Answer:
B. Rational functions
Step-by-step explanation:
The partial fraction decomposition is used for functions there are described by fractions, and for which the substitution method is not possible. These are rational functions, in which both the numerator and the denominator are polynomials.
So the correct answer is:
B. Rational functions
Partial fraction decomposition is used to simplify complex rational functions to make them easier to integrate. Rational function is the correct answer.
Explanation:The correct answer is B. Rational functions. Partial fraction decomposition is a mathematical technique used essentially to simplify complex rational functions. A rational function is a function that can be defined as the ratio of two polynomials. By breaking down a complex rational function into simpler fractions (which is what partial fraction decomposition does), integration becomes more manageable. For example, it's easier to integrate simple fractions like 1/x or 2/x^2, which would be the result of a partial fraction decomposition, than complex, intertwined expressions.
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Suppose that a random sample of 10 newborns had an average weight of 7.25 pounds and sample standard deviation of 2 pounds. a. Test the hypothesis that the average birthweights have decreased at a 5% significance level. Report the p-value of the test. b. Test the hypothesis that the variance of the birth weights have increased at a 1% significance level.
Answer:
[tex]z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565[/tex]
[tex]p_v =P(Z<-0.565)=0.286[/tex]
a) If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.
b) [tex]\chi^2 =\frac{10-1}{1.96} 4 =18.367[/tex]
[tex]p_v =P(\chi^2 >18.367)=0.0311[/tex]
If we compare the p value and the significance level provided we see that [tex]p_v >\alpha[/tex] so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.
Step-by-step explanation:
Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"
1) Data given and notation
[tex]\bar X=7.25[/tex] represent the sample mean
[tex]s=1.2[/tex] represent the sample standard deviation
[tex]\sigma=1.4[/tex] represent the population standard deviation
[tex]n=10[/tex] sample size
[tex]\mu_o =7.5[/tex] represent the value that we want to test
[tex]\alpha=0.05,0.01[/tex] represent the significance level for the hypothesis test.
z 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 true mean is less than 7.5, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 7.5[/tex]
Alternative hypothesis:[tex]\mu < 7.5[/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{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565[/tex]
4)P-value
Since is a left tailed test the p value would be:
[tex]p_v =P(Z<-0.565)=0.286[/tex]
5) Conclusion
Part a
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.
Part b
A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"
[tex]n=10[/tex] represent the sample size
[tex]\alpha=0.01[/tex] represent the confidence level
[tex]s^2 =4 [/tex] represent the sample variance obtained
[tex]\sigma^2_0 =1.96[/tex] represent the value that we want to test
Null and alternative hypothesis
On this case we want to check if the population variance increase, so the system of hypothesis would be:
Null Hypothesis: [tex]\sigma^2 \leq 1.96[/tex]
Alternative hypothesis: [tex]\sigma^2 >1.96[/tex]
Calculate the statistic
For this test we can use the following statistic:
[tex]\chi^2 =\frac{n-1}{\sigma^2_0} s^2[/tex]
And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.
[tex]\chi^2 =\frac{10-1}{1.96} 4 =18.367[/tex]
Calculate the p value
In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:
[tex]p_v =P(\chi^2 >18.367)=0.0311[/tex]
In order to find the p value we can use the following code in excel:
"=1-CHISQ.DIST(18.367,9,TRUE)"
Conclusion
If we compare the p value and the significance level provided we see that [tex]p_v >\alpha[/tex] so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.
Why is the absolute value used to find distances on a coordinate plane? Choose the correct answer below.
A. Absolute value is the distance between 2 points on a number line, so it gives the distance between any 2 points.
B. Absolute value is the distance from 0 to a point on a number line, so it gives the distance relative to 0 on the coordinate plane.
C. Absolute value is the distance between 2 points on a number line, so it gives the distance relative to 0 on the coordinate plane.
D. Absolute value is the distance from 0 to a point on a number line, so it gives the distance between any 2 points.
Answer: B
Step-by-step explanation:
Answer:
the real answer is2 or d
A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages are labeled as 8 ounces, the company wants the packages to contain a mean of 8.17 ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is selected perodically, and the packaging process is stopped if there is evidence that the mean amount packaged is different from 8.17 ounces. Suppose that in a particular sample of 50 packages the mean amount dispensed is 8.159 ounces, with a sample standard deviation of 0.051 ounce.
a. Is there evidence that the population mean amount is different from *.17 ounces? (Use a 0.05 level of significance.)
b. Determine the p-value and interpret its meaning.
Answer:
a) If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly different from 8.17 at 5% of signficance.
b) Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{(49)}<-1.525)=0.067[/tex]
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=8.159[/tex] represent the mean weight for the sample
[tex]s=0.051[/tex] represent the sample standard deviation
[tex]n=50[/tex] sample size
[tex]\mu_o =8.17[/tex] represent the value that we want to test
[tex]\alpha=0.05[/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 different from 8.17, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 8.57[/tex]
Alternative hypothesis:[tex]\mu \neq 8.57[/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{8.159-8.17}{\frac{0.051}{\sqrt{50}}}=-1.525[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=50-1=49[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{(49)}<-1.525)=0.067[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly different from 8.17 at 5% of signficance.
A major airline is concerned that the waiting time for customers at its ticket counter may be exceeding its target average of 190 seconds. To test this, the company has selected a random sample of 100 customers and times them from when the customer first arrives at the checkout line until he or she is at the counter being served by the ticket agent. The mean time for this sample was 202 seconds with a standard deviation of 28 seconds. Given this information and the desire to conduct the test using an alpha level of 0.02, which of the following statements is true?
A) The chance of a Type II error is 1 - 0.02 = 0.98.
The test to be conducted will be structured as a two-tailed test.
The test statistic will be approximately t = 4.286, so the null hypothesis should be rejected.
The sample data indicate that the difference between the sample mean and the hypothesized population mean should be attributed only to sampling error.
Answer:
a) Statement is true
b) Statement incorrect. Company goal is to attend customer at 190 second at the most. So test should be one tail test (right)
c) We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error
Step-by-step explanation:
a) as the significance level is = 0,02 that means α = 0,02 (the chance of error type I ) and the β (chance of type error II is
1 - 0.02 = 0,98
b) The company establishe in 190 seconds time for customer at its ticket counter (if this time is smaller is excellent ) company is concerned about bigger time because that could be an issue for customers. Therefore the test should be a one test-tail to the right
c) test statistic
Hypothesis test should be:
null hypothesis H₀ = 190
alternative hypothesis H₀ > 190
t(s) = ( μ - μ₀ ) / s/√n ⇒ t(s) = ( 202 - 190 )/(28/√100 )
t(s) = 12*10/28
t(s) = 4.286
That value is far away of any of the values found for 99 degree of fredom and between α ( 0,025 and 0,01 ). We have to reject H₀, and differences between sample mean and the hypothesized population mean should not be attributed only to sampling error
If we look table t-student we will find that for 99 degree of freedom and α = 0.02.
In a recent survey, 80% of the community favored building a police substation in their neighborhood. If 15 citizens are chosen, what is the mean number favoring the substation?
Answer:
The mean number favoring the substation is 12 citizens.
Step-by-step explanation:
For each citizen surveyed, there are only two possible outcomes. EIther they favored building a police substation in their neighborhood, or they opposed. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, with p probability, and X can only have two outcomes.
Has an expected value of:
[tex]E(X) = np[/tex]
In this problem, we have that:
[tex]n = 15, p = 0.8[/tex]
E(X) = 15*0.8 = 12.
The mean number favoring the substation is 12 citizens.
Final answer:
The mean number of citizens out of 15 who favor the construction of a police substation is 12, calculated by multiplying the sample size (15) by the probability of favoring the substation (0.80).
Explanation:
To find the mean number favoring the substation, we'll use the probability of a citizen favoring the substation to calculate the expected value in a sample of 15 citizens. Since 80% of the community favored building a police substation, the mean number favoring the substation can be computed by multiplying the sample size by the probability of favoring the substation.
Mean = Sample Size × Probability of Favoring the Substation
Mean = 15 × 0.80
Mean = 12
Therefore, the mean number of citizens out of 15 who favor the construction of the police substation is 12.
A curve in polar coordinates is given by: r=9+3cosθ. Point P is at θ=21π/18 .?
a. Find polar coordinate r for P , with r>0 and π<θ<3π/2 .
b. Find Cartesian coordinates for point P.
c. How may times does the curve pass through the origin when 0<θ<2π?
Answer:
Step-by-step explanation:
Given that a curve in polar coordinates is given by:
r=9+3cosθ
a) At point P, we have
[tex]θ=\frac{21\pi}{18}[/tex]
Substitute to get
[tex]r=9+cos \frac{21\pi}{18}\\=9.3932[/tex]
b) Cartesian coordinate is
[tex]x= rcos \theta=3.6934\\y =r sin \tjeta =8.6365[/tex]
c) At the origin r =0
when r =0
we have
[tex]9+3cos\theta=0\\cos\theta =-3[/tex]
Since cos cannot take values as -3 it doe snot pass through origin.
In summary, the polar coordinate r for point P is approximately 9.59. The Cartesian coordinates of P are approximately (0.99, -9.43). The given curve passes through the origin twice in the interval 0<θ<2π.
Explanation:This question is about a curve in polar coordinates. The curve is given by r=9+3cosθ. Let's explore the three sub-questions in order:
a. Find polar coordinate r for P
Because Point P is at θ=21π/18, we can plug this into the equation r=9+3cosθ. This gives r = 9 + 3cos(21π/18) = 9.59.
b. Find Cartesian coordinates for point P.
To convert from polar to Cartesian coordinates, we use the equations x = rcosθ and y = rsinθ. Plugging in our values, we get x = 9.59cos(21π/18) and y = 9.59sin(21π/18). Calculating these gives us x approximately equal to 0.99 and y approximately equal to -9.43. So the Cartesian coordinates for point P are (0.99, -9.43).
c. How many times does the curve pass through the origin when 0<θ<2π?
The curve reaches the origin when r is 0. From the curve equation r = 9 + 3cosθ = 0, we can see the curve crosses the origin twice, at θ = 2π/3 and θ = 4π/3.
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When consumers apply for credit, their credit is rated using FICO (Fair, Isaac, and Company) scores. Credit ratings are given below for a sample of applicants for car loans.
661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706
a. Use the sample data to construct a 99% confidence interval for the mean FICO score of all applicants for credit.
b. If one bank requires a credit rating of at least 620 for a car loan, does it appear that almost all applicants will have suitable credit ratings? Why or why not?
Answer:
a) The 99% confidence interval would be given by (589.588;731.038)
b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.
Step-by-step explanation:
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".
The data is:
661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706
Part a
Compute the sample mean and sample standard deviation.
In order to calculate the mean and the sample deviation we need to have on mind the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex]
=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)
On this case the average is [tex]\bar X= 660.313[/tex]
=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)
The sample standard deviation obtained was s=95.898
Find the critical value t* Use the formula for a CI to find upper and lower endpoints
In order to find the critical value we need to take in count that our sample size n =16<30 and on this case 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 99% of confidence, our significance level would be given by [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2 =0.005[/tex]. The degrees of freedom are given by:
[tex]df=n-1=16-1=15[/tex]
We can find the critical values in excel using the following formulas:
"=T.INV(0.005,15)" for [tex]t_{\alpha/2}=-2.95[/tex]
"=T.INV(1-0.005,15)" for [tex]t_{1-\alpha/2}=2.95[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
And if we find the limits we got:
[tex]660.313- 2.95\frac{95.898}{\sqrt{16}}=589.588[/tex]
[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]
So the 99% confidence interval would be given by (589.588;731.038)
Part b
If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.
Suppose we have two bags with the numbers. Each bag has a total of 100 numbers. In the first bag there are 31 lucky numbers, in the second bag there are 18 lucky numbers. We want to add one more bag with 100 numbers to decrease the probability that a randomly selected number from a random bag is the lucky number. How many lucky numbers should be in the third bag?
Answer:
There should be at most 24 lucky numbers in the third bag.
Step-by-step explanation:
Initially, there are 200 numbers. Two bags with 100 each. There are 31+18 = 49 lucky numbers. So there is a 49/200 = 0.245 probability that a randomly selected number from a random bag is the lucky number.
Now with 300 numbers, we want this probability to be lower than 24.5%. So we should solve the following rule of three:
200 - 49
300 - x
[tex]200x = 300*49[/tex]
[tex]x = 1.5*49[/tex]
[tex]x = 73.5[/tex]
With the third bag, the probability will be the same if 73.5-49 = 24.5 lucky numbers are added. So there should be at most 24 lucky numbers in the third bag.
A random sample of 11fields of spring wheat has a mean yield of 20.2bushels per acre and standard deviation of 5.19 bushels per acre. Determine the 99% confidence interval for the true mean yield. Assume the population is approximately normal.Step 1 of 2:Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.Step 2 of 2:Construct the 99% confidence interval. Round your answer to one decimal place
Answer:
1. [tex]t_{0.01/2,(10)}=3.169[/tex]
2. Confidence interval = [15.2, 25.2]
Step-by-step explanation:
X = 20.2
S = 5.19
n = 11
Step 1 of 2: Critical value
α = 1 - 0.99 = 0.01
df = 11 - 1 = 10
Looking at t distribution table with α = 0.01 and df = 10, we find
[tex]t_{0.01/2,(10)}=3.169[/tex]
Step 2 of 2: 99% confidence interval
[tex]std-err=\frac{S}{\sqrt{n}}=\frac{5.19}{\sqrt{11}}=1.5648[/tex]
[tex]X+t_{0.01/2,10}*std-err[/tex]
20.2 + 3.169*1.5648
20.2 + 4.9595
25.2
[tex]X-t_{0.01/2,10}*std-err[/tex]
20.2 - 3.169*1.5648
20.2 - 4.9595
15.2
Confidence interval = [15.2, 25.2]
Hope this helps!
How are the degrees of freedom computed for each test listed below?(a) a chi-square goodness-of-fit test df = k − 1 df = N − 1 df = n − 1 df = (n1 − 1)(n2 − 1) df = (k1 − 1)(k2 − 1)(b) a chi-square test for independence df = k − 1 df = n − 1 df = N − 1 df = (k1 − 1)(k2 − 1) df = (n1 − 1)(n2 − 1)
Answer:
a) [tex]df= k-1[/tex]
b) [tex]df= (k_1-1)(k_2-1)[/tex]
Step-by-step explanation:
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Solution to the problem
Let's define some notation:
N= Total number of individuals for a population
n = Total of individuals selected from a sample
n1= number of objects/individuals with characteristic 1
n2= number of objects/individuals with characteristic 2
k1= number of levels for the variable or factor 1
k2= number of levels for the variable or factor 2
Part a
For a chi square goodness of fit test the degrees of freedom are given by:
[tex]df= k-1[/tex]
Where k represent the total number of categories on the godness of fit test.
Part b
For a chi square test of independence the degrees of freedom are given by:
[tex]df= (k_1-1)(k_2-1)[/tex]
Where k1= number of levels for the variable or factor 1, k2= number of levels for the variable or factor 2 for the chi square test for independence.
In statistical analysis, degrees of freedom (df) are calculated differently based on the type of test. For a chi-square goodness-of-fit test, df equals the number of categories minus one. For a chi-square test for independence, df equals the product of subtracting one from the number of columns and rows.
Explanation:The degrees of freedom (df) play an important role in many statistical tests, including the chi-square tests for goodness-of-fit and independence. These degrees of freedom are calculated differently based on the type of test. In a chi-square goodness-of-fit test, the degrees of freedom is calculated as df = k - 1, where k represents the number of categories or groups in the dataset.
On the other hand, for a chi-square test for independence, the degrees of freedom are calculated using the formula df = (number of columns - 1)(number of rows - 1), where columns and rows represent the different categories or groups present in a contingency table used in the test of independence. In this context, the null hypothesis for this test states that the two factors being considered are independent.
It's important to remember that degrees of freedom often represent the quantity of the 'free' or varying elements in your dataset and directly impact the test results and conclusions. So, having an accurate calculation is crucial in your statistical analysis.
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A sample of 41 observations is selected from one population with a population standard deviation of 3.4. The sample mean is 100.0. A sample of 45 observations is selected from a second population with a population standard deviation of 5.6. The sample mean is 98.4. Conduct the following test of hypothesis using the 0.05 significance level. H0 : μ1 = μ2 H1 : μ1 ≠ μ2
a. One or two tail?
b. State the decision rule. The decision rule is to reject H0 if z is (Outside, Inside) the interval (____,____).
c. Compute value of Test statistic.
d. Reject or Do not reject?
e. what is the p value?
Answer:
a) If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.
b) (-1.9886, 1.9886)
c) [tex]t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617[/tex]
d) [tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.
e) [tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]
We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"
Step-by-step explanation:
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 = \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2\neq 0[/tex]
Our notation on this case :
[tex]n_1 =41[/tex] represent the sample size for group 1
[tex]n_2 =45[/tex] represent the sample size for group 2
[tex]\bar X_1 =100[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =98.4[/tex] represent the sample mean for the group 2
[tex]s_1=3.4[/tex] represent the sample standard deviation for group 1
[tex]s_2=5.6[/tex] represent the sample standard deviation for group 2
Part a
If we see the alternative hypothesis we see that we are conducting a bilateral test or two tail.
Part b
On this case since the significance level is 0.05 and we are conducting a bilateral test we have two critical values, and we need on each tail of the distribution [tex]\alpha/2 = 0.025[/tex] of the area.
The distribution on this cas since we don't know the population deviation for both samples is the t distribution with [tex]df=n_1+n_2 -2= 41+45-2=84[/tex] degrees of freedom.
We can use the following excel codes in order to find the critical values:
"=T.INV(0.025,84)", "=T.INV(1-0.025,84)"
And we got: (-1.9886, 1.9886)
Part c
The statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}[/tex]
And now we can calculate the statistic:
[tex]t=\frac{(100 -98.4)-(0)}{\sqrt{\frac{3.4^2}{41}}+\frac{5.6^2}{45}}=1.617[/tex]
The degrees of freedom are given by:
[tex]df=41+45-2=84[/tex]
Part d
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 is NOT significantly different than the mean for the group 2.
Part e
[tex]p_v =2*P(t_{84}>1.617) =0.1096[/tex]
We can use the following excel code to calculate it: "=2*(1-T.DIST(1.617;84;TRUE))"
A researcher wants to know if the average time in jail for robbery has increased from what it was several years ago when the average sentence was 7 years. He obtains data on 400 recent robberies and finds an average time served of 7.5 years. If we assume the standard deviation is 3 years, perform a significance test (calculating the P-value) at the 0.05 level.
Answer:
So, the interval is : (7.206,7.794)
Step-by-step explanation:
The mean μ = 7.5
Standard deviation σ=3
n = 400
At 95% confidence interval, the z score is 1.96
[tex]7.5+1.96(\frac{3}{\sqrt{400} } )[/tex]
And [tex]7.5-1.96(\frac{3}{\sqrt{400} } )[/tex]
7.5+0.294 and 7.5-0.294
So, the interval is : (7.206,7.794)
dentify the type I error and the type II error that correspond to the given hypothesis. The percentage of college students who own cars is equal to 35 %. Identify the type I error. Choose the correct answer below. A. Reject the null hypothesis that the percentage of college students who own cars is equal to 35 % when that percentage is actually different from 35 %. B. Reject the null hypothesis that the percentage of college students who own cars is equal to 35 % when that percentage is actually equal to 35 %. C. Fail to reject the null hypothesis that the percentage of college students who own cars is equal to 35 % when that percentage is actually different from 35 %. D. Fail to reject the null hypothesis that the percentage of college students who own cars is equal to 35 % when the percentage is actually equal to 35 %.
Answer:
B
Step-by-step explanation:
Type I error is basically rejection of the null hypothesis when the null hypothesis is true. In the given scenario the null hypothesis consists of the percentage of students who own cars is 35%. Hence the type I error would be rejection of null hypothesis that the percentage of students who own cars is 35% while the percentage is 35%.
Final answer:
A Type I error is rejecting the null hypothesis when it's true, and a Type II error is failing to reject the null hypothesis when it's false. In this case, Type I error is option B and Type II error is option C.
Explanation:
When performing hypothesis testing, there is potential to make two types of errors: Type I error and Type II error. A Type I error occurs when we reject the null hypothesis even though the null hypothesis is actually true. In the context of the question, the Type I error would correspond to option B: Reject the null hypothesis that the percentage of college students who own cars is equal to 35% when that percentage is actually equal to 35%.
In contrast, a Type II error happens when we fail to reject the null hypothesis whereas the null hypothesis is false. It is correctly identified by option C: Fail to reject the null hypothesis that the percentage of college students who own cars is equal to 35% when that percentage is actually different from 35%.
The equation of a circle is (x + 6)^2 + (y - 4)^2 = 16. The point (-6, 8) is on the circle.
What is the equation of the line that is tangent to the circle at (-6, 8)?
Answer:
y = 8 is the equation of tangent.
Step-by-step explanation:
The equation of the tangent to the circle at (-6,8) is of the form:
y = mx + c
where m is the slope of the tangent and c is the y-intercept.
The point (-6,8) lies on the circle and the tangent line as well.
Hence (-6,8) satisfies the line equation:
8 = m(-6) + c ⇒ c-6m = 8 -------------1
We know that slope of two perpendicular lines are related as:
[tex]m_{1}\times m_{2}=-1[/tex]
At any point on the circle, the normal line at a point is always perpendicular to the tangent line at that point.
Hence :
[tex]m_{normal} \times m_{tangent}=-1[/tex]
We can find the slope of the normal at point (-6,8) as it passes through the centre of the circle (-6,4) by using the two-points formula for slope.
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
[tex]=\frac{8-4}{-6+6}[/tex]
= ∞
Slope of the normal is infinity and hence slope of tangent is -1/∞ = 0
Hence m=0
Putting m=0 in equation 1 we get:
c = 8
The equation of tangent line at (-6,8) is:
y = 8
Answer:
y = 8
Step-by-step explanation:
The equation of the tangent to the circle at (-6,8) is of the form:
y = mx + c
where m is the slope of the tangent and c is the y-intercept.
The point (-6,8) lies on the circle and the tangent line as well.
Hence (-6,8) satisfies the line equation:
8 = m(-6) + c ⇒ c-6m = 8 -------------1
Emma's square patio below has been area=31 sq ft. She decides to decrease one dimension by 1 foot and decreases the other dimension by 4 feet. DO NOT USE DECIMAL APPROXIMATIONS. What are the dimensions?
Answer:
4.57ft by 1.57 ft
Step-by-step explanation:
We are given that
Emma's square patio has been area=31 sq.ft
One dimension decrease by 1 foot and other dimension decrease by 4 feet.
We have to find the new dimensions of Emma's patio.
Let x be the side of Emma's square patio
We know that
Area of square=[tex]x^2[/tex]
[tex]x^2=31[/tex]
[tex]x=\sqrt{31}=5.57[/tex] ft
One dimension=[tex]5.57-1=4.57[/tex] ft
other dimension=[tex]5.57-4=1.57 ft [/tex]
Hence, the new dimension of Emma's patio is given by
4.57ft by 1.57 ft
g A coffee-dispensing machine is supposed to deliver eight ounces of liquid into each paper cup, but a consumer believes that the actual mean amount is less. The critical value for z for a one-tailed test with the tail in the left end is minus1.645 and the obtained value is minus1.87. The appropriate decision is ________.
Answer: Reject the eight- ounces claim.
Step-by-step explanation:
For left tailed test , On a normal curve the rejection area lies on the left side of the critical value.
It means that if the observed z-value is less than the critical value then it will fall into the rejection region other wise not.
As per given ,
Objective : A coffee-dispensing machine is supposed to deliver eight ounces of liquid or less.
Then ,
[tex]H_0: \mu=8\\\\H_a: \mu<8[/tex] , since alternative hypothesis is left-tailed thus the test is an left-tailed test.
the critical value for z for a one-tailed test with the tail in the left end is -1.645 and the obtained value is -1.87.
Clearly , -1.87 < -1.645
⇒ -1.87 falls under rejection region.
⇒ Decision : Reject null hypothesis.
i.e. we reject the eight- ounces claim.
Leroy and Fred play chess at a club every Wednesday. The probability that Leroy will lose is .3, that he will stalemate is .5, and that he will win is .2. The probability that Fred will lose is .25, that he will stalemate is .4, and that he will win is .35. What is the probability that at least one of them wins on a given Wednesday?
A .55B .07C .45D .48E .62
Answer:
A .55
Step-by-step explanation:
Given that Leroy and Fred play chess at a club every Wednesday. The probability that Leroy will lose is .3, that he will stalemate is .5, and that he will win is .2. The probability that Fred will lose is .25, that he will stalemate is .4, and that he will win is .35.
lose stale win
Leroy 0.3 0.5 0.2
Fred 0.25 0.4 0.35
Probability that atleast one wins = Prob Leroy wins + Prob fred wins - prob both wins
(using addition theorem on probability)
But prob of both winning is 0
Hence required prob = 0.2 + 0.35 = 0.55
Option a is right
Answer:
0.48
Step-by-step explanation:
A sports researcher is interested in determining if there is a relationship between the number of home team and visiting team wins and different sports. A random sample of 526 games is selected and the results are given below. Find the critical value X 2/0 to test the claim that the number of home team and visiting team wins is independent of the sport. Use a = 0.01.Football basketball soccer baseballHome wins 39 156 25 83Visitor wins 31 98 19 75
Answer:
give me a second to research the answer
Step-by-step explanation:
The National Center for Education Statistics reported that 47% of college students work to pay for tuition and living expenses. Assume that a sample of 450 college students was used in the study. a. Provide a 95% confidence interval for the population proportion of college students who work to pay for tuition and living expenses (to 4 decimals).
Provide a 99% confidence interval for the population proportion of college students who work to pay for tuition and living expenses (to 4 decimals).
The 95% confidence interval for the population proportion of college students who work to pay for tuition is (0.4228, 0.5172) and the 99% confidence interval is (0.4086, 0.5314).
Explanation:To calculate a confidence interval for a population proportion, we use the formula p ± Z*(√((p(1-p))/n)), where p is the sample proportion, Z is the Z-score associated with our desired confidence level, and n is the sample size.
In this case, p = 0.47 (47%), n = 450, and the Z-scores are 1.96 for a 95% confidence interval and 2.58 for a 99% confidence interval.
For the 95% confidence interval, we calculate:
0.47 ± 1.96*(√((0.47*(1-0.47))/450)) = 0.47 ± 0.0472, so the 95% confidence interval is (0.4228, 0.5172).
For the 99% confidence interval, we calculate:
0.47 ± 2.58*(√((0.47*(1-0.47))/450)) = 0.47 ± 0.0614, so the 99% confidence interval is (0.4086, 0.5314).
Learn more about Confidence Interval here:https://brainly.com/question/34700241
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What is the x-intercept of the line described by the equation?
12x - 10y = 60
Write your answer as an ordered pair.
What is the y-intercept of the line described by the equation?
12x - 10y = 60
Write your answer as an ordered pair
Answer:
(0,-6)
Step-by-step explanation:
recall that the y-intercept is simply the point where the line crosses the y-axis at x = 0.
to find the y intercept, we simply substitute x=0 into the equation and solve for y
12x - 10y = 60, when x = 0,
12(0) - 10y = 60
-10y = 60
y = 60 / (-10)
y = -6
hence the coordinate of the y - intercept is (0,-6)
For a field trip the school bought 39 sandwiches for 4.35 each and 42 bags of chips for 15 cent each how much did the school spend in all
Answer:
Step-by-step explanation:
For a field trip, the school bought 39 sandwiches and 42 bags of chips. Each sandwich cost 4.35. That means the cost of 39 sandwiches would be
39 × 4.35 = 169.65
Each bag of chip costs 15 cents. Converting 15 cents to dollars, it becomes
15/100 =$ 0.15
Therefore, the cost of 42 bags of chips would be
42 × 0.15 = 6.3
Total cost of 39 sandwiches and 42 bags of chips would be
169.65 + 6.3 = 175.95
A sample of 60 items from population 1 has a sample variance of 8 while a sample of 40 items from population 2 has a sample variance of 10. If we test whether the variances of the two populations are equal, the test statistic will have a value of
a. 0.8
b. 1.56
c. 1.5
d. 1.25
Answer: a. 0.8
Step-by-step explanation:
When we test whether the variances of the two populations are equal we use F- test.
Test statistic : [tex]F=\dfrac{s_1^2}{s_2^2}[/tex]
, where [tex]s_1^2[/tex]= sample variance from population 1.
[tex]s_2^2[/tex]= sample variance from population 2.
As per given , we have
[tex]s_1^2=8[/tex] and [tex]s_2^2=10[/tex]
Then, If we test whether the variances of the two populations are equal, the test statistic will be [tex]F=\dfrac{8}{10}=0.8[/tex]
Hence, the test statistic will have a value of 0.8 .
Thus , the correct answer is a. 0.8 .
If $x^5 - x^4 x^3 - px^2 qx 4$ is divisible by $(x 2)(x - 1),$ find the ordered pair $(p,q).$
Answer: The required ordered pair (p, q) is (-7, -12).
Step-by-step explanation: Given that (x+2)(x-1) divides the following polynomial f(x) :
[tex]f(x)=x^5-x^4+x^3-px^2+qx+4.[/tex]
We are to find the ordered pair (p,q).
We have the following theorem :
Factor theorem : If (x-a) divides a polynomial h(x), then h(a) = 0.
According to the given information, we can say that (x+2) divides f(x). So, we get
[tex]f(-2)=0\\\\\Rightarrow (-2)^5-(-2)^4+(-2)^3-p(-2)^2+q(-2)+4=0\\\\\Rightarrow -32-16-8-4p-2q+4=0\\\\\Rightarrow 2p+q=-26~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
Also, (x-1) is a factor of f(x). So,
[tex]f(1)=0\\\\\Rightarrow (1)^5-(1)^4+(1)^3-p(1)^2+q(1)+4=0\\\\\Rightarrow 1-1+1-p+q+4=0\\\\\Rightarrow p-q=5~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
Adding equations (i) and (ii), we get
[tex]3p=-21\\\\\Rightarrow p=-7.[/tex]
From equation (ii), we get
[tex]-7-q=5\\\\\Rightarrow q=-12.[/tex]
Thus, the required ordered pair (p, q) is (-7, -12).
The producer of a certain bottling equipment claims that the variance of all its filled bottles is .027 or less. A sample of 30 bottles showed a standard deviation of .2. The p-value for the test is _____.
a. .025
b. between .025 and .05
c. .05
d. between .05 and .01
Answer:
[tex]p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459[/tex]
b. between .025 and .05
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 = 30 sample size
s= 0.2 represent the sample deviation
[tex]\sigma_o =\sqrt{0.027}=0.164[/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=[/tex] significance level
State the null and alternative hypothesis
On this case we want to check if the population standard deviation is less than 0.027, so the system of hypothesis are:
H0: [tex]\sigma \leq 0.027[/tex]
H1: [tex]\sigma >0.027[/tex]
In order to check the hypothesis we need to calculate the 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 value of your test statistic?
Now we have everything to replace into the formula for the statistic and we got:
[tex] t=(30-1) [\frac{0.2}{0.164}]^2 =42.963[/tex]
What is the approximate p-value of the test?
The degrees of freedom are given by:
[tex] df=n-1= 30-1=29[/tex]
For this case since we have a right tailed test the p value is given by:
[tex]p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459[/tex]
And the best option would be:
b. between .025 and .05
An industrial expert claims that the average useful lifetime of a typical car transimssion which comes with ten years warranty is significantly more than 10 years. In order to test this claim, 9 car transmissions are randomly selected and their useful lifetimes are recorded. The sample mean lifetime is 13.5 years and the sample standard deviation is 3.2 years. Assuming that the useful lifetime of a typical car transmission has a normal distribution, based on these sample result, the correct conclusion at 1% significance level for this testing hypotheses problem is:
a. none of these answers.b. Data provides sufficient evidence, at 1% significance level, to reject the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T < -3.281).c. Data provides insufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( Z > 2.896).d. Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.355).e. Data provides insufficient evidence, at 1% significance level, to support the researcher's claim. In addition the p-value (or the observed significance level) is equal to P(Z > 2.896).
Answer:
a. none of these answers
Data provides sufficient evidence, at 1% significance level, to support the expert's claim. In addition the p-value (or the observed significance level) is equal to P( T >3.281)
Step-by-step explanation:
Data given and notation
[tex]\bar X=13.5[/tex] represent the mean height for the sample
[tex]s=3.2[/tex] represent the sample standard deviation for the sample
[tex]n=9[/tex] sample size
[tex]\mu_o =10[/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)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is higher than 10 years, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 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)
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{13.5-10}{\frac{3.2}{\sqrt{9}}}=3.28[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=9-1=8[/tex]
Since is a one side right tailed test the p value would be:
[tex]p_v =P(t_{(8)}>3.281)=0.00558[/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 reject the null hypothesis, so we can't conclude that we have a mean higher than 10 years at 1% of significance.
a. none of these answers
The geowall team measured the strength of paper strips by applying a force until they broke. They tested 5 strips and found an average strength of 12.9 pounds with a standard deviation of 1 pound. Because of the small sample size, we cannot assume a normal distribution. What is the probability that a strip breaks with only 11 pounds of applied force? (a) 0.065 (b) 0.029 (c) 1.07 × 10?5 (d) 0.058 (e) none of the above
Answer:
Step-by-step explanation:
Given
mean [tex]\mu =12.9 Pounds[/tex]
Standard deviation [tex]\sigma =1\ Pound[/tex]
[tex]P(x<11)=P\left ( \frac{x-\mu }{\sigma }< \frac{11-12.9}{1}\right )[/tex]
[tex]P(x<11)=P\left ( z< -1.9\right )[/tex]
From z table
[tex]P(x<11)=0.0289\approx 0.029[/tex]
A popcorn company builds a machine to fill 1 kg bags of popcorn. They test the first hundred bags filled and find that the bags have an average weight of 1,040 grams with a standard deviation of 25 grams. 1.) Fill out the normal distribution curve for this situation. 2.) What percentage of people would receive a bag that had a weight greater than 1115 grams?
To fill out the normal distribution curve, use the average weight and standard deviation. To find the percentage of people who would receive a bag with a weight greater than a certain amount, calculate the Z-score and use a Z-table.
Explanation:To fill out the normal distribution curve for this situation, we can use the given information. The bags have an average weight of 1,040 grams with a standard deviation of 25 grams. This means that the mean of the distribution is 1,040 and the standard deviation is 25. We can plot the normal distribution curve using these values.
To find the percentage of people who would receive a bag that had a weight greater than 1,115 grams, we need to find the area under the normal distribution curve to the right of 1,115. We can calculate this using a Z-score calculator or a Z-table. The Z-score for 1,115 is (1,115 - 1,040) / 25 = 3. From the Z-table, we can find that the area to the right of Z-score 3 is approximately 0.0013. This means that approximately 0.13% of people would receive a bag that weighs more than 1,115 grams.
Maria class recycled 2 2/8 boxes of paper in a month if they recycled another 10 4/8 boxes the next month was in the total amount they recycled
Answer:
12 6/8 or 12 3/4
Step-by-step explanation:
Answer:the total amount that they recycled is 12 3/4 boxes
Step-by-step explanation:
Maria's class recycled 2 2/8 boxes of paper in a month. Converting 2 2/8 boxes of paper in a month to improper fraction, it becomes
18/8 boxes of paper in a month.
They recycled another 10 4/8 boxes the next month. Converting 10 4/8 boxes to improper fraction, it becomes 84/8 boxes.
Therefore, the total amount of boxes that they recycled would be
18/8 + 84/8 = 102/8 boxes
Converting to whole number, it becomes
12 3/4 boxes
A research report claims that mice with an average life span of 32 months will live to be about 40 months old when 40% of the calories in their diet are replaced by vitamins and protein. Is there any reason to believe that muμless than<40 if 70 mice that are placed on this diet have an average life of 3939 months with a standard deviation of 8.8 months? Use a P-value in your conclusion.
Answer:
[tex]t=\frac{39-40}{\frac{8.8}{\sqrt{70}}}=-0.9508[/tex]
[tex]p_v =P(t_{69}<-0.9508)=0.1725[/tex]
If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly less than 40 months.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=39[/tex] represent the sample mean
[tex]s=8.8[/tex] represent the standard deviation for the sample
[tex]n=70[/tex] sample size
[tex]\mu_o =40[/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 determine if the mean is less than 40 months, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 40[/tex]
Alternative hypothesis:[tex]\mu < 40[/tex]
We don't know the population deviation, so for this case 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{39-40}{\frac{8.8}{\sqrt{70}}}=-0.9508[/tex]
Calculate the P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=70-1=69[/tex]
Since is a one-side lower test the p value would be:
[tex]p_v =P(t_{69}<-0.9508)=0.1725[/tex]
Conclusion
If we compare the p value and the significance level given for example [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly less than 40 months.
What is the number of subsets of S= {1, 2, 3…10} that contain five element include 3 or 4 but not both?
Answer:
140
Step-by-step explanation:
To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.
First, let's count the number of subsets that contain the element 3.
Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is [tex]{}_8C_4=70[/tex].
Now, let's count the number of subsets that contain the element 4.
4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in [tex]{}_8C_4=70[/tex] ways.
We conclude that there are 70+70=140 required subsets of S.