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.
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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.
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).
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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.
Consider the case0502 data from Sleuth3. <<< This is the data. Sleuth3 is preloaded into R studio.
Dr Benjamin Spock was tried in Boston for encouraging young men not to register for the draft. It was conjectured that the judge in Spock’s trial did not have appropriate representation of women. The jurors were supposed to be selected by taking a random sample of 30 people (called venires), from which the jurors would be chosen. In the data case0502, the percent of women in 7 judges’ venires are given.
a. Create a boxplot of the percent women for each of the 7 judges. Comment on whether you believe that Spock’s lawyers might have a point.
b. Determine whether there is a significant difference in the percent of women included in the 6 judges’ venires who aren’t Spock’s judge.
c. Determine whether there is a significant difference in the percent of women incuded in Spock’s venires versus the percent included in the other judges’ venires combined. (Your answer to a. should justify doing this.)
Answer:
Consider the following calculations
Step-by-step explanation:
The complete R snippet is as follows
install.packages("Sleuth3")
library("Sleuth3")
attach(case0502)
data(case0502)
## plot
# plots
boxplot(Percent~ Judge, data=case0502,ylab="Values",
main="Boxplots of the Data",col=c(2:7,8),horizontal=TRUE)
# perform anova analysis
a<- aov(lm(Percent~ Judge,data=case0502))
#summarise the results
summary(a)
### we can use the independent sample t test here
sp<-case0502[which(case0502$Judge=="Spock's"),]
nsp<-case0502[which(case0502$Judge!="Spock's"),]
## perform the test
t.test(sp$Percent,nsp$Percent)
The results are CHECK THE IMAGE ATTACHED
b)
> summary(a)
Df Sum Sq Mean Sq F value Pr(>F)
Judge 6 1927 321.2 6.718 6.1e-05 *** as the p value is less than 0.05 , hence there is a significant difference in the percent of women included in the 6 judges’ venires who aren’t Spock’s judge
Residuals 39 1864 47.8
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
c)
t.test(sp$Percent,nsp$Percent)
Welch Two Sample t-test
data: sp$Percent and nsp$Percent
t = -7.1597, df = 17.608, p-value = 1.303e-06 ## as the p value is less than 0.05 , hence we reject the null hypothesis in favor of alternate hypothesis and conclude that there is a significant difference in the percent of women incuded in Spock’s venires versus the percent included in the other judges’ venires combined
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-19.23999 -10.49935
sample estimates:
mean of x mean of y
14.62222 29.49189
Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. An article reports that for a sample of n = 75 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.81 min and the standard deviation was 8.4 min.
Does this data suggest that true average lateral recumbency time under these conditions is less than 20 min? Test the appropriate hypotheses at level of significance 0.10. State the appropriate null and alternative hypotheses.
Answer:
[tex]p_v =P(t_{74}<-1.227)=0.112[/tex]
If we compare the p value and a significance level for example [tex]\alpha=0.1[/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 it's not significantly less than 20 min.
Step-by-step explanation:
Data given and notation
[tex]\bar X=18.81[/tex] represent the average lateral recumbency for the sample
[tex]s=8.4[/tex] represent the sample standard deviation
[tex]n=75[/tex] sample size
[tex]\mu_o =20[/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 left tailed test.
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu \geq 20[/tex]
Alternative hypothesis :[tex]\mu < 20[/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{18.81-20}{\frac{8.4}{\sqrt{75}}}=-1.227[/tex]
The degrees of freedom are given by:
[tex]df=n-1=75-1=74[/tex]
Give the appropriate conclusion for the test
Since is a one side left tailed test the p value would be:
[tex]p_v =P(t_{74}<-1.227)=0.112[/tex]
Conclusion
If we compare the p value and a significance level for example [tex]\alpha=0.1[/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 it's not significantly less than 20 min.
use the intermediate value theorem to determine whether the following equation has a solution or not x^3-3x-1
Answer:
Yes, this equation has a solution. According to Intermediate Value Theorem at least one solution for [0,2]
Step-by-step explanation:
Hi there!
1) Remember a definition.
Intermediate Value Theorem:
If [tex]f[/tex] is continuous on a given closed interval [a,b], and f(a)≠f(b) and f(a)<k<f(b) then there has to be at least one number 'c' between 'a' and 'b', such that f(c)=k
----
(Check the first graph as an example)
2) The Intermediate Value Theorem can be applied to determine whether there is a solution on a given interval.
Let's choose the interval [tex][0,2][/tex]
[tex]f(x)=x^{3}-3x-1\\f(0)=(0)^{3}-3(0)-1\\f(0)=-1\\f(0)<0\\[/tex]
Proceed to the other point: 2
[tex]f(x)=x^{3}-3x-1\\f(2)=(2)^{3}-3(2)-1\\f(2)=1\\f(2)>0\\[/tex]
3) Check the 2nd Graph for a the Visual answer, of it. And the 3rd graph for all solutions of this equation.
If the sample size is n = 75, what are the degrees of freedom for the appropriate chi-square distribution when testing for independence of two variables each with three categories? a. 4 b. 69 c. 74
Final answer:
The degrees of freedom for a chi-square test of independence with two variables each having three categories and a sample size of n = 75 is 4.
Explanation:
The question at hand involves determining the degrees of freedom (df) for a chi-square test of independence where two variables each have three categories, and the sample size is n = 75. To calculate the degrees of freedom for this scenario, you use the formula df = (r - 1)(c - 1) where r represents the number of rows (categories of one variable) and c represents the number of columns (categories of the other variable).
In this case, with both variables having three categories, we have r = 3 and c = 3, which gives us:
df = (3 - 1)(3 - 1) = (2)(2) = 4.
Therefore, the correct answer is a. 4 degrees of freedom for the chi-square distribution when testing for independence with a sample size of n = 75.
A computer can be classified as either cutting-edge or ancientancient. Suppose that 91% of computers are classified as ancient.
a) Two computers are chosen at random. What is the probability that both computers are ancient?
b) Eight computers are chosen at random. What is the probability that all eight computers are ancient?
c) What is the probability that at least one of eight randomly selected computers is cutting-edge? Would it be unusual that at least one of eighteight randomly selected computers is cutting-edge?
Answer:
a)P(X=2) = (2C2)(0.91)^2 (1-0.91)^{2-2}=0.8281[/tex]
b) P(X=8)=(8C8)(0.91)^8 (1-0.91)^{8-8}=0.4703[/tex]
c) [tex]P(X \geq 1)=1-P(X<1)=1-P(X=0)=1-0.4703=0.5297[/tex]
Step-by-step explanation:
A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]
a. Two computers are chosen at random. What is the probability that both computers are ancient?
[tex]P(X=2)=(2C2)(0.91)^2 (1-0.91)^{2-2}=0.8281[/tex]
b. Eight computers are chosen at random. What is the probability that all eight computers are ancient?
On this case we are looking for this probability:
[tex]P(X=8)=(8C8)(0.91)^8 (1-0.91)^{8-8}=0.4703[/tex]
c. What is the probability that at least one of eight randomly selected computers is cutting-edge? Would it be unusual that at least one of eighteight randomly selected computers is cutting-edge?
Since we are interested on the cutting edge class the new probability of success would be p=1-0.91=0.09. And we want to find this probability:
[tex]P(X \geq 1)=1-P(X<1)=1-[P(X=0)][/tex]
And we can find the indiviudal probabilitiy like this:
[tex]P(X=0)=(8C0)(0.09)^0 (1-0.09)^{8-0}=0.4703[/tex]
And if we replace we got:
[tex]P(X \geq 1)=1-P(X<1)=1-P(X=0)=1-0.4703=0.5297[/tex]
The probability of both computers being ancient is about 82.81%, all eight being ancient is about 43.05%, and at least one of eight being cutting-edge is roughly 56.95%, which is not unusual.
Explanation:Probability of Selecting Ancient Computers
When dealing with probabilities of independent events, such as selecting computers at random, we can calculate the probability of multiple events occurring in sequence by multiplying the probabilities of each individual event.
The probability that both computers are ancient when two are chosen is 0.91 × 0.91 or about 0.8281 (82.81%).The probability that all eight computers are ancient when eight are chosen is 0.918 or about 0.43046721 (43.05%).To find the probability that at least one of eight computers is cutting-edge, we first calculate the probability that none are cutting-edge (all ancient) and subtract it from 1. This is 1 – 0.918 or about 0.5695 (56.95%). It would not be unusual that at least one is cutting-edge as this probability is higher than 50%.Note that these calculations assume that each computer's classification as ancient or cutting-edge is independent of the others.
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
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)
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%.
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 owner of Limp Pines Resort wanted to know the average age of its clients. A random sample of 25 tourists is taken. It shows a mean age of 46 years with a standard deviation of 5 years. The margin of error of a 98 percent CI for the true mean client age is approximately:
Step-by-step explanation:
Since we have given that
Sample size = 25
Mean = 46 years
Standard deviation = 5 years
We need to find the margin of error of a 98% confidence interval for the true mean client age.
So, Margin of error is given by
[tex]z\times \dfrac{\sigma}{\sqrt{n}}\\\\=2.33\times \dfrac{5}{\sqrt{25}}\\\\=2.33\times \dfrac{5}{5}\\\\=2.33[/tex]
Hence, margin of error is 2.33.
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!
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
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.
EC bisects <BED, m<ARV =11x- 12 and m<CED =4x +1. Find m<AEC
HELP!!
Answer:Angle AEC is 139 degrees.
Step-by-step explanation:
Since line EC bisects angle BED, it divides angle BED equally into 2. This means that
Angle BEC = angle CED
If angle CED = 4x + 1
Therefore,
Angle BED = 2 × angle CED
= 2(4x + 1) = 8x + 2
The sum of the angles in a straight line is 180 degrees. Therefore
Angle AEB + angle BED = 180
Angle AEB = 11x - 12. Therefore
11x - 12 + 8x + 2 = 180
19x - 10 = 180
19x = 180 + 10 = 190
x = 190/19 = 10
Angle BEC = 4x + 1 = 4×10 + 1 = 41
Angle AEB = 11x - 12 = 11×10 - 12 = 98
Angle AEC = 41 + 98 = 139
Answer: AEC = 139
Step-by-step explanation:
You first have to find x.
To find x, we need to add all the angles together to get 180°.
Since EC bisects BED, we know angle BEC and CED equal the same measure.
AEB + BEC + CED = 180°
(11x - 12) + (4x + 1) + (4x + 1) = 180°
Add like terms and solve
19x - 10 = 180
19x = 190
x = 10
Now, we substitute 'x' in AEB and BEC
AEB = 11x - 12
11 (10) - 12
AEB = 98
BEC = 4x + 1
4 (10) + 1
BEC = 41
98 + 41 = 139
AEC = 139
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
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.
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.
Learn more about Polar Coordinates here:https://brainly.com/question/33601587
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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:
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 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
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 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)
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.
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.
Why must integration be used to find the work required to pump water out of a tank?
A. Different volumes of water are moved different distances.
B. Water from the same horizontal planes is lifted different distances.
C. Integration is necessary because the acceleration of gravity changes at each level.
D. Integration is necessary because W = mgy.
Answer:
A. Different volumes of water are moved different distances.
Step-by-step explanation:
Integration is used to find work required to pump water out of a tank because different volumes of water are moved different distances and to sum it all we the tool required is integration. Moreover, work done is a path function or an inexact differential. It does depend upon the path followed by the process.
hence the correct answer is A.
The mean annual tuition and fees for a sample of 12 private colleges was 36,800 with a standard deviation of 5,000 . A dotplot shows that it is reasonable to assume that the population is approximately normal. You wish to test whether the mean tuition and fees for private colleges is different from . Compute the value of the test statistic and state the number of degrees of freedom.
Answer:
The value of the test statistic and degrees of freedom is 2.148 and 11 respectively.
Step-by-step explanation:
Consider the provided information.
The mean annual tuition and fees for a sample of 12 private colleges was 36,800 with a standard deviation of 5,000 .
Thus, n = 12, [tex]\bar x=36800[/tex] σ = 5000
degrees of freedom = n-1 = 12-1 = 11
[tex]H_0: \mu = 33700\ and\ H_a: \mu \neq 33700[/tex]
Formula to find the value of z is: [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Where [tex]\bar x[/tex] is mean of sample, μ is the mean of population, σ is the standard deviation of population and n is number of observations.
[tex]z=\frac{36800-33700}{\frac{5000}{\sqrt{12}}}[/tex]
[tex]z=2.148[/tex]
Hence, the value of the test statistic and degrees of freedom is 2.148 and 11 respectively.