H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0 is the competing hypotheses to test the claim that there is an increase in the proportion of people who marry outside their race or ethnicity. The correct option is option A.
A hypothesis is a put forth explanation of a phenomenon (plural: hypotheses). A hypothesis must be testable according to the scientific method for it to be considered a scientific hypothesis. Scientific hypotheses are typically based on prior observations that cannot be adequately explained by the current body of knowledge.
H0:
p1 − p2 = 0;
HA: p1 −
p2 ≠ 0
H0:
p1 − p2 ≥ 0;
HA: p1 −
p2 < 0
H0:
p1 − p2 ≤ 0;
HA: p1 −
p2 > 0
H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0
Therefore, the correct option is option A.
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please help!!!! I don't have alot of time!! 100 points and will mark brainliest
At x = -2, there is a blue dot, which is a local minimum, because it is below the x axis.
The dot is located at y = -4
The answer is Yes there is a local minimum at x = -2
The local minimum is y = -4
Answer:
The answer is Yes there is a local minimum at x = -2
The local minimum is y = -4
Step-by-step explanation:
Found answer online so no work!!!
The Department of Education would like to test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000. A random sample of 34 students had an average debt load of $18,200. It is believed that the population standard deviation for student debt load is $4,200. The Department of Education would like to set α = 0.05. Which one of the following statements is true?
Because the p-value is greater than α, we fail to reject the null hypothesis and conclude that the average debt load is equal to $17,000.
Because the p-value is less than α, we reject the null hypothesis and conclude that the average debt load is equal to $17,000.
Because the p-value is less than α, we fail to reject the null hypothesis and conclude that the average debt load is not equal to $17,000.
Because the p-value is greater than α, we fail to reject the null hypothesis and cannot conclude that the average debt load is not equal to $17,000.
To test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000, a hypothesis test needs to be conducted using the given sample data. The p-value needs to be calculated and compared to the significance level α to determine the conclusion. Without the p-value, we cannot make a definitive conclusion.
Explanation:To test the hypothesis that the average debt load of graduating students with a Bachelor's degree is equal to $17,000, the Department of Education would conduct a hypothesis test using the given sample data. The null hypothesis, denoted as H0, states that the average debt load is equal to $17,000, while the alternative hypothesis, denoted as Ha, states that the average debt load is not equal to $17,000.
Based on the given information, the sample mean debt load is $18,200, the population standard deviation is $4,200, and the sample size is 34. Using these values, we can calculate the p-value, which represents the probability of obtaining a sample mean as extreme as $18,200 or more extreme, assuming the null hypothesis is true. The p-value is the probability of observing a sample mean of $18,200 or more extreme, given that the average debt load is $17,000.
To determine the conclusion of the hypothesis test, we compare the p-value to the significance level α. In this case, the given α is 0.05. If the p-value is less than α, we reject the null hypothesis and conclude that the average debt load is not equal to $17,000. If the p-value is greater than or equal to α, we fail to reject the null hypothesis and cannot conclude that the average debt load is not equal to $17,000.
In this scenario, the p-value is not given, so we cannot make a definitive conclusion. We would need to calculate the p-value based on the given information to determine which statement is true.
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The correct option is A. [tex]\text{Because the p-value is greater than}[/tex] [tex]\alpha,[/tex] [tex]\text{ we fail to reject the null hypothesis and conclude that the average debt load is equal to }[/tex] [tex]\$17,000.[/tex]
To determine the correct statement based on the hypothesis test conducted:
Given data:
Population standard deviation[tex](\( \sigma \)): \$4,200[/tex]
Sample size [tex](\( n \)): 34[/tex]
Sample mean [tex](\( \bar{x} \)): \$18,200[/tex]
Hypothesized population mean [tex](\( \mu_0 \))[/tex]: [tex]\$17,000.[/tex]
Significance level [tex](\alpha): 0.05[/tex]
Hypotheses
We are testing whether the average debt load [tex](\( \mu \))[/tex] of graduating students with a Bachelor's degree is equal to [tex]\$17,000.[/tex]
Null hypothesis [tex](\( H_0 \)): \( \mu = 17,000 \)[/tex]
Alternative hypothesis [tex](\( H_1 \)): \( \mu \neq 17,000 \)[/tex]
This is a two-tailed test because [tex]\( H_1 \)[/tex] states that [tex]\( \mu \)[/tex] is not equal to [tex]\$17,000.[/tex]
Test Statistic
Since the population standard deviation [tex](\( \sigma \))[/tex] is known and the sample size [tex](\( n \))[/tex] is greater than [tex]30[/tex], we use a z-test.
The test statistic [tex]\( z \)[/tex] is calculated as:
[tex]\[ z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \][/tex]
Calculate [tex]\( z \)[/tex]
[tex]\[ z = \frac{18,200 - 17,000}{\frac{4,200}{\sqrt{34}}} \][/tex]
[tex]\[ z = \frac{1,200}{\frac{4,200}{5.83}} \][/tex]
[tex]\[ z = \frac{1,200 \times 5.83}{4,200} \][/tex]
[tex]\[ z = 1.66 \][/tex]
P-Value Calculation
The p-value is the probability of observing a sample mean at least as extreme as [tex]18,200[/tex] if the null hypothesis [tex](\( \mu = 17,000 \))[/tex] is true. Since this is a two-tailed test, we consider both tails of the normal distribution.
From the standard normal distribution table:
The area to the right of [tex]\( z = 1.66 \)[/tex] (since [tex]\( z \)[/tex] is positive) is approximately [tex]0.0485.[/tex]
Therefore, the p-value for the two-tailed test is approximately [tex]\( 2 \times 0.0485 = 0.097 \).[/tex]
Conclusion
Compare the p-value [tex](0.097)[/tex] with the significance level [tex]\alpha = 0.05)\( \text{p-value} (0.097) > \alpha (0.05) \)[/tex]
Since the p-value is greater than the significance level ([tex]\alpha[/tex]), we fail to reject the null hypothesis [tex]\( H_0 \)[/tex]. This means we do not have enough evidence to conclude that the average debt load is different from [tex]\$17,000.[/tex]
The logistic growth function at right describes the number of people, f(t), who have become ill with influenza t weeks after its initial outbreak in a particular community. f (t )equals StartFraction 107 comma 000 Over 1 plus 4200 e Superscript negative t EndFraction.
a. How many people became ill with the flu when the epidemic began?
b. How many people were ill by the end of the fourth week?
c. What is the limiting size of the population that becomes ill?
Answer:
Part (a): 25 people became ill with the flu when the epidemic began.
Part (b): 1373 people were ill by the end of the fourth week.
Part (c): The limiting size of the population that becomes ill is 107,000.
Step-by-step explanation:
Consider the provided logistic growth function
[tex]f (t )=\frac{ 107,000}{1 +4200e^{-t}}[/tex]
Part (a) How many people became ill with the flu when the epidemic began?
Substitute t = 0 in above function.
[tex]f (t )=\frac{ 107,000}{1 +4200e^{-0}}[/tex]
[tex]f (t )=\frac{ 107,000}{4201}[/tex]
[tex]f (t )=25.47[/tex]
Hence, 25 people became ill with the flu when the epidemic began.
Part (b) How many people were ill by the end of the fourth week?
Substitute t = 4 in above function.
[tex]f (t )=\frac{ 107,000}{1 +4200e^{-4}}[/tex]
[tex]f (t )=1373.10[/tex]
Hence, 1373 people were ill by the end of the fourth week.
Part (c) What is the limiting size of the population that becomes ill?
Substitute t = ∞ in above function.
[tex]f (t )=\frac{ 107,000}{1 +4200e^{-\infty}}[/tex]
[tex]f (t )=\frac{ 107,000}{1 +0}[/tex]
[tex]f (t )=107,000[/tex]
Hence, the limiting size of the population that becomes ill is 107,000.
Using the logistic growth function, the number of sick people at the start of the epidemic and at the end of the fourth week can be found by substituting the appropriate values for time into the function. The limiting size of the potentially ill population is the maximum value of the function, which is 107,000.
Explanation:The logistic growth function provided describes the number of people, represented by f(t), who have become ill with influenza, t weeks after its initial outbreak. The function is given as f(t) = 107000/(1 + 4200 * e-t).
a. At the beginning of the epidemic (i.e., when t=0), the number of ill people, f(0), can be calculated by substituting 0 for t in the equation, resulting in 107,000/(1+4200) people.
b. By the end of the fourth week (i.e., when t=4), the number of ill people, f(4), can be calculated by substituting 4 for t in the equation.
c. The limiting size of the population that becomes ill is represented by the maximum value the growth function can take, which in this case is 107,000 (as seen by the numerator of the function). This means that illness will eventually spread to reach a potential maximum of 107,000 people in this specific community.
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In the slideshow Simplifying Radicals, under media in Chapter 9, one specific condition to satisfy is to use the quotient rule.
A. True
B. False
Answer:
True
Step-by-step explanation:
In Mathematics, A radical symbol is defined as √
It is an expression that is uses a root, such as the square root (√ ) or cube root ( ∛ )
To fully show a radical expression:
[tex]\sqrt{A}[/tex]
A is the radicand
√ is the radical symbol
Here, the degree is 2.
Quotient Rule for Radicals
For non- negative real numbers, x and y:
[tex]\frac{\sqrt[n]{x}}{\sqrt[n]{y} }[/tex] = [tex]\sqrt[n]{\frac{x}{y}}[/tex]
Example :
[tex]\sqrt{\frac{36}{9}}[/tex]= [tex]\frac{\sqrt{36}}{\sqrt{9} }[/tex] = [tex]\frac{6}{3}[/tex] = 2
So, in simplifying radicals, quotient rule is used as demonstrated above. So the condition is true
In an advertisement, a pizza shop claims its mean delivery time is 30 minutes. A consumer group believes that the mean delivery time is greater than the pizza shop claims. A random sample of 41 delivery times has a mean of 31.5 minutes with standard deviation of 3.5 minutes. Does this indicate at a 5% significance level that the mean delivery time is longer than what the pizza shop claims?
Answer:
We conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 30 minutes
Sample mean, [tex]\bar{x}[/tex] = 31.5 minutes
Sample size, n = 41
Alpha, α = 0.05
Sample standard deviation, s = 3.5 minutes
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 30\text{ minutes}\\H_A: \mu > 30\text{ minutes}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{31.5 - 30}{\frac{3.5}{\sqrt{41}} } = 2.744[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that at a 5% significance level that the mean delivery time is longer than what the pizza shop claims.
The reported average cost per workbook at a large college is $27.50. A professor claims that the actual average cost per workbook is higher than $27.50. A sample of 44 different workbooks has an average cost of $28.90. The population standard deviation is known to be $5.00. Can the null hypothesis be rejected at alpha= 0.05?
Answer:
We conclude that the actual average cost per workbook is higher than $27.50.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = $27.50
Sample mean, [tex]\bar{x}[/tex] = $28.90
Sample size, n = 44
Alpha, α = 0.05
Population standard deviation, σ = $5.00
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 27.50\text{ dollars}\\H_A: \mu > 27.50\text{ dollars}[/tex]
We use one-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{28.90 - 27.50}{\frac{5.00}{\sqrt{44}} } = 1.8573[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
We reject the null hypothesis and accept the alternate hypothesis.
Thus, we conclude that the actual average cost per workbook is higher than $27.50.
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.
a. .0500 b. .0228 c. .0456 d. .0250
Answer:
The correct option is B) 0.0228
Step-by-step explanation:
Consider the provided information.
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.
Thus, the value of n = 100, [tex]\bar x=3.1[/tex] and σ = 0.5
We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.
The null and alternative hypotheses are,
[tex]H_0:\mu\leq3[/tex] and [tex]H_a:\mu>3[/tex]
Compute the test statistic [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z=\frac{3.1-3}{\frac{0.5}{\sqrt{100}}}[/tex]
[tex]z=\frac{0.1}{0.05}[/tex]
[tex]z=2[/tex]
By using the table.
P value = P(Z>2) = 0.0228
Hence, the correct option is B) 0.0228
A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of awriting machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined,and a normal probability plot of the resulting data supports the use of a one-sample t test.a. What hypotheses should be tested if the investigators believe a priori that the design specification has been satisfied?b. What conclusion is appropriate if the hypotheses of part (a) are tested,t = —23, and a = .05?c. What conclusion is appropriate if the hypotheses of part (a) are tested, t = —1.8, and u = .01?d. What should be concluded if the hypotheses of part (a) are tested and t = —3.6 ?
Answer:
a) Null hypothesis:[tex]\mu \geq 10[/tex]
Alternative hypothesis:[tex]\mu < 10[/tex]
b) [tex]p_v =P(t_{17}<-2.3)=0.017[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
c) [tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]
And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.
d) [tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X[/tex] represent the sample mean
[tex]s[/tex] represent the sample deviation
[tex]n=18[/tex] sample size
[tex]\mu_o =10[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is at least 10 hours, the system of hypothesis would be:
Part a
Null hypothesis:[tex]\mu \geq 10[/tex]
Alternative hypothesis:[tex]\mu < 10[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Part b
For this case we have t=-2.3 , [tex]\alpha=0.05[/tex]
First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]
Now since we are conducting a left tailed test the p value is given by:
[tex]p_v =P(t_{17}<-2.3)=0.017[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
Part c
For this case we have t=-1.8 , [tex]\alpha=0.01[/tex]
First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]
Now since we are conducting a left tailed test the p value is given by:
[tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]
And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.
Part d
For this case we have t=-3.6 , [tex]\alpha=0.05[/tex]
First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]
Now since we are conducting a left tailed test the p value is given by:
[tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
The student is learning about conducting a one-sample t-test in a hypothesis testing problem related to the lifetime of a pen. The null and alternative hypothesis are defined, and conclusions are drawn based on various test statistics and significance levels.
Explanation:This situation relates to hypothesis testing in statistics, in particular the usage of a one-sample t-test.
The null hypothesis (H0) and alternative hypothesis (Ha) to be tested would be:
H0: μ ≥ 10 hours (The average lifetime of the pen is at least 10 hours)Ha: μ < 10 hours (The average lifetime of the pen is less than 10 hours)The t value represents the test statistic for conducting the hypothesis test, and the 'α' and 'u' are the significance levels.
With t = -23 and α = 0.05, this falls within the rejection region, as the t value is less than the critical value for a one-tailed test at significance level α = 0.05. Hence, we reject the null hypothesis. There is evidence to suggest that the pen's lifetime is less than 10 hours.
Similarly, with t = -1.8 and u = 0.01, the null hypothesis would be retained, as this value is within the acceptance region.The conclusion would be that there is not sufficient evidence to suggest that the pen's lifetime is less than 10 hours.
Finally, with t = -3.6, the null hypothesis is rejected, suggesting that the pen's lifetime is less than 10 hours. However, this is only applicable if the significance level is above the calculated p-value for t = -3.6.
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Directions: Estimate the sum or difference using front end estimation.
1) 97 – 39
2) 812,344 – 187,675
3) 321 + 79
Directions: Estimate the product using front end estimation.
4) 315 x 821
5) 562 x 791
6) 82 x 156
7) 711 x 884
8) 126 x 952
9) 824 x 541
10) 4027 x 78
11) 796 x 123
Directions: Estimate the quotient using front end estimation.
12) 817 19
13) 3615 72
14) 232 64
15) 559 81
16) 2986 222
17) 10275 232
18) 7428 286
19) 7143 369
20) 628,597 1525
Answer:
Step-by-step explanation:
1) 1) 97 – 39. It becomes
100 - 40 = 60
2)2) 812,344 – 187,675. It becomes
800000 - 200000 = 600000
3)321 + 79. It becomes
300 + 80 = 410
4) 315 x 821. It becomes
300 + 800 = 1100
5) 562 x 791. It becomes
600× 800 = 480000
6) 82 x 156. It becomes
80×200 = 16000
7) 711 x 884. It becomes
700×900 = 630000
8) 126 x 952. It becomes
100×1000 = 100000
9) 824 x 541. It becomes
800× 500 = 400000
10) 4027 x 78. It becomes
4000×80 = 320000
11) 796 x 123. It becomes
800×100 = 80000
12) 817 19. It becomes
800/20 = 40
13) 3615 72. It becomes
4000/70 = 57.1 = 60
14) 232 64. It becomes
200/60 = 33.3 = 30
15) 559 81. It becomes
600/80 = 7.5 = 8
16) 2986 222. It becomes
3000/200 = 15
17) 10275 232. It becomes
10000/200 = 50
18) 7428 286. It becomes
7000/300 = 23.3 = 20
19) 7143 369. It becomes
7000/400 = 17.5 = 18
20) 628,597 1525, it becomes
600000 - 2000 = 300
Answer:
1) 100 - 40 = 60
2) 800000 - 200000 = 600000
3) 300 + 80 = 410
4) 3000 + 800 = 1100
5) 600× 800 = 480000
6) 80×200 = 16000
7) 700×900 = 630000
8) 100×1000 = 100000
9) 800× 500 = 400000
10) 4000×80 = 320000
11) 800×100 = 80000
12) 800/20 = 40
13) 4000/70 = 57.1 = 60
14) 200/60 = 33.3 = 30
15) 600/80 = 7.5 = 8
16) 3000/200 = 15
17) 10000/200 = 50
18) 7000/300 = 23.3 = 20
19) 7000/400 = 17.5 = 18
20) 600000 / 2000 = 300
Step-by-step explanation:
1) 1) 97 – 39. It becomes
100 - 40 = 60
2)2) 812,344 – 187,675. It becomes
800000 - 200000 = 600000
3)321 + 79. It becomes
300 + 80 = 410
4) 315 x 821. It becomes
300 + 800 = 1100
5) 562 x 791. It becomes
600× 800 = 480000
6) 82 x 156. It becomes
80×200 = 16000
7) 711 x 884. It becomes
700×900 = 630000
8) 126 x 952. It becomes
100×1000 = 100000
9) 824 x 541. It becomes
800× 500 = 400000
10) 4027 x 78. It becomes
4000×80 = 320000
11) 796 x 123. It becomes
800×100 = 80000
12) 817 19. It becomes
800/20 = 40
13) 3615 72. It becomes
4000/70 = 57.1 = 60
14) 232 64. It becomes
200/60 = 33.3 = 30
15) 559 81. It becomes
600/80 = 7.5 = 8
16) 2986 222. It becomes
3000/200 = 15
17) 10275 232. It becomes
10000/200 = 50
18) 7428 286. It becomes
7000/300 = 23.3 = 20
19) 7143 369. It becomes
7000/400 = 17.5 = 18
20) 628,597 1525, it becomes
600000 - 2000 = 300
The radius of a sphere increases at a constant rate of 2 com/min at the time when the volume of the sphere is 40 cm^3ç What is the rate of increase of the volume in cm^3/min?
Answer:
[tex]\frac{dV}{dt}=525 cm^{3}/min[/tex]
Step-by-step explanation:
The volume of a sphere is:
[tex]V=\frac{4}{3}\pi r^{3}[/tex] (1)
We know that:
dr/dt = 2 cm/min (increasing rate of radius)V = 40 cm³If we take the derivative of (1) with respect of time t, we ca n find the rate of increase of the volume.
[tex]\frac{dV}{dt}=\frac{4}{3}\pi 3r^{2}\frac{dr}{dt}=4\pi r^{2}\frac{dr}{dt}[/tex] (2)
We also know that the volume is 40 cm³, then using the (1) we can get the radius at this value.
Solving (1) for r, we have:
[tex]r=\left(\frac{3\cdot V}{4\pi}\right)^{1/3}=\left(\frac{3\cdot 400}{4\pi}\right)^{1/3}=4.57 cm[/tex]
Finally dV/dt will be:
[tex]\frac{dV}{dt}=4\pi (4.57)^{2}\cdot 2=525 cm^{3}/min[/tex]
I hope it helps you!
A manufacturer of cereal has a machine that, when working properly, puts 20 ounces of cereal on average into a box with a standard deviation of 1 ounce. Every morning workers weigh 25 filled boxes. If the average weight is off by more than 1 percent from the desired 20 ounces per box, company policy requires them to recalibrate the machine. In a sample of 100 days where the machine is working properly all day, on how many of the days is it expected the machine will be recalibrated?
Answer:
Step-by-step explanation:
Given
mean [tex]\mu =20\ ounces[/tex]
standard deviation [tex]\sigma =1\ ounce[/tex]
The no of sample boxes weigh Every morning is 25
Average weight is 1 % more than average
i.e. [tex]20+20\times 0.1=20.2 ounce[/tex]
The company re-calibrates the machine
[tex]P\left ( \bar{x}> 20.2\right )=P\left ( z-\frac{20.2-20}{\frac{1}{\sqrt{25}}}\right )[/tex]
[tex]=P\left ( z> 1\right )=1-P\left ( z\leq 1\right )[/tex]
[tex]=1-0.8413[/tex]
[tex]=0.1587[/tex]
Therefore the Probability that the average weight of box is more than 20.2 ounce is 0.1587
No of days the machine is expected to re-calibrate is [tex]100\times 0.1587=16\ days[/tex]
when a tow truck is called, the cost of the service is given by the linear function y = 3x +75, when y is in dollars and x is the number of miles the car is towed. Find and interpret the slope and y intercept of the linear equation.
Answer:
Slope: 3
Y intercept: y = 75
Step-by-step explanation:
The slope of a linear function is the coefficient in front of x. In this case, 3 is being multiplied by x, indicating the slope is 3.
The y-intercept is the y value where the graph of the function crosses the y-axis. In other words, it's the y value when x = 0. To find the y-intercept, simply substitute 0 for x:
[tex]y = 3(0) + 75 \\ y = 75[/tex]
We find that the y-intercept is y = 75.
In the linear function y=3x+75, the slope 3 represents the rate of cost per mile to tow a car ($3/mile), and the y-intercept 75 is the basic fee for the towing service. So, the total cost is a fixed fee of $75 plus $3 per mile towed.
Explanation:In the linear function y = 3x + 75, the slope, represented by '3', is the cost in dollars per mile to tow a car. This means for each mile the car is towed, the cost increases by $3.
The y-intercept, represented by '75', is the basic fee for the service, or the cost in dollars that one must pay even if the car isn't towed any miles.
So, to interpret the equation function: the total cost (y) of having a car towed is composed of a fixed fee of $75 plus an additional $3 for each mile (x) the car is towed.
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A recent report states the variance of the fuel consumption of a certain car is equal to 16.81 miles per gallon (mpg). A researcher claims the actual variance is lower. A sample of 21 cars produced a variance of 6.25 mpg. Test the claim at alpha = 0.05, use the P-value method. For question #29, tell what type of test his is (right, left, or two-tailed). For question #30, select from choices given with regard to identifying the alternate hypothesis. For question #31, fill in the blank with your Critical Value(s) for Chi-square. For question #32, fill in the space provided with your calculated value of Chi-square. For question #33, use your answer from #32 to select between which two areas your P-value lies. For question #34, based on your answer for #33, answer True or False as to whether the null hypothesis will be rejected.
Answer:
H0: [tex]\sigma \geq 4.1[/tex]
H1: [tex]\sigma <4.1[/tex]
[tex]\Chi^2_{crit} =10.851[/tex]
[tex] t=(21-1) [\frac{2.5}{4.1}]^2 =7.436[/tex]
[tex]p_v = P(\Chi^2_{20}<7.436)=0.0050[/tex]
True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)
Step-by-step explanation:
Previous concepts and notation
The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".
[tex]\bar X [/tex] represent the sample mean
n = 21 sample size
[tex]s^2 =6.25[/tex] represent the sample variance
[tex]s=2.5[/tex] represent the sample deviation
[tex]\sigma^2_0 =16.81[/tex] represent the target variance
[tex]\sigma_o =4.1[/tex] the value that we want to test
[tex]p_v [/tex] represent the p value for the test
t represent the statistic
[tex]\alpha=0.05[/tex] significance level
State the null and alternative hypothesis
On this case we want to check if the population variance is lower (or if the deviation is lower than 4.1) than 16.81, so the system of hypothesis are:
H0: [tex]\sigma \geq 4.1[/tex]
H1: [tex]\sigma <4.1[/tex]
In order to check the hypothesis we need to calculate th statistic given by the following formula:
[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]
This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.
What is the critical value for the test statistic at an α = 0.05 significance level?
Since is a left tailed test the critical zone it's on the left tail of the distribution. On this case we need a quantile on the chi square distribution with 20 degrees of freedom that accumulates 0.05 of the area on the left tail and 0.95 on the right tail.
We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.05,20)". And our critical value would be [tex]\Chi^2_{crit} =10.851[/tex]
What is the value of your test statistic?
Now we have everything to replace into the formula for the statistic and we got:
[tex] t=(21-1) [\frac{2.5}{4.1}]^2 =7.436[/tex]
What is the approximate p-value of the test?
For this case since we have a left tailed test the p value is given by:
[tex]p_v = P(\Chi^2_{20}<7.436)=0.0050[/tex]
Based on your answer for the p value, answer True or False as to whether the null hypothesis will be rejected.
True, since our p value is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance. And we can conclude that the population standard dviation is less than 4.1 (Or the variance is less than 16.81)
You go to a local mechanic to get your tires changed. The tires cost $300. There is a 6% sales tax, but you get a 10% discount. There is also a $10 non-taxable disposal fee for your old tires, which the mechanic tells you is not subject to discount.a. Write a function, t(x) for the total purchase amount after taxes but before discounts and fees.b. Write a function, d(x) to account for the total after discounts on purchase amount x.c. Does it matter whether the mechanic adds the tax first or takes the discount first?
Answer:
a) [tex]T(x)=300x+18x=318x[/tex]
b) [tex]D(x)=318x-(0.1)(318)+10x\\=296.2x[/tex]
c) Yes.
Step-by-step explanation:
We have that:
[tex]Tires=$300\\Tax=0.006\\Discount=0.1\\[/tex]
And we have $10 free of taxes.
Making x= number of tires to buy, then we have that the total cost of tires is:
[tex]Total_{Tires}=300x[/tex]
So, what we pay for taxes is given by:
[tex]Taxes=(300x)(0.06)=18x[/tex]
a) Then, according to the above, we can write down the total cost before the discount as:
[tex]T(x)=300x+18x=318x[/tex]
b) And the total cost after discounts, is then given by:
[tex]D(x)=318x-(0.1)(318)+10x\\=296.2x[/tex]
c) If the discount is added first, then less tax will be paid because the amount on which it is paid is lower. If the discount is added later, then the taxes will have been taxed on a higher amount, so it does matter whether they are added first or later.
find an equation of the line satisfying the given conditions. write the equation using function notation. Through (-20,5) and (-36, 9) And can you show me the steps?
Answer:
The equation of the line using function notation is
[tex]y=f(x)=-\frac{1}{4}x[/tex]
Step-by-step explanation:
Given points are (-20,5) and (-36, 9)
Now to find the equation of the line passes through these points
Let [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] be the two given points (-20,5) and (-36, 9) respectively.
To find slope
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]m=\frac{9-5}{-36-(-20)}[/tex]
[tex]m=\frac{4}{-36+20}[/tex]
[tex]m=\frac{4}{-16}[/tex]
[tex]m=-\frac{1}{4}[/tex]
Therefore [tex]m=-\frac{1}{4}[/tex]
The equation of the line is of thr form y=mx+c
The point (-20,5) passes through the above line and [tex]m=-\frac{1}{4}[/tex]
[tex]5=-\frac{1}{4}(-20)+c[/tex]
[tex]5=5+c[/tex]
[tex]c=0[/tex]
[tex]y=-\frac{1}{4}x+0[/tex]
Therefore [tex]y=-\frac{1}{4}x[/tex]
Therefore the equation of the line using function notation is
[tex]y=f(x)=-\frac{1}{4}x[/tex]
A production manager at a wall clock company wants to test their new wall clocks. The designer claims they have a mean life of 16 years with a standard deviation of 4 years. If the claim is true, in a sample of 42 wall clocks, what is the probability that the mean clock life would be greater than 15.1 years? Round your answer to four decimal places.
Answer:
92.79% probability that the mean clock life would be greater than 15.1 years.
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].
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 16, \sigma = 4, n = 42, s = \frac{4}{\sqrt{42}} = 0.6172[/tex]
What is the probability that the mean clock life would be greater than 15.1 years?
This probability is 1 subtracted by the pvalue of Z when [tex]X = 15.1[/tex]. So:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{15.1 - 16}{0.6172}[/tex]
[tex]Z = -1.46[/tex]
[tex]Z = -1.46[/tex] has a pvalue of 0.0721.
So there is a 1-0.0721 = 0.9279 = 92.79% probability that the mean clock life would be greater than 15.1 years.
Final answer:
Using the Central Limit Theorem and the given standard deviation and mean, the standard error was calculated, and then the z-score was determined. After finding the z-score, the area to the right of it on the standard normal distribution curve gives the probability that the mean clock life is greater than 15.1 years, which is 0.9274.
Explanation:
To find the probability that the mean clock life of a sample of 42 wall clocks is greater than 15.1 years, we can use the Central Limit Theorem which tells us that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough. Since we are dealing with a sample of 42, which is generally considered sufficiently large, the sampling distribution of the sample mean is normally distributed. Given the population mean (μ) is 16 years, the standard deviation (σ) is 4 years, the sample size (n) is 42, and we want to find the probability of the sample mean (μ_x) > 15.1 years, we first need to find the standard error (SE) which is σ/sqrt(n). Then we can find the z-score which is (15.1 - μ) / SE. Lastly, we use the z-score to find the probability from z-tables or a calculator with normal distribution functions.
The standard error (SE) is:
SE = 4 / sqrt(42) ≈ 0.6172
The z-score (z) is:
z = (15.1 - 16) / 0.6172 ≈ -1.4573
To find the probability of the sample mean being greater than 15.1 years, we want the area to the right of the z-score on the standard normal distribution curve. Looking up the z-score of -1.4573 on the z-table or using a normal distribution calculator gives us the probability to the left of the z-score. We subtract this value from 1 to find the probability to the right:
P(X > 15.1) = 1 - P(Z < -1.4573) ≈ 1 - 0.0726 = 0.9274
Therefore, the probability that the mean clock life is greater than 15.1 years is 0.9274 when rounded to four decimal places.
Random samples of 50 women and 50 men are taken at Norwich University. They are asked their reaction to increased tuition fees. Of the women, 23 favored the increase. Of the men, 19 favor the increase. At a 10% significance level, does this indicate that a larger proportion of women favor the increase than men?
Answer:
[tex]z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)(\frac{1}{50}+\frac{1}{50})}}=0.8104[/tex]
[tex]p_v =P(Z>0.8104)=0.209[/tex]
If we compare the p value and using any significance level for example [tex]\alpha=0.1[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.
Step-by-step explanation:
1) Data given and notation
[tex]X_{1}=23[/tex] represent the number of women favored the increase
[tex]X_{2}=19[/tex] represent the number of men favored the increase
[tex]n_{1}=50[/tex] sample 1 selected
[tex]n_{2}=50[/tex] sample 2 selected
[tex]p_{1}=\frac{23}{50}=0.46[/tex] represent the proportion of women favored the increase
[tex]p_{2}=\frac{19}{50}=0.38[/tex] represent the proportion of men favored the increase
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.1[/tex] represent the significance level
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if that a larger proportion of women favor the increase than men, the system of hypothesis would be:
Null hypothesis:[tex]p_{1} \leq p_{2}[/tex]
Alternative hypothesis:[tex]p_{1} > p_{2}[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{23+19}{50+50}=0.42[/tex]
3) Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.46-0.38}{\sqrt{0.42(1-0.42)(\frac{1}{50}+\frac{1}{50})}}=0.8104[/tex]
4) Statistical decision
We can calculate the p value for this test.
Since is a one right side test the p value would be:
[tex]p_v =P(Z>0.8104)=0.209[/tex]
If we compare the p value and using any significance level for example [tex]\alpha=0.1[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of women favored the increase is not significantly higher than the proportion of men favored the increase at 10% of significance.
Given is a finite set of spherical planets, all of the same radius and no two intersecting. On the surface of each planet consider the set of points not visible from any other planet. Prove that the total area of these sets is equal to the surface area of one planet.
Answer:
Step-by-step explanation:
To answer this question, First, we define a direction in space, which defines
the North Pole of each planet.
We also assume that the direction is not at 90° perpendicular to the axis of any two planet.
Assume we now define an order on the set of planet by saying that planet A ≥ B, when removing all the other planet from space, the north pole B, is visible from A.
If we refer to the direction of the planet in the diagram, we discover that,
1, A ≥ A
2, If A ≥ B and B≥ A,
Then A = B
3, If A ≥ B, and B ≥ C , then A ≥ C
4, Either A ≥ B or B ≥ A
It should also be noted that the array of order above has a unique maximal element (M). This is the only planet whose North Pole is not visible from another.
If we now consider a sphere of the same radius as the planets. Remove
from it all North Poles defined by directions that are perpendicular
to the axis of two of the planets. This is a set of area zero.
For every other point on this sphere, there exists a direction in
space that makes it the North Pole, and for that direction, there
exists a unique North Pole on one of the planets which is not visible
from the others. Hence the total area of invisible points is equal
to the area of this sphere, which in turn is the area of one of the planets.
The spinner for a board game has eight colors the arrow can land on. What are the degrees of freedom for a goodness-of-fit test of the fairness of this spinner?
A. 9
B. 8
C. 7
D. 6
E. We can't determine the degrees of freedom without knowing the sample size.
Answer: Option 'c' is correct.
Step-by-step explanation:
Since we have given that
Number of colors that the arrow can land on = 8
We need to find the degrees of freedom for a goodness of fit test of the fairness of this spinner.
As we know that degree of freedom is 1 less than the size of sample.
So, Mathematically, it is expressed as below:
So, Degree of freedom would be
[tex]v=n-1\\\\v=8-1\\\\v=7[/tex]
Hence, Option 'c' is correct.
equation form for
slope =3
point (2, 17)
Answer:
The equation of the line with slope is 3 and point (2,17) is [tex]y=3x+11[/tex]
Step-by-step explanation:
Given slope is 3 ie., m=3 and point (2,17)
To find the equation of the line with slope and a point:
Using slope point formula
[tex]y-y_{1}=m(x-x_{1})[/tex]
Let the point (2,17) be [tex](x_{1},y_{1})[/tex]
Substitute the values of m=3 and point (2,17)
[tex]y-y_{1}=m(x-x_{1})[/tex]
[tex]y-17=3(x-2)[/tex]
[tex]y-17=3x-6[/tex]
[tex]y=3x-6+17[/tex]
[tex]y=3x+11[/tex]
Therefore the equation of the line is [tex]y=3x+11[/tex]
1. Alexander’s grandparents get him a puppy for his thirteenth birthday, and he names the puppy Boomer. When Boomer is 1 month old, he weighs 4 pounds. When he is 2 months old, he weighs 7 pounds. In this relationship, x represents Boomer’s age (in months), and y represent her weight (in pounds). 1a.) Graph the two points for this relationship and the line passing through them on the coordinate plane. (2 points). 1b). How do you know the graph shows a proportional or non-proportional relationship? (1pt)
Answer:
1a) Plot on the graph.
1b) Its proportional if y (weight) varies directly with variation in x(age) and fits the form y=k*x
Step-by-step explanation:
1a)
draw graph with x axis in increments of 1 month
draw y-axis in increments of 1 pound.
put 1 dot at (1, 4)
put 1 dot at (2,7)
connect the 2 dots, and continue the line to hit one of the axes.
It will not hit the origin.
1b) show that y=kx
4=k*1 , so k=4
7=k*2 , so k=3.5
Since we have 2 different values for k (4 & 3.5), it's not proportional.
The probability a guest at a downtown hotel arrived via taxi is 0.42. The probability a guest arriving at a downtown hotel requested parking for a car is 0.29. Assume a guest arriving by taxi is mutually exclusive of a guest requesting parking for a car. What is the probability of a guest arriving by taxi and requesting parking for a car?
Answer:
0% probability of a guest arriving by taxi and requesting parking for a car.
Step-by-step explanation:
Two events are said to be mutually exclusive if one event precludes the other. That means that if A and B are mutually exclusive, the probability of A and B happening at the same time is 0%.
In this problem, we have that a guest arriving via taxi and requesting parking at the hotel are mutually exclusive.
So there is a 0% probability of a guest arriving by taxi and requesting parking for a car.
A political scientist believes that she can show that the average age of the voters in a presidential election is less than 40 years. To this end an exit poll of 100 voters is obtained. The data obtained in the poll has a sample average age of 39.2 years with a sample standard deviation of 8 years. We establish a 1% level of significance.
a. Referring to Political Scientist, give the alternative hypothesis.
b. Referring to Political Scientist, describe the rejection region. Be sure to label the endpoint(s).
c. Referring to Political Scientist, what is the conclusion?
Answer:
a) Null hypothesis:[tex]\mu \geq 40[/tex]
Alternative hypothesis:[tex]\mu < 40[/tex]
b) The rejection zone for this test would be:
[tex] (-\infty, -2.36)[/tex]
c) If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=39.2[/tex] represent the sample mean
[tex]s=8[/tex] represent the sample standard deviation
[tex]n=100[/tex] sample size
[tex]\mu_o =40[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
Part a
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is less than 40, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 40[/tex]
Alternative hypothesis:[tex]\mu < 40[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{39.2-40}{\frac{8}{\sqrt{100}}}=-1[/tex]
Part b: Rejection zone
On this case we are conducting a left tailed test so we need to find first the degrees of freedom for the distribution given by:
[tex]df=n-1=100-1=99[/tex]
Now we need to find a critical value on th t distribution with 99 degrees of freedom that accumulates 0.01 of the area on the right tail and this value is given by [tex]t_{crit}=-2.36[/tex]
And we can find it using the following excel code: "=T.INV(0.01,99)"
So then the rejection zone for this test would be:
[tex] (-\infty, -2.36)[/tex]
Part c: P-value and conclusion
Since is a one left side test the p value would be:
[tex]p_v =P(t_{(99)}<-1)=0.160[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly less than 40 years at 1% of signficance.
In a certain city, projections for the next year indicate there is a 20% chance that electronics jobs will increase by 1300, a 50% chance that they will increase by 300, and a 30% chance they will decrease by 900. Find the expected change in the number of electronics jobs in that city in the next year. The expected change is an increase of nothing jobs.
Answer:
110
Step-by-step explanation:
%20 increase = [tex]\frac{20}{100} 1300 = 260\\[/tex]
%50 decrease = [tex]\frac{50}{100}300 = 150[/tex]
%30 no change = [tex]\frac{30}{100}0 = 0[/tex]
Expected change = Increase - Decrease + No Change
Expected change = 260 - 150 + 0
Expected change = 110
Extreme cold and hot temperatures are known to affect the operation of electronic components. Winter is approaching and you are interested in the mean minimum temperature that will damage an iPod. As a research project, 9 iPods were tested to see what minimum temperature damaged the electronic component. The damaging temperature averaged 5 ºF with a standard deviation of 3 ºF. Find a 95% confidence interval for the true mean minimum temperature that will damage an iPod. (Disclaimer: Apple recommends that to "keep iPod comfy," the iPod’s battery should be kept between temperatures of 32 ºF and 95 ºF.)If the question requires a test of significance, your solution should clearly show the four steps to the test. In addition, you should show that you have checked that the conditions for inference are met.Step 1: State the null and alternative hypothesis.Step 2: Calculate the test statistic.Step 3: Find the p-value.Step 4: State your conclusion in the context of the problem. (Do not just say "Reject H0" or "Do not reject H0.")If the question requires a confidence interval, be sure to make a statement about the confidence interval in context of the problem.
Answer:
A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F
Test statistic(Z) = 2.31
P-value = 0.0104
Step-by-step explanation:
Step 1
Null hypothesis: The average damaging temperature of nine iPods is 5°F
Alternate hypothesis: The average damaging temperature of nine iPods differs from 5°F
Step 2
Mean=5°F, Sd=3, df=n-1=9-1=8
The t-value corresponding to 8 degrees of freedom and 95% confidence level (5% significance level) is 2.306
Confidence Interval(CI) = (mean + or - t×sd/√n)
CI = (5 + 2.306×3/√8) = 5 + 6.918/2.828= 5+2.45=7.45°F
CI = (5 - 2.306×3/√8) = 5-2.45 = 2.55°F
Z = (sample mean - population mean)/(sd÷√n) = (5-2.55)/(3÷√8) = 2.45/1.061 = 2.31
Step 3
Using the standard distribution table, the cumulative area to the left of Z = 2.31 is 0.9896
P-value = 1 - 0.9896 = 0.0104
Step 4
Conclusion: A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F
Can someone help me with this!
Answer:
Please read the answers below.
Step-by-step explanation:
Let's recall that in a square all its sides are equal length and all the four internal angles measure 90 °
5. If LN = 46, then we have:
OM = 46 (Same length than LN)
PN = LN/2 = 46/2 = 23
ON = √LN²/2 = √ 46²/2 = √ 2,116/2 = √ 1,058 = 32.53 (Rounding to two decimal places)
MN = ON = 32.53
6. m ∠EFG = 90°
m ∠GDH = ∠GDH/2 = 90/2 = 45°
m ∠FEG = ∠DEF/2 = 90/2 = 45°
m ∠DHG = 180 - (∠GDH + ∠DGH) = 180 - (45 + 45)= 180 - 90 = 90°
7. Solve for x
6x - 21 = ∠PQR/2
6x - 21 = 90/2
6x - 21 = 45
6x = 45 +21
6x = 66
x = 11
In a recent public opinion poll of 750 adults, 40% said that, if they had to decide between watching a rerun of Gilligan's Island and watching the State of the Union address, they'd choose the rerun. The poll reported a margin of error of 5%. Which of the following best describes what is meant by a 5% margin of error?A. Of those polled, 5% were unclear as to which broadcast they would rather watch.B. The true population percentage is likely to be within 5% of the sample percentage.C. The 750 adults are not a random sample from the adult population.D. Of those polled, 40% will not watch the State of the Union address.E. Of those polled, 5% will probably change their minds and watch the State of the Union address.
Answer:
B. The true population percentage is likely to be within 5% of the sample percentage.
Step-by-step explanation:
These samples have a confidence level of x%, which leads us to a confidence interval with a margin of error is M%.
This means that we are x% sure that the true population percentage is within M% of the sample percentage.
In this problem, the true population percentage is likely to be within 5% of the sample percentage.
So the correct answer is:
B. The true population percentage is likely to be within 5% of the sample percentage.
Here are summary statistics for randomly selected weights of newborn girls: n=235, x=30.5 hg, s=6.7 hg. Construct a confidence interval estimate of the mean. Use a 95% confidence level. Are these results very different from the confidence interval 28.9 hg< μ < 31.9 hg with only 12 sample values, x=30.4 hg, and s=2.3 hg?
What is the confidence interval for the population mean μ?
Answer:
a) The 95% confidence interval would be given by (29.639;31.361 )
And the results are not very different from 28.9 hg< μ < 31.9 hg
b)The 95% confidence interval would be given by (28.939;31.861)
And the results are not very different from 28.9 hg< μ < 31.9 hg
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".
[tex]\bar X=30.5[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=6.7 represent the sample standard deviation
n=235 represent the sample size
2) First confidence interval
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=235-1=234[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,234)".And we see that [tex]t_{\alpha/2}=1.97[/tex]
Now we have everything in order to replace into formula (1):
[tex]30.5-1.97\frac{6.7}{\sqrt{235}}=29.639[/tex]
[tex]30.5+1.97\frac{6.7}{\sqrt{235}}=31.361[/tex]
So on this case the 95% confidence interval would be given by (29.639;31.361 )
And the results are not very different from 28.9 hg< μ < 31.9 hg
3) Second confidence interval
[tex]\bar X=30.4[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=2.3 represent the sample standard deviation
n=12 represent the sample size
[tex]df=n-1=12-1=11[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,11)".And we see that [tex]t_{\alpha/2}=2.20[/tex]
Now we have everything in order to replace into formula (1):
[tex]30.4-2.20\frac{2.3}{\sqrt{12}}=28.939[/tex]
[tex]30.4+2.20\frac{2.3}{\sqrt{12}}=31.861[/tex]
So on this case the 95% confidence interval would be given by (28.939;31.861)
And the results are not very different from 28.9 hg< μ < 31.9 hg
A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of funds' returns shows an average rate of 12%. If the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is
A. 0.037773 to 0.202227
B. 3.7773% to 20.2227%
C. 59.98786% to 61.01214%
D. 51.7773% to 68.2227%
E. 10.988% to 13.012%
A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of 60 funds' returns shows an average rate of 12%. If the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is
A. 0.037773 to 0.202227
B. 3.7773% to 20.2227%
C. 59.98786% to 61.01214%
D. 51.7773% to 68.2227%
E. 10.988% to 13.012%
Answer: E. 10.988% to 13.012%
Step-by-step explanation:
Given;
Mean x= 12%
Standard deviation r = 4%
Number of samples tested n = 60
Confidence interval is 95%
Z' = t(0.025)= 1.96
Confidence interval = x +/- Z'(r/√n)
= 12% +/- 1.96(4%/√60)
= 12% +/- 0.01214%
Confidence interval= (10.988% to 13.012%)
What is the recommended frequency for updating lean budget distribution?
Answer:
On demand
Step-by-step explanation:
On demand (fund Value Streams, not projects).
Lean Budgets is a collection of methods that significantly reduce overhead by financing and encouraging Value Streams instead of projects while preserving financial and user-fit governance. This is done by comprehensive work program assessment, active management of Epic finances, and complex budget changes.
Hence, the recommended frequency for updating lean budget distribution on demand that is fund value streams.
Lean budget distribution should be updated at least once per quarter, or as often as required by the specific circumstances of your organization. The goal of a lean budget is to optimize resource use and reduce waste to deliver more value to customers.
Explanation:The recommended frequency for updating lean budget distribution is not fixed and largely depends on the specific needs, context, and circumstances of your organization. However, a common practice is to review and adjust the lean budget allocation at least once per quarter. This allows you to take into account any changes in strategy, project progress, and business environment. It's important to highlight that these revisions should be strategic and informed by data and not just random shifts in resource allocation. Remember, the main goal of a lean budget is to minimize waste and optimize resource utilization to bring more value to customers.
Learn more about Lean Budget Distribution here:https://brainly.com/question/14314990
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