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 Department of Transportation of the State of New York claimed that it takes an average of 200 minutes to travel by train from New York to Buffalo. To test if the average travel time differs from 200 minutes, a random sample of 40 trains was taken and the average time required to travel from New York to Buffalo was 188 minutes, with a standard deviation of 28 minutes. What is the p-value for this test?
Answer:
[tex]t=\frac{188-200}{\frac{28}{\sqrt{40}}}=-2.7105[/tex]
[tex]p_v =2*P(t_{39}<-2.7105)=0.004967[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=188[/tex] represent the sample mean
[tex]s=28[/tex] represent the sample standard deviation
[tex]n=40[/tex] sample size
[tex]\mu_o =2-0[/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 apply a two tailed tailed test.
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu = 200[/tex]
Alternative hypothesis :[tex]\mu \neq 200[/tex]
Compute the test statistic
The statistic for this case 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{188-200}{\frac{28}{\sqrt{40}}}=-2.7105[/tex]
Give the appropriate conclusion for the test
First we need to find the degrees of freedom given by:
[tex]df=n-1=40-1=39[/tex]
Since is a two tailed test the p value would be:
[tex]p_v =2*P(t_{39}<-2.7105)=0.004967[/tex]
Conclusion
If we compare the p value and a significance level assumed for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can concldue that the true mean is significantly different from 200 minutes at 5% of significance.
a one-parameter family of solutions of the de p' = p(1 − p) is given below.
P =
c1et
1 + c1et
Does any solution curve pass through the point (0, 4)? Through the point (0, 1)? (If yes, give the corresponding value of
c1.
If not, enter DNE.)
(0, 4) __________
(0, 1) _______________
Answer:
A solution curve pass through the point (0,4) when [tex]c_{1} = -\frac{4}{3}[/tex].
There is not a solution curve passing through the point(0,1).
Step-by-step explanation:
We have the following solution:
[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]
Does any solution curve pass through the point (0, 4)?
We have to see if P = 4 when t = 0.
[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]
[tex]4 = \frac{c_{1}}{1 + c_{1}}[/tex]
[tex]4 + 4c_{1} = c_{1}[/tex]
[tex]c_{1} = -\frac{4}{3}[/tex]
A solution curve pass through the point (0,4) when [tex]c_{1} = -\frac{4}{3}[/tex].
Through the point (0, 1)?
Same thing as above
[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]
[tex]1 = \frac{c_{1}}{1 + c_{1}}[/tex]
[tex]1 + c_{1} = c_{1}[/tex]
[tex]0c_{1} = 1[/tex]
No solution.
So there is not a solution curve passing through the point(0,1).
Salmon Weights: Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh 17 such salmon. The mean weight from your sample is 19.2pounds with a standard deviation of 4.4 pounds. You want to construct a 90% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River.
(a) What is the point estimate for the mean weight of all spawning Chinook salmon in the Columbia River?
pounds
(b) Construct the 90% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River. Round your answers to 1 decimal place.
< ? <
(c) Are you 90% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 18 pounds and why?
No, because 18 is above the lower limit of the confidence interval.
Yes, because 18 is below the lower limit of the confidence interval.
No, because 18 is below the lower limit of the confidence interval.
Yes, because 18 is above the lower limit of the confidence interval.
(d) Recognizing the sample size is less than 30, why could we use the above method to find the confidence interval?
Because the sample size is greater than 10.
Because we do not know the distribution of the parent population.
Because the parent population is assumed to be normally distributed.
Because the sample size is less than 100.
Answer:
a) [tex]\bar X= 19.2[/tex] represent the sample mean. And that represent the best estimator for the population mean since [tex]\hat \mu =\bar X=19.2[/tex] b) The 90% confidence interval is given by (17.3;21.1)
c) No, because 18 is above the lower limit of the confidence interval.
d) Because the parent population is assumed to be normally distributed.
The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.
Step-by-step explanation:
1) Notation and definitions
n=17 represent the sample size
Part a
[tex]\bar X= 19.2[/tex] represent the sample mean. And that represent the best estimator for the population mean since [tex]\hat \mu =\bar X=19.2[/tex] Part b
[tex]s=4.4[/tex] represent the sample standard deviation
m represent the margin of error
Confidence =90% or 0.90
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".
2) Calculate the critical value tc
In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. The degrees of freedom are given by:
[tex]df=n-1=17-1=16[/tex]
We can find the critical values in excel using the following formulas:
"=T.INV(0.05,16)" for [tex]t_{\alpha/2}=-1.75[/tex]
"=T.INV(1-0.05,16)" for [tex]t_{1-\alpha/2}=1.75[/tex]
The critical value [tex]tc=\pm 1.75[/tex]
3) Calculate the margin of error (m)
The margin of error for the sample mean is given by this formula:
[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]
[tex]m=1.75 \frac{4.4}{\sqrt{17}}=1.868[/tex]
4) Calculate the confidence interval
The interval for the mean is given by this formula:
[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]
And calculating the limits we got:
[tex]19.2 - 1.75 \frac{4.4}{\sqrt{17}}=17.332[/tex]
[tex]19.2 + 1.75 \frac{4.4}{\sqrt{17}}=21.068[/tex]
The 90% confidence interval is given by (17.332;21.068) and rounded would be: (17.3;21.1)
Part c
No, because 18 is above the lower limit of the confidence interval.
Part d
Because the parent population is assumed to be normally distributed.
The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.
The U.S. Bureau of Labor Statistics reports that 11.3% of U.S. workers belong to unions (BLS website, January 2014). Suppose a sample of 400 U.S. workers is collected in 2014 to determine whether union efforts to organize have increased union membership. a. Formulate the hypotheses that can be used to determine whether union membership increased in 2014. H 0: p H a: p b. If the sample results show that 52 of the workers belonged to unions, what is the p-value for your hypothesis test (to 4 decimals)? c. At = .05, what is your conclusion?
Answer:
There is not enough evidence to support the claim that union membership increased.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 400
p = 11.3% = 0.113
Alpha, α = 0.05
Number of women belonging to union , x = 52
First, we design the null and the alternate hypothesis
[tex]H_{0}: p = 0.113\\H_A: p > 0.113[/tex]
The null hypothesis sates that 11.3% of U.S. workers belong to union and the alternate hypothesis states that there is a increase in union membership.
Formula:
[tex]\hat{p} = \dfrac{x}{n} = \dfrac{52}{400} = 0.13[/tex]
[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
Putting the values, we get,
[tex]z = \displaystyle\frac{0.13-0.113}{\sqrt{\frac{0.113(1-0.113)}{400}}} = 1.073[/tex]
now, we calculate the p-value from the table.
P-value = 0.141636
Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.
Thus, there is not enough evidence to support the claim that union membership increased.
The evidence isn't sufficient enough to support the claim that union membership increased.
What is p-value?This is a statistical measurement used to validate a hypothesis against observed data.
ParametersSample size, n = 400
p = 11.3% = 0.113
Alpha, α = 0.05
Number of women belonging to union = 52
H₀ : p = 0.113
Hₐ : p > 0.113
This means 11.3% of U.S. workers belong to union and there was an increase.
p = x / n
z = p - p /(( √p(1-p) /n
= 53/400 = 0.13
z = p - p /(( √p(1 - p) /n)).
Substitute the values into the equation.
z = 0.13 - 0.113 / ((√0.113(1-0.113)/400)) = 1.073
P-value = 0.141636 from the table which is greater than the significance level, hence we accept the null hypothesis.
The evidence is therefore not sufficient enough to support the claim that union membership increased.
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Brian Vanecek, VP of Operations at Portland Trust Bank, is evaluating the service level provided to walk-in customers. Accordingly, his staff recorded the waiting times for 64 randomly selected walk-in customers and determined that their mean waiting time was 15 minutes. Assume that the population standard deviation is 4 minutes. The 95% confidence interval for the population mean of waiting times is ________.A. 14.02 to 15.98B. 7.16 to 22.84C. 14.06 to 15.94D. 8.42 to 21.58E. 19.80 to 23.65
Answer: A. 14.02 to 15.98
Step-by-step explanation:
Let [tex]\mu[/tex] denotes the mean waiting time for population.
Given : Sample size : n= 64
Sample mean : [tex]\overline{x}=15[/tex] (minutes)
Population standard deviation = [tex]\sigma= 4[/tex]
Confidence level : 95%
By z-table , the critical values for 95% confidence = z*=1.96
Confidence interval for population mean : [tex]\overline{x}\pm z^* \dfrac{\sigma}{\sqrt{n}}[/tex]
The 95% confidence interval for the population mean of waiting times will be :
[tex]15\pm (1.96)\dfrac{4}{\sqrt{64}}[/tex]
[tex]15\pm (1.96)\dfrac{4}{8}[/tex]
[tex]15\pm (1.96)(0.5)[/tex]
[tex]15\pm 0.98[/tex]
[tex](15-0.98,\ 15+0.98)=(14.02,\ 15.98)[/tex]
Hence, the 5% confidence interval for the population mean of waiting times is 14.02 to 15.98.
Thus , the correct answer is Option A.
A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If this is to happen, it will have to move one of its most viewed shows to another slot. The network conducts a survey asking its viewers which show they would rather watch. The network receives 827 responses, of which 438 indicate that they would like to see the new show in the lineup. The test statistic for this hypothesis would be:
a. 2.05
b. 1.71
c. 2.25
d. 1.01
Answer:
b. 1.71
[tex]z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71[/tex]
Step-by-step explanation:
1) Data given and notation
n=827 represent the random sample taken
X=438 represent the people that indicate that they would like to see the new show in the lineup
[tex]\hat p=\frac{438}{827}=0.5296[/tex] estimated proportion of people that indicate that they would like to see the new show in the lineup
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:
Null hypothesis:[tex]p \leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>1.71)=0.044[/tex]
If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.5.
We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys walk is approximately normally distributed with a standard deviation of 2.5 months. A random sample of 25 boys has a mean of 19.2 months. Which of the following statements are correct?
I. This finding is significant for a two-tailed test at .05.
II. This finding is significant for a two-tailed test at .01.
III. This finding is significant for a one-tailed test at .01.
a. I only
b. II only
c. III only
d. II and III only
e. I and III only
Answer:
II. This finding is significant for a two-tailed test at .01.
III. This finding is significant for a one-tailed test at .01.
d. II and III only
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=19.2[/tex] represent the battery life sample mean
[tex]\sigma=2.5[/tex] represent the population standard deviation
[tex]n=25[/tex] sample size
[tex]\mu_o =18[/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)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:
Null hypothesis:[tex]\mu = 18[/tex]
Alternative hypothesis:[tex]\mu \neq 18[/tex]
And for part III we have a one tailed test with the following hypothesis:
Null hypothesis:[tex]\mu \leq 18[/tex]
Alternative hypothesis:[tex]\mu > 18[/tex]
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-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".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4[/tex]
4) P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=25-1=24[/tex]
Since is a two tailed test for parts I and II, the p value would be:
[tex]p_v =2*P(t_{(24)}>2.4)=0.0245[/tex]
And for part III since we have a one right tailed test the p value is:
[tex]p_v =P(t_{(24)}>2.4)=0.0122[/tex]
5) Conclusion
I. This finding is significant for a two-tailed test at .05.
Since the [tex]p_v <\alpha[/tex]. We reject the null hypothesis so we don't have a significant result. FALSE
II. This finding is significant for a two-tailed test at .01.
Since the [tex]p_v >\alpha[/tex]. We FAIL to reject the null hypothesis so we have a significant result. TRUE.
III. This finding is significant for a one-tailed test at .01.
Since the [tex]p_v >\alpha[/tex]. We FAIL to reject the null hypothesis so we have a significant result. TRUE.
So then the correct options is:
d. II and III only
Answer:
E. I and III only
Step-by-step explanation:
I. .05
III. one-tailed at .01
Tell whether the number is evenly divisible by 2, 3, 4, or 6.
6) 44
7) 38
8) 726
9) 2112
10) 1221
Answer:
Step-by-step explanation:
If a number is evenly divisible by another number, it means that the number divides it completely without a remainder.
6) 44 is evenly divisible by 2 and 4. 44 divided by 3 and 6 would have remainders.
7) 38 is evenly divisible by 2. There would be remainders if 38 divides 3, 4 or 6
8) 726 is evenly divisible by 2, 3 and 6. It is not evenly divisible by 4
9) 2112 is evenly divisible by 2, 3 4 and 6
10) 1221 is evenly divisible by 3. It is not evenly divisible by 2 4 and 6
Human body temperatures are normally distributed with a mean of 98.20oF and a standard deviation of 0.62oF If 19 people are randomly selected, find the probability that their mean body temperature will be less than 98.50oF. Your answer should be a decimal rounded to the fourth decimal place
Answer:
Step-by-step explanation:
Since the human body temperatures are normally distributed, the formula for normal distribution is expressed as
z = (x - u)/s
Where
x = human body temperatures
u = mean body temperature
s = standard deviation
From the information given,
u = 98.20oF
s = 0.62oF
We want to find the probability that their mean body temperature will be less than 98.50oF. It is expressed as
P(x lesser than 98.50)
For x = 98.50,
z = (98.50 - 98.20)/0.62 = 0.48
Looking at the normal distribution table, the corresponding probability to the z score is 0.6844
P(x lesser than 98.50) = 0.6844
A sample of 161children was selected from fourth and fifth graders at elementary schools in Philadelphia. In addition to recording the grade level, the researchers determined whether each child had a previously undetected reading disability. Sixty-six children were diagnosed with a reading disability. Of these children, 32 were fourth graders and 34 were fifth graders. Similarly, of the 95 children with normal reading achievement, 55 were fourth graders and 40 were fifth graders.
a. Identify the two qualitative variables (and corresponding levels) measured in the study.
b. From the information provided, form a contigency table.
c. Assuming that the two variables are independent, calculate the expected cell counts.
Answer:
Step-by-step explanation:
Given that a sample of 161children was selected from fourth and fifth graders at elementary schools in Philadelphia. In addition to recording the grade level, the researchers determined whether each child had a previously undetected reading disability
a) The two qualitative variables are disability and not having disability and secondly the grades of children
b) Contingency table:
Grade 4 5 Total
Normal read. 32 34 66
Not normal read. 23 6 29
Total 55 40 95
H0: Reading disability is independent of grade.
Ha: There is association between the two
c) 4 5 Total
Nor read 38.21052632 27.78947368 66
Not norm 16.78947368 12.21052632 29
Expected cells are obtained using the formula
row total*col total/grand total
Each lap around pavia park is 1 7/8 miles. Ellen rode her bike for 3 1/2 laps before leaving the park. How many total miles did ellen ride her bike in pavia park?
Answer:
Step-by-step explanation:
The distance of each lap around pavia park is 1 7/8 miles. Converting
1 7/8 miles to improper fraction, it becomes 15/8 miles.
Ellen rode her bike for 3 1/2 laps before leaving the park. Converting
3 1/2 laps to improper fraction, it becomes 7/2 laps.
The total number of miles that Ellen rode her bike in pavia park would be the product of the distance of each lap and the number of laps that he covered. It becomes
15/8 × 7/2 = 105/16 = 6.5626 miles
Use the general slicing method to find the volume of the following solid.
The solid with a semicircular base of radius 11 whose cross sections perpendicular to the base and parallel to the diameter are squares. Place the semicircle on the xy-plane so that its diameter is on the x-axis and it is centered on the y-axis. Set up the integral that gives the volume of the solid. Use increasing limits of integration.
The integral that gives the volume of the solid is
∫[from -11 to 11] 4(121 - y²) dy, which equals 21395 cubic units.
We have,
To find the volume of the given solid using the general slicing method, we need to integrate the areas of the individual slices perpendicular to the base.
Each cross-section perpendicular to the base and parallel to the diameter is a square.
Let's set up the integral to calculate the volume:
First, let's consider a vertical slice at a distance "y" from the x-axis.
This slice would be a square with a side length "2x," where "x" is the horizontal distance from the y-axis to the rightmost edge of the square.
Since the diameter of the semicircle is on the x-axis and the semicircle is centred on the y-axis, we have a right triangle formed by the radius (11), the distance from the y-axis (x), and the distance from the x-axis (y).
Using the Pythagorean theorem: x² + y² = 11²
Solving for "x": x² = 11² - y²,
so x = √(121 - y²)
Now, the area of the square slice is (side length)² = (2x)² = 4x².
Since the limits of integration are determined by the range of y values, which go from -11 to 11 (the radius of the semicircle), the integral for the volume is:
V = ∫[from -11 to 11] 4x² dy
Substitute the expression for "x" in terms of "y":
V = ∫[from -11 to 11] 4(121 - y²) dy
Simplify:
V = 4 ∫[from -11 to 11] (484 - 4y²) dy
Integrate:
V = 4 [484y - (4/3)y³] | from -11 to 11
V = 4 [484(11) - (4/3)(11)³] - [484(-11) - (4/3)(-11)³]
V = 4 [5344 - (4/3) * 1331] + [5344 + (4/3) * 1331]
V = 14276 + 7119
= 21395 cubic units
Thus,
The integral that gives the volume of the solid is
∫[from -11 to 11] 4(121 - y²) dy, which equals 21395 cubic units.
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The volume of the solid is found by integrating the area of the square cross sections, which are determined by the y-coordinate on the semicircle, along the x-axis. The equation for the volume integral is V = ∫ from -11 to 11 of 4(11² - x²) dx.
Explanation:To find the volume of the solid, we need to integrate the area of the square cross sections along the length of the semicircle. The side length of each square is equal to the y-coordinate at that point (which ranges from -11 to 11). The general form of our semicircle in this position is [tex]y = \sqrt (11^2 - x^2),[/tex] where -11 ≤ x ≤ 11. The equation for the area of each square, then, is [tex]A(x) = (2y)^2 = 4y^2 = 4(11^2 - x^2).[/tex]
Following the disk method for volumes of revolution, we need to integrate the cross-sectional area along the x-axis, from -11 to 11. So the total volume V is given by V = ∫ from -11 to 11 of A(x) dx = ∫ from -11 to 11 of 4(11² - x²) dx.
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5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the general population. To determine the aptitude of her district's children, a study was conducted. The results of her district's test scores were: 105, 109, 115, 112, 124, 115, 103, 110, 125, 99. If the mean of the general population of school children is 106, what could be said about her claim? Use alpha = .05
Answer:
We conclude that children in district are brighter, on average, than the general population.
Step-by-step explanation:
We are given the following data set:
105, 109, 115, 112, 124, 115, 103, 110, 125, 99
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{1117}{10} = 111.7[/tex]
Sum of squares of differences = 642.1
[tex]S.D = \sqrt{\frac{642.1}{49}} = 8.44[/tex]
We are given the following in the question:
Population mean, μ = 106
Sample mean, [tex]\bar{x}[/tex] = 111.7
Sample size, n = 10
Alpha, α = 0.05
Sample standard deviation, s = 8.44
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 106\\H_A: \mu > 106[/tex]
We use one-tailed(right) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833[/tex]
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
We conclude that children in district are brighter, on average, than the general population.
13 gallons of gas cost $24.31 what is the cost per gallon
Answer:
The cost per gallon is US$ 1.87
Step-by-step explanation:
1. Let's review the information provided to us to answer the question correctly:
Number of gallons of gas = 13
Cost of the gallons of gas = US$ 24.31
2. What is the cost per gallon?
Cost per gallon = Cost of the gallons of gas/Number of gallons of gas
Replacing with the real values, we have:
Cost per gallon = 24.31/13
Cost per gallon = 1.87
The cost per gallon is US$ 1.87
Determine whether the variable is qualitative or quantitative.
Street name of address
is the variable qualitative or quantitative?
A. The variable is quantitative because it is an attribute characteristic
B. The variable is qualitative because it is a numerical measure
C. The variable is quantitative because it is a numerical measure
D. The variable is qualitative because it is an attribute characteristic.
Answer:
D. The variable is qualitative because it is an attribute characteristic.
Step-by-step explanation:
In an address, the street name is an attribute of the address.
An attribute is a qualitative variable.
So the correct answer is:
D. The variable is qualitative because it is an attribute characteristic.
The variable is qualitative because it is an attribute characteristic.
Option D is correct
A qualitative variable is a variable whose values are varied by attributes or characteristics. Examples are hair color, course done in school, gender, etc.
A quantitative variable is a variable whose values are varied by actual measurement. Examples are number of odd numbers, number of students in a class, the population of a country, etc.
The description of the given variable is:
Street name of address
This description represents the attribute or characteristic of a location. Therefore, it is a qualitative variable
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A biologist observed that a certain bacterial colony obeys the population growth law and that the colony triples every 4 hours.
If the colony occupied 2 square centimeters initially, find:
(a) An expression for the size P(t) of the colony at any time t.
(b) The area occupied by the colony after 12 hours.
(c) The doubling time for the colony?
Answer:
a) [tex]P(t) = 2e^{0.275t}[/tex]
b) 54.225 square centimeters.
c) 2.52 hours
Step-by-step explanation:
The population growth law is:
[tex]P(t) = P_{0}e^{rt}[/tex]
In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the growth rate, as a decimal.
In this problem, we have that:
The colony occupied 2 square centimeters initially, so [tex]P_{0} = 2[/tex]
The colony triples every 4 hours. So
[tex]P(4) = 3P_{0} = 6[/tex]
(a) An expression for the size P(t) of the colony at any time t.
We have to find the value of r. We can do this by using the P(4) equation.
[tex]P(t) = P_{0}e^{rt}[/tex]
[tex]6 = 2e^{4r}[/tex]
[tex]e^{4r} = 3[/tex]
Applying ln to both sides, we get:
[tex]4r = 1.1[/tex]
[tex]r = 0.275[/tex]
So
[tex]P(t) = 2e^{0.275t}[/tex]
(b) The area occupied by the colony after 12 hours.
[tex]P(t) = 2e^{0.275t}[/tex]
[tex]P(12) = 2e^{0.275*12}[/tex]
[tex]P(12) = 54.225[/tex]
(c) The doubling time for the colony?
t when [tex]P(t) = 2P_{0} = 2*2 = 4[/tex].
[tex]P(t) = 2e^{0.275t}[/tex]
[tex]4 = 2e^{0.275t}[/tex]
[tex]e^{0.275t} = 2[/tex]
Applying ln to both sides
[tex]0.275t = 0.6931[/tex]
[tex]t = 2.52[/tex]
Please help me with these 2 questions! 50 points!
Answer:x = 9
Step-by-step explanation:
The attached photo is that of the given diagram. b represents the angle adjacent 75 degrees.
If line m is parallel to line n, it means that angle b degrees and angle (10x + 15) are corresponding angles. Corresponding angles are equal.
Therefore,
b = 10x + 15
The sum of angles on a straight line is 180 degrees. It means that
b + 75 = 180
b = 180 - 75 = 105
Therefore
10x + 15 = 105
10x = 105 - 15 = 90
x = 90/10 = 9
Answer:
x = 9°
Step-by-step explanation:
105° must be equal to 10x + 15 ° for lines to be parallel.
> 105° = 10x + 15°
> 10x = 90°
> x = 9°
The following information regarding a portfolio of two stocks are given: w1 = .25, w2 = .75, E(R1) = .08, and E(R2) = .15.
Which of the following regarding the portfolio expected return, E(Rp), is correct?
-.3640
-.2300
-.1325
-.1699
Answer:
0.1325
Step-by-step explanation:
Weight of the first stock (w1) = .25
Weight of the second stock (w2) = .75
Expected return for the first stock (E(R1)) = .08
Expected return for the second stock (E(R2)) = .15
The expected return of the portfolio is given by the weighted average of the expected return of each stock:
[tex]E(R_p)=w_1*E(R_1)+w_2*E(R_2)\\E(R_p)=0.25*.08 +0.75*.15\\E(R_p)=0.1325[/tex]
The portfolio expected return, E(Rp), is 0.1325
A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the 0.01 significance level, can we conclude that the single-earner couples on average spend more time watching television together?
We can see here that at the 0.01 significance level, we can actually conclude that the single-earner couples on average spend more time watching television together.
How we arrived at the solution?To determine whether we can conclude that single-earner couples spend more time watching television together on average than dual-earner couples, we can perform a hypothesis test.
The null hypothesis (H₀) assumes that there is no difference in the mean time spent watching television between the two groups, while the alternative hypothesis (H₁) suggests that single-earner couples spend more time together watching television.
Let's set up the hypotheses:
Null Hypothesis (H₀): μ₁ ≤ μ₂ (The mean time spent together watching television for single-earner couples is less than or equal to the mean time for dual-earner couples.)
Alternative Hypothesis (H₁): μ₁ > μ₂ (The mean time spent together watching television for single-earner couples is greater than the mean time for dual-earner couples.)
Where:
μ₁ = population mean time spent watching television for single-earner couples
μ₂ = population mean time spent watching television for dual-earner couples
Next, we will use a two-sample t-test to test the hypotheses. Since we are trying to determine if single-earner couples spend more time watching television, this will be a one-tailed t-test.
Given the sample means, sample standard deviations, and sample sizes, we can calculate the t-statistic and compare it to the critical t-value at the 0.01 significance level (α = 0.01) with degrees of freedom d f = n₁ + n₂ - 2, where n₁ and n₂ are the sample sizes of single-earner and dual-earner couples, respectively.
Let's assume the sample sizes are n₁ = n₂ = 30 (the actual sample sizes from the study are not given in the question, but this is just for demonstration purposes).
Now, we can calculate the t-statistic:
t = (x₁ - x₂) / √((s₁²/n₁) + ([tex]s_{2}[/tex]² /n₂))
where:
x₁ = sample mean time for single-earner couples
x₂ = sample mean time for dual-earner couples
s₁ = sample standard deviation for single-earner couples
[tex]s_{2}[/tex] = sample standard deviation for dual-earner couples
n₁ = sample size for single-earner couples
n₂ = sample size for dual-earner couples
Using the provided values:
x₁ = 61 minutes
x₂ = 48.4 minutes
s₁ = 15.5 minutes
= 18.1 minutes
n₁ = n₂ = 30 (sample sizes assumed for demonstration)
Calculating the t-statistic:
t = (61 - 48.4) / √((15.5²/30) + (18.1²/30))
t ≈ 4.083
Next, we need to find the critical t-value from the t-distribution table at α = 0.01 significance level and df = 30 + 30 - 2 = 58 (degrees of freedom).
The critical t-value at α = 0.01 with d f = 58 is approximately 2.660.
Since the calculated t-statistic (4.083) is greater than the critical t-value (2.660), we reject the null hypothesis (H₀).
Therefore, at the 0.01 significance level, we can conclude that single-earner couples, on average, spend more time watching television together than dual-earner couples based on the data provided in the study.
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To determine whether single-earner couples spend more time watching television together than dual-earner couples, an independent samples t-test must be conducted at the 0.01 significance level. Using the provided means and standard deviations, we would calculate a t-statistic and compare it against critical values to either reject the null or fail to reject it.
The question asks whether single-earner couples spend more time watching television together than dual-earner couples, based on a study with provided mean values and standard deviations for both groups. To determine if there is a statistically significant difference between the two means, we would conduct a hypothesis test, specifically an independent samples t-test, at the 0.01 significance level. The null hypothesis (0) would state that there is no difference in the mean television watching times between the groups, while the alternative hypothesis (A) would claim that there is a difference, specifically that single-earner couples watch more television.
Given that the mean time spent watching television for single-earner couples is 61 minutes with a standard deviation of 15.5 minutes, and for dual-earner couples it is 48.4 minutes with a standard deviation of 18.1 minutes, we would calculate the t-statistic and compare it against the t-distribution critical values for the given degrees of freedom. If the calculated t-statistic exceeds the critical value for a one-tailed test at the 0.01 level, we would reject the null hypothesis and conclude that there is a significant difference, supporting the claim that single-earner couples spend more time watching television together.
A job shop consists of three machines and two repairmen. The amount of time a machine works before breaking down is exponentially distributed with mean 10. If the amount of time it takes a single repairman to fix a machine is exponentially distributed with mean 8, then(a) what is the average number of machines not in use?(b) what proportion of time are both repairmen busy?
Answer:
Step-by-step explanation:
Let X(t) denote the number of machines breakdown at time t.
The givenn problem follows birth-death process with finite space
S={0, 1, 2, 3} with
[tex] \lambda_0=\frac{3}{10}, \mu_1=\frac{1}{8}\\\\ \lambda_1=\frac{2}{10}, \mu_2=\frac{2}{8}\\\\ \lambda_2=\frac{1}{10}, \mu_3=\frac{2}{8}[/tex]
The birth-death process having balance equations [tex]\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2[/tex]
since, state rate at which leave = rate at which enter
0 [tex]\lambda_0P_0=\mu_1P_1[/tex]
1 [tex](\lambda_1+\mu_1)P_1= \mu_2P_2 + \lambda_0P_0[/tex]
2 [tex](\lambda_2+\mu_2)P_2= \mu_3P_3 + \lambda_1P_1[/tex]
[tex]P_1=\frac{12}{5}P_0=P_0=\frac{5}{12}P_1\\\\P_2=\frac{48}{25}P_0=P_0=\frac{25}{48}P_2\\\\P_3=\frac{192}{250}P_0=P_0=\frac{250}{192}P_3[/tex]
Since [tex]\sum\limits^3_{i=0} {P_i=1}\\\\p_0=[1+\frac{5}{12}+\frac{48}{25}+\frac{192}{250}]^{-1}=\frac{250}{1522}[/tex]
a)
Average number not in use equals the mean of the stationary distribution [tex]P_1+2P_2+3P_3=\frac{2136}{751}[/tex]
b)
Proportion of time both repairmen are busy [tex]P_2+P_3=\frac{672}{1522}=\frac{336}{761}[/tex]
The average number of machines not in use is 0.5, and the repairmen are both busy 64% of the time. This has been found under the assumption of exponential distribution for both longevity of machines and repair time. The scenario represents a M/M/2 queue in operations research.
Explanation:In this scenario, the life of the machines and the repair time are governed by exponential distributions. Exponential distribution is often used to model the amount of time until an event occurs, such as machine failure in this case.
(a) To find the average number of machines not in use, we need to consider the rate of machine breakdown and repair. A machine works an average of 10 hours before failure, which translates to a failure rate of 1/10. A single repairman can fix a machine in an average of 8 hours, meaning a rate of 1/8 per repairman, or 1/4 for two repairmen combined. As there are three machines, the average number of machines in use is the ratio of the arrival rate to the service rate: (1/10) / (1/4) = 2.5 machines. This implies that on average, 0.5 machines are not in use.
(b) Both repairmen will be busy when there are at least two machines that require fixing. The proportion of time in which this is the case is obtained by calculating the probability that the number of machines failed is two or more. This is a problem of queueing theory, in particular an M/M/2 queue. The formula for this probability is P(X >= k) = (1 - rho) * rho^(k) / (1 - rho^(c+1)), where rho = arrival rate / service rate, k = 2, and c = 2 (service channels). Substituting rho = 2.5, we obtain P(X >= 2) = 0.64, meaning that the repairmen are both busy 64% of the time.
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A researcher wants to find a 90% confidence interval for the population proportion of those who support additional handgun control. She collects an SRS of 80 people, 50 of whom say they support additional controls. Which of these is the correct confidence interval?a. (.52, .73)b. (.54, .71)c. (.49, .76)d. (.51, .75)e. (.58, .68)
Answer: b. (.54, .71)
Step-by-step explanation:
Confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
,where [tex]\hat{p}[/tex] = sample proportion
z= Critical z-value
n= sample size.
Let p be the proportion of people who support additional handgun control.
As per given , we have
n= 80
[tex]\hat{p}=\dfrac{50}{80}=0.625[/tex]
Critical z-value for 90% confidence interval is 1.645
Now , a 90% confidence interval for the population proportion of those who support additional handgun control will become:
[tex]0.625\pm (1.645)\sqrt{\dfrac{0.625(1-0.625)}{80}}[/tex]
[tex]=0.625\pm (1.645)\sqrt{0.0029296875}[/tex]
[tex]=0.625\pm 0.089\\\\=(0.625-0.089, 0.625+0.089)\\\\=(0.536,\ 0.714)\approx(0.54,\ 0.71)[/tex]
So the correct answer is : b. (.54, .71)
A factory produces plate glass with a mean thickness of 4 mm and a standard deviation of 1.1 mm. A simple random sample of 100 sheets of glass is to be measured, and the mean thickness of the 100 sheets is to be computed. What is the probability that the average thickness of the 100 sheets is less than 3.83 mm? . Round your answers to 5 decimal places.
Answer:
[tex]P(\bar X<3.83)=0.06117[/tex]
Step-by-step explanation:
1) 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 thickness of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(4,1.1)[/tex]
Where [tex]\mu=4[/tex] and [tex]\sigma=1.1[/tex]
And 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(4,\frac{1.1}{\sqrt{100}})[/tex]
2) Solution to the problem
We are interested on this probability
[tex]P(\bar X<3.83)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X<3.83)=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{3.83-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]
[tex]=P(Z<\frac{3.83-4}{\frac{1.1}{\sqrt{100}}})=P(Z<-1.545)[/tex]
And in order to find this probability we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(Z<-1.545)=0.06117[/tex]
And the excel formula to calculate it would be:
"=NORM.DIST(-1.545,0,1,TRUE)"
The probability is 49.20%
The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation.\\\\For\ a\ sample\ size\ n:\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }[/tex]
Given that n = 100, μ = 4 mm, σ = 1.1 mm
For x < 3.83 mm:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\z=\frac{3.83-4}{1.1/\sqrt{100} } =-0.0187[/tex]
P(x < 3.83) = P(z < -0.0187) = 0.4920 = 49.20%
From the normal distribution table, the probability that the average thickness of the 100 sheets is less than 3.83 mm is 49.20%
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Student scores on exams given by a certain instruc-tor have mean 74 and standard deviation 14. Thisinstructor is about to give two exams, one to a classof size 25 and the other to a class of size 64.(a)Approximate the probability that the averagetest score in the class of size 25 exceeds 80.(b)Repeat part (a) for the class of size 64.(c)Approximate the probability that the averagetest score in the larger class exceeds that ofthe other class by over 2.2 points.(d)Approximate the probability that the averagetest score in the smaller class exceeds that ofthe other class by over 2.2 points.
Answer:
Step-by-step explanation:
Given that Student scores on exams given by a certain instruc-tor have mean 74 and standard deviation 14.
Group I X Group II Y
Sample mean 74 74
n 25 64
Std error (14/sqrtn) 2.8 1.75
a) P(X>80) =[tex]1-0.9839\\= 0.0161[/tex]
b) P(Y>80) = [tex]1-0.9997\\=0.0003[/tex]
c) X-Y is Normal with mean = 0 and std deviation = [tex]\sqrt{2.8^2+1.75^2} \\=3.302[/tex]
P(X-Y>2.2) = [tex]1-0.8411\\=0.1589[/tex]
d) [tex]P(\bar x -\bar Y>2.2) = 0.1589[/tex]
Records at the UH library show that 12% of all UH students check out books on history, 28% of all UH students check out books on science, and 6% check out books on both history and science. What is the probability that a randomly selected UH student checks out a history book or a science book or both?
Answer:
There is a 34% probability that a randomly selected UH student checks out a history book or a science book or both.
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a UH student checks out books on history.
B is the probability that a UH students checks out books on science.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a UH student checks a book on history but not on science and [tex]A \cap B[/tex] is the probability that a UH student checks books both on history and science.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
What is the probability that a randomly selected UH student checks out a history book or a science book or both?
[tex]P = a + b + (A \cap B)[/tex]
We start finding these values from the intersection.
6% check out books on both history and science. So [tex]A \cap B = 0.06[/tex]
28% of all UH students check out books on science. So [tex]B = 0.28[/tex]
[tex]B = b + (A \cap B)[/tex]
[tex]0.28 = b + 0.06[/tex]
[tex]b = 0.22[/tex]
12% of all UH students check out books on history
[tex]A = a + (A \cap B)[/tex]
[tex]0.12 = a + 0.06[/tex]
[tex]a = 0.06[/tex]
So
[tex]P = a + b + (A \cap B) = 0.06 + 0.22 + 0.06 = 0.34[/tex]
There is a 34% probability that a randomly selected UH student checks out a history book or a science book or both.
A researcher determines thatχ2 = 3.76to test for significance for a phi correlation coefficient. What was the decision for this phi correlation test?a) Retain the null hypothesis.b) Reject the null hypothesis c) There is not enough information to answer this question.
Answer:
C. There is not enough information to answer this question
Step-by-step explanation:
Conclusion cannot be made whether to retain or reject the null hypothesis because the information given is not sufficient
A marketing research company desires to know the mean consumption of milk per week among males over age 25. They believe that the milk consumption has a mean of 2.5 liters, and want to construct a 85% confidence interval with a maximum error of 0.07 liters. Assuming a variance of 1.21 liters, what is the minimum number of males over age 25 they must include in their sample? Round your answer up to the next integer.
In this exercise we have to use the knowledge of variance to calculate the value of n, so we have that:
the sample is n=306
Organizing the information given in the statement we have that:
Mean of milk consumption = 2.5litresMaximum error E = 0.07Variance S = 1.21 litresConfidence interval of 85%So given by the equation we have:
[tex]Z' = t(0.075)= 1.44\\n = (Z'*S/E)^2\\n = ( 1.44 * 0.85/0.07)^2\\n = (17.4857)^2\\n = 305.75\\n = 306[/tex]
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The minimum number of males over age 25 they must include in their sample is 512.
Given, Desired confidence level: 85%
Maximum error (E): 0.07 liters
Variance ([tex]\sigma^{2}[/tex]): 1.21 liters
Standard deviation ([tex]\(\sigma\)[/tex]): [tex]\(\sigma\)[/tex] = [tex]\sqrt{1.21}[/tex] = 1.1
Z-value for 85% confidence level (lookup Z-value for 0.425 in the standard normal distribution): [tex]\[ Z \approx 1.44 \][/tex]
n= [tex]\left(\frac{Z \sigma}{E}\right)^2[/tex]
[tex]\[ n = \left(\frac{1.44 \times 1.1}{0.07}\right)^2 \][/tex]
n= (1.584/0.07)² = 511.986
n = 512
Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test of mu = 100 vs not = 100 Variable N Mean StDev SE Mean 95% CI (Lower) 95% CI (Upper) T X 19 98.77 4.77 Enter your answer; SE Mean Enter your answer; 95% CI (Lower) Enter your answer; 95% CI (Upper) Enter your answer; T (a) How many degrees of freedom are there on the t-statistic? Enter your answer in accordance to the item a) of the question statement (b) What is your conclusion if ? Choose your answer in accordance to the item b) of the question statement(c) What is your conclusion if the hypothesis is versus ? Choose your answer in accordance to the item c) of the question statement
Final answer:
The degrees of freedom for the t-statistic are 18. The conclusion cannot be determined without the p-value. If the null hypothesis is true, there is not enough evidence to support the alternative hypothesis.
Explanation:
(a) The degrees of freedom for the t-statistic are found by subtracting 1 from the sample size. In this case, the sample size is 19 so the degrees of freedom would be 19 - 1 = 18.
(b) If the p-value is less than the alpha level (typically 0.05), we reject the null hypothesis. In this case, the p-value is not provided, so we cannot determine the conclusion.
(c) If the null hypothesis is true, we would not reject it and conclude that there is not enough evidence to support the alternative hypothesis.
Solve for x.
x + 8 = 12
Answer:
Step-by-step explanation:
move constant to the right and change the sign
X=12-8
X=4
With respect to the number of categories, k, when would a multinomial experiment be identical to a binomial experiment?
a. k = 2
b. k = 3
c. k = 4
d. k = 1
Answer:
Option A) k = 2
Step-by-step explanation:
Multimonial Experiment
A multimonial experiment is an experiment with n repeated trials and each trial has a discrete number of possible outcomes.
Binomial Experiment
Binomial experiment is an experiment with n repeated trials and each trial has only two possible outcomes.
Thus, if k represents the number of possible outcomes, then for k = 2, a multimonial experiment will become a binomial experiment.
Option A) k = 2
Use Green's Theorem to calculate the circulation of F =2xyi around the rectangle 0≤x≤8, 0≤y≤3, oriented counterclockwise.
Green's theorem says the circulation of [tex]\vec F[/tex] along the rectangle's border [tex]C[/tex] is equal to the integral of the curl of [tex]\vec F[/tex] over the rectangle's interior [tex]D[/tex].
Given [tex]\vec F(x,y)=2xy\,\vec\imath[/tex], its curl is the determinant
[tex]\det\begin{bmatrix}\frac\partial{\partial x}&\frac\partial{\partial y}\\2xy&0\end{bmatrix}=\dfrac{\partial(0)}{\partial x}-\dfrac{\partial(2xy)}{\partial y}=-2x[/tex]
So we have
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_D-2x\,\mathrm dx\,\mathrm dy=-2\int_0^3\int_0^8x\,\mathrm dx\,\mathrm dy=\boxed{-192}[/tex]
The circulation of the vector field 2xyi around the given rectangle, as computed via Green's Theorem, is 0 due to the curl of F being 0.
Explanation:To use Green's Theorem to calculate the circulation around a rectangle, first we should realize that Green's Theorem states that the line integral around a simple closed curve C of F.dr is equal to the double integral over the region D enclosed by C of the curl of F. Here, F is the vector field defined as 2xyi. The given rectangle is oriented counterclockwise and the values of x and y are given as 0≤x≤8 and 0≤y≤3 respectively. The line integral denotes the circulation of the field.
The circulation is thus the double integral over the rectangle of ∇ x F. But in this case, since F = 2xyi, we get ∇ x F = 0. Hence, the circulation of F around the given rectangle is 0.
Learn more about Green's Theorem here:https://brainly.com/question/35137715
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