3 seniors and 6 juniors), hence we have 1×6 options for this choice. So, the total number of ways to assign the coders is the product of these options: 10×3×6 = 180 different ways.
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