Answer:
Step-by-step explanation:
Given
There are 45 numbers out of which 5 numbers are required to win an enormous sum of money
No of ways in which 5 numbers can be selected out of 45 numbers.
[tex]^nC_r=\frac{n!}{(n-r)!r!}[/tex]
here n=45, r=5
[tex]^{45}C_5=\frac{45!}{40!5!}[/tex]
[tex]^{45}C_5=1221759[/tex]
out of which there is only combination foe contest winning
[tex]P=\frac{1}{1221759}[/tex]
When 100 different tickets are bought then
Probability of winning[tex]=\frac{100}{1221759}[/tex]
The probability of winning with one combination of five numbers is [tex]\( \frac{1}{1,221,759} \)[/tex] .
The probability of winning if 100 different lottery tickets are purchased is [tex]\( 1 - \left(1 - \frac{1}{1,221,759}\right)^{100} \)[/tex], approximately [tex]\( 0.0000182 \)[/tex].
Step 1
The probability of winning the lottery with one combination of five numbers can be calculated using the formula for combinations:
[tex]\[ P(\text{win with one ticket}) = \frac{1}{\binom{45}{5}} \][/tex] Where [tex]\( \binom{45}{5} \)[/tex]represents the number of ways to choose 5 numbers from 45 without regard to the order. Calculating [tex]\( \binom{45}{5} \)[/tex] :
[tex]\[ \binom{45}{5} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1} = 1,221,759 \][/tex]
So, the probability [tex]\( P(\text{win with one ticket}) \)[/tex] is:
[tex]\[ P(\text{win with one ticket}) = \frac{1}{1,221,759} \approx 8.19 \times 10^{-7} \][/tex]
If 100 different lottery tickets are purchased, the probability of winning at least once is calculated using the complement rule:
[tex]\[ P(\text{win with 100 tickets}) = 1 - \left( 1 - \frac{1}{\binom{45}{5}} \right)^{100} \][/tex]
Step 2
Substituting [tex]\( \binom{45}{5} \)[/tex] :
[tex]\[ P(\text{win with 100 tickets}) = 1 - \left( 1 - \frac{1}{1,221,759} \right)^{100} \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 1 - (1 - 8.19 \times 10^{-7})^{100} \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 1 - (0.999999181)^{100} \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 1 - 0.9999818 \][/tex]
[tex]\[ P(\text{win with 100 tickets}) \approx 0.0000182 \][/tex]
The probability of winning the lottery with one ticket is approximately [tex]\( 8.19 \times 10^{-7} \)[/tex], while the probability of winning if 100 different tickets are purchased increases to approximately 0.0000182. Purchasing more tickets improves the chances of winning, but it remains a very low probability event due to the large number of possible combinations.
Find the point on the line 5x + y = 7 that is closest to the point (−3, 2).
To find the closest point on a line to another point, convert the problem into finding a minimum of a distance function, then use calculus (derivatives) to find the minimum.
Explanation:To solve this problem, you need to identify a goal and an approach using calculus. The goal is to find a point (x, y) on the given line 5x + y = 7 that is closest to the point (−3, 2). This translates into minimizing the distance between (x, y) and (-3, 2) with respect to (x, y).
The distance formula between two points (x1, y1) and (x2, y2) is: √[(x2-x1)²+(y2-y1)²]. For our purposes x1 = -3, y1 = 2 and x2 = x, y2 = y. However, since y is from our given line (5x + y = 7), we can substitute for y = 7 - 5x. Thus, our new distance will become a function in terms of x: d(x) = √[(x - -3)²+(7 - 5x - 2)²].
To minimize it, we should take the derivative of this function, and set it equal to zero, which will give an extreme point (either minimum or maximum). Using the chain-rule results in the minimizing x value, and substituting it back to the line gives us the y value. This resultant point (x, y) would be our answer to the problem.
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The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200. The value of the test statistic is:
The p-value is, (round to 4 decimal places):
Answer:
Null hypothesis:[tex]\mu \leq 8000[/tex]
Alternative hypothesis:[tex]\mu > 8000[/tex]
[tex]z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2[/tex]
[tex]p_v =P(Z>2)=0.0228[/tex]
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=8300[/tex] represent the sample mean
[tex]\sigma=1200[/tex] represent the population standard deviation
[tex]n=64[/tex] sample size
[tex]\mu_o =800[/tex] represent the value that we want to test
[tex]\alpha[/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 mean is higher than 8000, the system of hypothesis are :
Null hypothesis:[tex]\mu \leq 8000[/tex]
Alternative hypothesis:[tex]\mu > 8000[/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{8300-8000}{\frac{1200}{\sqrt{64}}}=2[/tex]
4) P-value
Since is a one-side upper test the p value would given by:
[tex]p_v =P(Z>2)=0.0228[/tex]
5) Conclusion
If we compare the p value and the significance level assumed, for example [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.
Records at the UH library show that 12% of all UH students check out books on history, 28% of all UH students check out books on science, and 6% check out books on both history and science. What is the probability that a randomly selected UH student checks out a history book or a science book or both?
Answer:
There is a 34% probability that a randomly selected UH student checks out a history book or a science book or both.
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a UH student checks out books on history.
B is the probability that a UH students checks out books on science.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a UH student checks a book on history but not on science and [tex]A \cap B[/tex] is the probability that a UH student checks books both on history and science.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
What is the probability that a randomly selected UH student checks out a history book or a science book or both?
[tex]P = a + b + (A \cap B)[/tex]
We start finding these values from the intersection.
6% check out books on both history and science. So [tex]A \cap B = 0.06[/tex]
28% of all UH students check out books on science. So [tex]B = 0.28[/tex]
[tex]B = b + (A \cap B)[/tex]
[tex]0.28 = b + 0.06[/tex]
[tex]b = 0.22[/tex]
12% of all UH students check out books on history
[tex]A = a + (A \cap B)[/tex]
[tex]0.12 = a + 0.06[/tex]
[tex]a = 0.06[/tex]
So
[tex]P = a + b + (A \cap B) = 0.06 + 0.22 + 0.06 = 0.34[/tex]
There is a 34% probability that a randomly selected UH student checks out a history book or a science book or both.
Use Green's Theorem to calculate the circulation of F =2xyi around the rectangle 0≤x≤8, 0≤y≤3, oriented counterclockwise.
Green's theorem says the circulation of [tex]\vec F[/tex] along the rectangle's border [tex]C[/tex] is equal to the integral of the curl of [tex]\vec F[/tex] over the rectangle's interior [tex]D[/tex].
Given [tex]\vec F(x,y)=2xy\,\vec\imath[/tex], its curl is the determinant
[tex]\det\begin{bmatrix}\frac\partial{\partial x}&\frac\partial{\partial y}\\2xy&0\end{bmatrix}=\dfrac{\partial(0)}{\partial x}-\dfrac{\partial(2xy)}{\partial y}=-2x[/tex]
So we have
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_D-2x\,\mathrm dx\,\mathrm dy=-2\int_0^3\int_0^8x\,\mathrm dx\,\mathrm dy=\boxed{-192}[/tex]
The circulation of the vector field 2xyi around the given rectangle, as computed via Green's Theorem, is 0 due to the curl of F being 0.
Explanation:To use Green's Theorem to calculate the circulation around a rectangle, first we should realize that Green's Theorem states that the line integral around a simple closed curve C of F.dr is equal to the double integral over the region D enclosed by C of the curl of F. Here, F is the vector field defined as 2xyi. The given rectangle is oriented counterclockwise and the values of x and y are given as 0≤x≤8 and 0≤y≤3 respectively. The line integral denotes the circulation of the field.
The circulation is thus the double integral over the rectangle of ∇ x F. But in this case, since F = 2xyi, we get ∇ x F = 0. Hence, the circulation of F around the given rectangle is 0.
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The probability of having a girl is 51%. A couple has 5 children, what is the expected number of girls? (round your answer to the nearest tenth)
Final answer:
The expected number of girls in a family with five children, given a 51% probability of having a girl, is 2.6 when rounded to the nearest tenth.
Explanation:
The question involves calculating the expected number of girls in a family with five children, given that the probability of having a girl is 51%. The expected value in probability theory is calculated by multiplying each possible outcome by its probability and then summing these values. In this case, since there are five children and each child has a 51% chance of being a girl, the expected number of girls can be found using the formula for the expectation:
Expected number of girls = Total number of children × Probability of having a girl
Therefore, the expected number of girls = 5 × 0.51 = 2.55. Rounding this to the nearest tenth, we get 2.6 girls.
A onesided test of a hypothesis, based on a sample of size 9, yields a Pvalue of .035. Which of the following best describes the possible range of t values that yields this Pvalue? .706 < t < .889 1.11 < t < 1.40 1.34 < t < 1.44 2.45 < t < 2.90 1.86 < t < 2.31
Answer:
1.86 < t < 2.31
Step-by-step explanation:
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean height actually is higher\lower than an specified value, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq \mu_o[/tex]
Alternative hypothesis:[tex]\mu > \mu_o[/tex]
Or like this:
Null hypothesis:[tex]\mu \geq \mu_o[/tex]
Alternative hypothesis:[tex]\mu < \mu_o[/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 don't know on this case the calculated value and with the p value we can find it.
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 test the p value would be:
[tex]p_v =P(t_{(8)}>t_o)=0.035[/tex]
And we can find the critical value with the following excel code:
" =T.INV(1-0.035,8)" and we got [tex]t_p =2.09[/tex]
And cannot be a lower tail test since all the options have positive values for the statistic. So on this case the best option is:
1.86 < t < 2.31
A factory cans fresh pineapple in sugar syrup. The manager in charge is interested in estimating the average amount of sugar per can to within 2 mg of the true mean. From previous experiments, he knows that the (true) standard deviation of the sugar content is 15 mg. Each test of measuring the amount of sugar in a can costs $5.00. One thousand dollars have been budgeted for this experiment. Does the manager have enough funds to estimate the average amount of sugar per can with 95% confidence
Answer:
No. He would need $1,080 to perform a test with this error.
Step-by-step explanation:
To answer this question, we have to calculate the sample size. This is the sample size that allow to estimate the average amount of sugar per can with 95% confidence (95% CI).
The difference between the upper and lower limit of the CI have to be equal or less to e=2 mg. The z value for a 95% CI is z=1.96.
[tex]e\leq z\sigma/\sqrt{n}[/tex]
[tex]e=z\sigma/\sqrt{n}\\\\\sqrt{n}=z\sigma/e=\\\\n=(z\sigma/e)^2=(1.96*15/2)^2=14.7^2=216\\\\n=216[/tex]
The minimum sample size needed for this error is 216. At a cost of $5/test, this sample size would cost [tex]n*p=216*5=\$ 1,080[/tex].
This is over the budget for this experiment ($1000).
Brian Vanecek, VP of Operations at Portland Trust Bank, is evaluating the service level provided to walk-in customers. Accordingly, his staff recorded the waiting times for 64 randomly selected walk-in customers and determined that their mean waiting time was 15 minutes. Assume that the population standard deviation is 4 minutes. The 95% confidence interval for the population mean of waiting times is ________.A. 14.02 to 15.98B. 7.16 to 22.84C. 14.06 to 15.94D. 8.42 to 21.58E. 19.80 to 23.65
Answer: A. 14.02 to 15.98
Step-by-step explanation:
Let [tex]\mu[/tex] denotes the mean waiting time for population.
Given : Sample size : n= 64
Sample mean : [tex]\overline{x}=15[/tex] (minutes)
Population standard deviation = [tex]\sigma= 4[/tex]
Confidence level : 95%
By z-table , the critical values for 95% confidence = z*=1.96
Confidence interval for population mean : [tex]\overline{x}\pm z^* \dfrac{\sigma}{\sqrt{n}}[/tex]
The 95% confidence interval for the population mean of waiting times will be :
[tex]15\pm (1.96)\dfrac{4}{\sqrt{64}}[/tex]
[tex]15\pm (1.96)\dfrac{4}{8}[/tex]
[tex]15\pm (1.96)(0.5)[/tex]
[tex]15\pm 0.98[/tex]
[tex](15-0.98,\ 15+0.98)=(14.02,\ 15.98)[/tex]
Hence, the 5% confidence interval for the population mean of waiting times is 14.02 to 15.98.
Thus , the correct answer is Option A.
A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If this is to happen, it will have to move one of its most viewed shows to another slot. The network conducts a survey asking its viewers which show they would rather watch. The network receives 827 responses, of which 438 indicate that they would like to see the new show in the lineup. The test statistic for this hypothesis would be:
a. 2.05
b. 1.71
c. 2.25
d. 1.01
Answer:
b. 1.71
[tex]z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71[/tex]
Step-by-step explanation:
1) Data given and notation
n=827 represent the random sample taken
X=438 represent the people that indicate that they would like to see the new show in the lineup
[tex]\hat p=\frac{438}{827}=0.5296[/tex] estimated proportion of people that indicate that they would like to see the new show in the lineup
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:
Null hypothesis:[tex]p \leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71[/tex]
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>1.71)=0.044[/tex]
If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true proportion is higher than 0.5.
Student scores on exams given by a certain instruc-tor have mean 74 and standard deviation 14. Thisinstructor is about to give two exams, one to a classof size 25 and the other to a class of size 64.(a)Approximate the probability that the averagetest score in the class of size 25 exceeds 80.(b)Repeat part (a) for the class of size 64.(c)Approximate the probability that the averagetest score in the larger class exceeds that ofthe other class by over 2.2 points.(d)Approximate the probability that the averagetest score in the smaller class exceeds that ofthe other class by over 2.2 points.
Answer:
Step-by-step explanation:
Given that Student scores on exams given by a certain instruc-tor have mean 74 and standard deviation 14.
Group I X Group II Y
Sample mean 74 74
n 25 64
Std error (14/sqrtn) 2.8 1.75
a) P(X>80) =[tex]1-0.9839\\= 0.0161[/tex]
b) P(Y>80) = [tex]1-0.9997\\=0.0003[/tex]
c) X-Y is Normal with mean = 0 and std deviation = [tex]\sqrt{2.8^2+1.75^2} \\=3.302[/tex]
P(X-Y>2.2) = [tex]1-0.8411\\=0.1589[/tex]
d) [tex]P(\bar x -\bar Y>2.2) = 0.1589[/tex]
With respect to the number of categories, k, when would a multinomial experiment be identical to a binomial experiment?
a. k = 2
b. k = 3
c. k = 4
d. k = 1
Answer:
Option A) k = 2
Step-by-step explanation:
Multimonial Experiment
A multimonial experiment is an experiment with n repeated trials and each trial has a discrete number of possible outcomes.
Binomial Experiment
Binomial experiment is an experiment with n repeated trials and each trial has only two possible outcomes.
Thus, if k represents the number of possible outcomes, then for k = 2, a multimonial experiment will become a binomial experiment.
Option A) k = 2
A marketing research company desires to know the mean consumption of milk per week among males over age 25. They believe that the milk consumption has a mean of 2.5 liters, and want to construct a 85% confidence interval with a maximum error of 0.07 liters. Assuming a variance of 1.21 liters, what is the minimum number of males over age 25 they must include in their sample? Round your answer up to the next integer.
In this exercise we have to use the knowledge of variance to calculate the value of n, so we have that:
the sample is n=306
Organizing the information given in the statement we have that:
Mean of milk consumption = 2.5litresMaximum error E = 0.07Variance S = 1.21 litresConfidence interval of 85%So given by the equation we have:
[tex]Z' = t(0.075)= 1.44\\n = (Z'*S/E)^2\\n = ( 1.44 * 0.85/0.07)^2\\n = (17.4857)^2\\n = 305.75\\n = 306[/tex]
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The minimum number of males over age 25 they must include in their sample is 512.
Given, Desired confidence level: 85%
Maximum error (E): 0.07 liters
Variance ([tex]\sigma^{2}[/tex]): 1.21 liters
Standard deviation ([tex]\(\sigma\)[/tex]): [tex]\(\sigma\)[/tex] = [tex]\sqrt{1.21}[/tex] = 1.1
Z-value for 85% confidence level (lookup Z-value for 0.425 in the standard normal distribution): [tex]\[ Z \approx 1.44 \][/tex]
n= [tex]\left(\frac{Z \sigma}{E}\right)^2[/tex]
[tex]\[ n = \left(\frac{1.44 \times 1.1}{0.07}\right)^2 \][/tex]
n= (1.584/0.07)² = 511.986
n = 512
A biologist observed that a certain bacterial colony obeys the population growth law and that the colony triples every 4 hours.
If the colony occupied 2 square centimeters initially, find:
(a) An expression for the size P(t) of the colony at any time t.
(b) The area occupied by the colony after 12 hours.
(c) The doubling time for the colony?
Answer:
a) [tex]P(t) = 2e^{0.275t}[/tex]
b) 54.225 square centimeters.
c) 2.52 hours
Step-by-step explanation:
The population growth law is:
[tex]P(t) = P_{0}e^{rt}[/tex]
In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the growth rate, as a decimal.
In this problem, we have that:
The colony occupied 2 square centimeters initially, so [tex]P_{0} = 2[/tex]
The colony triples every 4 hours. So
[tex]P(4) = 3P_{0} = 6[/tex]
(a) An expression for the size P(t) of the colony at any time t.
We have to find the value of r. We can do this by using the P(4) equation.
[tex]P(t) = P_{0}e^{rt}[/tex]
[tex]6 = 2e^{4r}[/tex]
[tex]e^{4r} = 3[/tex]
Applying ln to both sides, we get:
[tex]4r = 1.1[/tex]
[tex]r = 0.275[/tex]
So
[tex]P(t) = 2e^{0.275t}[/tex]
(b) The area occupied by the colony after 12 hours.
[tex]P(t) = 2e^{0.275t}[/tex]
[tex]P(12) = 2e^{0.275*12}[/tex]
[tex]P(12) = 54.225[/tex]
(c) The doubling time for the colony?
t when [tex]P(t) = 2P_{0} = 2*2 = 4[/tex].
[tex]P(t) = 2e^{0.275t}[/tex]
[tex]4 = 2e^{0.275t}[/tex]
[tex]e^{0.275t} = 2[/tex]
Applying ln to both sides
[tex]0.275t = 0.6931[/tex]
[tex]t = 2.52[/tex]
Find the percentages of people who said "yes, definitely", "yes, probably", "no, probably not", and "no, definitely not".
Answer:
yes,def=(2399/3713)x100=64.61%
yes,probably= (774/3713)x100=20.85%
yes,probably no= (322/3713)x100=8.67%
yes,def.no= (218/3713)x100=5.87%
Step-by-step explanation:
Data given
We assume the following frequency table:
Type Frequency
Yes, definitely 2399
Yes, probably 774
No, probably not 322
No, probably not 218
Total 3713
Solution to the problem
And for this case in order to find the percentages we can use the definition of percentage frequency with the following formula:
[tex] \% = \frac{Number of people with characteristic}{Total} x100[/tex]
And using this formula we can find the percentages like this:
yes,def=(2399/3713)x100=64.61%
yes,probably= (774/3713)x100=20.85%
yes,probably no= (322/3713)x100=8.67%
yes,def.no= (218/3713)x100=5.87%
HELP! I AM BEING TIMED!!!
According to the Fundamental Theorem of Algebra, which polynomial function has exactly 6 roots?
Answer:
f(x) = 7x⁶ + 3x³ + 12
Step-by-step explanation:
The number of roots a function has corresponds to the highest exponent value.
In the function f(x) = 7x⁶ + 3x³ + 12 ,
the number 7x⁶ has the highest exponent value of 6, so it'll have 6 roots.
a one-parameter family of solutions of the de p' = p(1 − p) is given below.
P =
c1et
1 + c1et
Does any solution curve pass through the point (0, 4)? Through the point (0, 1)? (If yes, give the corresponding value of
c1.
If not, enter DNE.)
(0, 4) __________
(0, 1) _______________
Answer:
A solution curve pass through the point (0,4) when [tex]c_{1} = -\frac{4}{3}[/tex].
There is not a solution curve passing through the point(0,1).
Step-by-step explanation:
We have the following solution:
[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]
Does any solution curve pass through the point (0, 4)?
We have to see if P = 4 when t = 0.
[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]
[tex]4 = \frac{c_{1}}{1 + c_{1}}[/tex]
[tex]4 + 4c_{1} = c_{1}[/tex]
[tex]c_{1} = -\frac{4}{3}[/tex]
A solution curve pass through the point (0,4) when [tex]c_{1} = -\frac{4}{3}[/tex].
Through the point (0, 1)?
Same thing as above
[tex]P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}[/tex]
[tex]1 = \frac{c_{1}}{1 + c_{1}}[/tex]
[tex]1 + c_{1} = c_{1}[/tex]
[tex]0c_{1} = 1[/tex]
No solution.
So there is not a solution curve passing through the point(0,1).
A job shop consists of three machines and two repairmen. The amount of time a machine works before breaking down is exponentially distributed with mean 10. If the amount of time it takes a single repairman to fix a machine is exponentially distributed with mean 8, then(a) what is the average number of machines not in use?(b) what proportion of time are both repairmen busy?
Answer:
Step-by-step explanation:
Let X(t) denote the number of machines breakdown at time t.
The givenn problem follows birth-death process with finite space
S={0, 1, 2, 3} with
[tex] \lambda_0=\frac{3}{10}, \mu_1=\frac{1}{8}\\\\ \lambda_1=\frac{2}{10}, \mu_2=\frac{2}{8}\\\\ \lambda_2=\frac{1}{10}, \mu_3=\frac{2}{8}[/tex]
The birth-death process having balance equations [tex]\lambda_sP_i=\mu_{s+1}P_{i+1},i=0,1,2[/tex]
since, state rate at which leave = rate at which enter
0 [tex]\lambda_0P_0=\mu_1P_1[/tex]
1 [tex](\lambda_1+\mu_1)P_1= \mu_2P_2 + \lambda_0P_0[/tex]
2 [tex](\lambda_2+\mu_2)P_2= \mu_3P_3 + \lambda_1P_1[/tex]
[tex]P_1=\frac{12}{5}P_0=P_0=\frac{5}{12}P_1\\\\P_2=\frac{48}{25}P_0=P_0=\frac{25}{48}P_2\\\\P_3=\frac{192}{250}P_0=P_0=\frac{250}{192}P_3[/tex]
Since [tex]\sum\limits^3_{i=0} {P_i=1}\\\\p_0=[1+\frac{5}{12}+\frac{48}{25}+\frac{192}{250}]^{-1}=\frac{250}{1522}[/tex]
a)
Average number not in use equals the mean of the stationary distribution [tex]P_1+2P_2+3P_3=\frac{2136}{751}[/tex]
b)
Proportion of time both repairmen are busy [tex]P_2+P_3=\frac{672}{1522}=\frac{336}{761}[/tex]
The average number of machines not in use is 0.5, and the repairmen are both busy 64% of the time. This has been found under the assumption of exponential distribution for both longevity of machines and repair time. The scenario represents a M/M/2 queue in operations research.
Explanation:In this scenario, the life of the machines and the repair time are governed by exponential distributions. Exponential distribution is often used to model the amount of time until an event occurs, such as machine failure in this case.
(a) To find the average number of machines not in use, we need to consider the rate of machine breakdown and repair. A machine works an average of 10 hours before failure, which translates to a failure rate of 1/10. A single repairman can fix a machine in an average of 8 hours, meaning a rate of 1/8 per repairman, or 1/4 for two repairmen combined. As there are three machines, the average number of machines in use is the ratio of the arrival rate to the service rate: (1/10) / (1/4) = 2.5 machines. This implies that on average, 0.5 machines are not in use.
(b) Both repairmen will be busy when there are at least two machines that require fixing. The proportion of time in which this is the case is obtained by calculating the probability that the number of machines failed is two or more. This is a problem of queueing theory, in particular an M/M/2 queue. The formula for this probability is P(X >= k) = (1 - rho) * rho^(k) / (1 - rho^(c+1)), where rho = arrival rate / service rate, k = 2, and c = 2 (service channels). Substituting rho = 2.5, we obtain P(X >= 2) = 0.64, meaning that the repairmen are both busy 64% of the time.
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Please help me with these 2 questions! 50 points!
Answer:x = 9
Step-by-step explanation:
The attached photo is that of the given diagram. b represents the angle adjacent 75 degrees.
If line m is parallel to line n, it means that angle b degrees and angle (10x + 15) are corresponding angles. Corresponding angles are equal.
Therefore,
b = 10x + 15
The sum of angles on a straight line is 180 degrees. It means that
b + 75 = 180
b = 180 - 75 = 105
Therefore
10x + 15 = 105
10x = 105 - 15 = 90
x = 90/10 = 9
Answer:
x = 9°
Step-by-step explanation:
105° must be equal to 10x + 15 ° for lines to be parallel.
> 105° = 10x + 15°
> 10x = 90°
> x = 9°
Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. In a manual on how to have a number one song, it is stated that a song must be no longer than 210 seconds. A simple random sample of 40 current hit songs results in a mean length of 234.1 sec and a standard deviation of 54.52 sec. Use a 0.05 significance level and the accompanying Minitab display to test the claim that the sample is from a population of songs with a mean greater than 210 sec. What do these results suggest about the advice given in the manual?
Answer:
There is enough evidence to conclude that the sample is from a population of songs with a mean greater than 210 sec.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 210 seconds
Sample mean, [tex]\bar{x}[/tex] = 234.1 sec
Sample size, n = 40
Sample standard deviation, s = 54.52 sec
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu \leq 210\text{ seconds}\\H_A: \mu > 210\text{ seconds}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{234.1 - 210}{\frac{54.52}{\sqrt{40}} } = 2.795[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.684[/tex]
We calculate the p-value with the help of standard normal table.
P-value = 0.004005
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to accept the null hypothesis and accept the alternate hypothesis.
Thus, there is enough evidence to conclude that the sample is from a population of songs with a mean greater than 210 sec.
The given data shows that the sample mean length of 234.1 sec results in a low p-value, it reject the claim that songs must be no longer than 210 seconds.
Explanation of hypothesis testing for mean song length compared to a specific value with given data.
Null hypothesis (H0): The mean length of songs is 210 seconds or less (μ ≤ 210 sec).
Alternative hypothesis (Ha): The mean length of songs is greater than 210 seconds (μ > 210 sec).
Test statistic: Calculate the z-score using the sample mean, population mean, standard deviation, and sample size.
P-value: Using the z-score, find the p-value from the standard normal distribution table.
Conclusion: If the p-value is less than the significance level (0.05), reject the null hypothesis. In this case, if the sample mean length of 234.1 sec results in a low p-value, we reject the claim that songs must be no longer than 210 seconds.
A sample of 161children was selected from fourth and fifth graders at elementary schools in Philadelphia. In addition to recording the grade level, the researchers determined whether each child had a previously undetected reading disability. Sixty-six children were diagnosed with a reading disability. Of these children, 32 were fourth graders and 34 were fifth graders. Similarly, of the 95 children with normal reading achievement, 55 were fourth graders and 40 were fifth graders.
a. Identify the two qualitative variables (and corresponding levels) measured in the study.
b. From the information provided, form a contigency table.
c. Assuming that the two variables are independent, calculate the expected cell counts.
Answer:
Step-by-step explanation:
Given that a sample of 161children was selected from fourth and fifth graders at elementary schools in Philadelphia. In addition to recording the grade level, the researchers determined whether each child had a previously undetected reading disability
a) The two qualitative variables are disability and not having disability and secondly the grades of children
b) Contingency table:
Grade 4 5 Total
Normal read. 32 34 66
Not normal read. 23 6 29
Total 55 40 95
H0: Reading disability is independent of grade.
Ha: There is association between the two
c) 4 5 Total
Nor read 38.21052632 27.78947368 66
Not norm 16.78947368 12.21052632 29
Expected cells are obtained using the formula
row total*col total/grand total
5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the general population. To determine the aptitude of her district's children, a study was conducted. The results of her district's test scores were: 105, 109, 115, 112, 124, 115, 103, 110, 125, 99. If the mean of the general population of school children is 106, what could be said about her claim? Use alpha = .05
Answer:
We conclude that children in district are brighter, on average, than the general population.
Step-by-step explanation:
We are given the following data set:
105, 109, 115, 112, 124, 115, 103, 110, 125, 99
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{1117}{10} = 111.7[/tex]
Sum of squares of differences = 642.1
[tex]S.D = \sqrt{\frac{642.1}{49}} = 8.44[/tex]
We are given the following in the question:
Population mean, μ = 106
Sample mean, [tex]\bar{x}[/tex] = 111.7
Sample size, n = 10
Alpha, α = 0.05
Sample standard deviation, s = 8.44
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 106\\H_A: \mu > 106[/tex]
We use one-tailed(right) t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833[/tex]
Since,
[tex]t_{stat} > t_{critical}[/tex]
We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
We conclude that children in district are brighter, on average, than the general population.
We're testing the hypothesis that the average boy walks at 18 months of age (H0: p = 18). We assume that the ages at which boys walk is approximately normally distributed with a standard deviation of 2.5 months. A random sample of 25 boys has a mean of 19.2 months. Which of the following statements are correct?
I. This finding is significant for a two-tailed test at .05.
II. This finding is significant for a two-tailed test at .01.
III. This finding is significant for a one-tailed test at .01.
a. I only
b. II only
c. III only
d. II and III only
e. I and III only
Answer:
II. This finding is significant for a two-tailed test at .01.
III. This finding is significant for a one-tailed test at .01.
d. II and III only
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=19.2[/tex] represent the battery life sample mean
[tex]\sigma=2.5[/tex] represent the population standard deviation
[tex]n=25[/tex] sample size
[tex]\mu_o =18[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:
Null hypothesis:[tex]\mu = 18[/tex]
Alternative hypothesis:[tex]\mu \neq 18[/tex]
And for part III we have a one tailed test with the following hypothesis:
Null hypothesis:[tex]\mu \leq 18[/tex]
Alternative hypothesis:[tex]\mu > 18[/tex]
Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]z=\frac{19.2-18}{\frac{2.5}{\sqrt{25}}}=2.4[/tex]
4) P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=25-1=24[/tex]
Since is a two tailed test for parts I and II, the p value would be:
[tex]p_v =2*P(t_{(24)}>2.4)=0.0245[/tex]
And for part III since we have a one right tailed test the p value is:
[tex]p_v =P(t_{(24)}>2.4)=0.0122[/tex]
5) Conclusion
I. This finding is significant for a two-tailed test at .05.
Since the [tex]p_v <\alpha[/tex]. We reject the null hypothesis so we don't have a significant result. FALSE
II. This finding is significant for a two-tailed test at .01.
Since the [tex]p_v >\alpha[/tex]. We FAIL to reject the null hypothesis so we have a significant result. TRUE.
III. This finding is significant for a one-tailed test at .01.
Since the [tex]p_v >\alpha[/tex]. We FAIL to reject the null hypothesis so we have a significant result. TRUE.
So then the correct options is:
d. II and III only
Answer:
E. I and III only
Step-by-step explanation:
I. .05
III. one-tailed at .01
Human body temperatures are normally distributed with a mean of 98.20oF and a standard deviation of 0.62oF If 19 people are randomly selected, find the probability that their mean body temperature will be less than 98.50oF. Your answer should be a decimal rounded to the fourth decimal place
Answer:
Step-by-step explanation:
Since the human body temperatures are normally distributed, the formula for normal distribution is expressed as
z = (x - u)/s
Where
x = human body temperatures
u = mean body temperature
s = standard deviation
From the information given,
u = 98.20oF
s = 0.62oF
We want to find the probability that their mean body temperature will be less than 98.50oF. It is expressed as
P(x lesser than 98.50)
For x = 98.50,
z = (98.50 - 98.20)/0.62 = 0.48
Looking at the normal distribution table, the corresponding probability to the z score is 0.6844
P(x lesser than 98.50) = 0.6844
A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the 0.01 significance level, can we conclude that the single-earner couples on average spend more time watching television together?
We can see here that at the 0.01 significance level, we can actually conclude that the single-earner couples on average spend more time watching television together.
How we arrived at the solution?To determine whether we can conclude that single-earner couples spend more time watching television together on average than dual-earner couples, we can perform a hypothesis test.
The null hypothesis (H₀) assumes that there is no difference in the mean time spent watching television between the two groups, while the alternative hypothesis (H₁) suggests that single-earner couples spend more time together watching television.
Let's set up the hypotheses:
Null Hypothesis (H₀): μ₁ ≤ μ₂ (The mean time spent together watching television for single-earner couples is less than or equal to the mean time for dual-earner couples.)
Alternative Hypothesis (H₁): μ₁ > μ₂ (The mean time spent together watching television for single-earner couples is greater than the mean time for dual-earner couples.)
Where:
μ₁ = population mean time spent watching television for single-earner couples
μ₂ = population mean time spent watching television for dual-earner couples
Next, we will use a two-sample t-test to test the hypotheses. Since we are trying to determine if single-earner couples spend more time watching television, this will be a one-tailed t-test.
Given the sample means, sample standard deviations, and sample sizes, we can calculate the t-statistic and compare it to the critical t-value at the 0.01 significance level (α = 0.01) with degrees of freedom d f = n₁ + n₂ - 2, where n₁ and n₂ are the sample sizes of single-earner and dual-earner couples, respectively.
Let's assume the sample sizes are n₁ = n₂ = 30 (the actual sample sizes from the study are not given in the question, but this is just for demonstration purposes).
Now, we can calculate the t-statistic:
t = (x₁ - x₂) / √((s₁²/n₁) + ([tex]s_{2}[/tex]² /n₂))
where:
x₁ = sample mean time for single-earner couples
x₂ = sample mean time for dual-earner couples
s₁ = sample standard deviation for single-earner couples
[tex]s_{2}[/tex] = sample standard deviation for dual-earner couples
n₁ = sample size for single-earner couples
n₂ = sample size for dual-earner couples
Using the provided values:
x₁ = 61 minutes
x₂ = 48.4 minutes
s₁ = 15.5 minutes
= 18.1 minutes
n₁ = n₂ = 30 (sample sizes assumed for demonstration)
Calculating the t-statistic:
t = (61 - 48.4) / √((15.5²/30) + (18.1²/30))
t ≈ 4.083
Next, we need to find the critical t-value from the t-distribution table at α = 0.01 significance level and df = 30 + 30 - 2 = 58 (degrees of freedom).
The critical t-value at α = 0.01 with d f = 58 is approximately 2.660.
Since the calculated t-statistic (4.083) is greater than the critical t-value (2.660), we reject the null hypothesis (H₀).
Therefore, at the 0.01 significance level, we can conclude that single-earner couples, on average, spend more time watching television together than dual-earner couples based on the data provided in the study.
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To determine whether single-earner couples spend more time watching television together than dual-earner couples, an independent samples t-test must be conducted at the 0.01 significance level. Using the provided means and standard deviations, we would calculate a t-statistic and compare it against critical values to either reject the null or fail to reject it.
The question asks whether single-earner couples spend more time watching television together than dual-earner couples, based on a study with provided mean values and standard deviations for both groups. To determine if there is a statistically significant difference between the two means, we would conduct a hypothesis test, specifically an independent samples t-test, at the 0.01 significance level. The null hypothesis (0) would state that there is no difference in the mean television watching times between the groups, while the alternative hypothesis (A) would claim that there is a difference, specifically that single-earner couples watch more television.
Given that the mean time spent watching television for single-earner couples is 61 minutes with a standard deviation of 15.5 minutes, and for dual-earner couples it is 48.4 minutes with a standard deviation of 18.1 minutes, we would calculate the t-statistic and compare it against the t-distribution critical values for the given degrees of freedom. If the calculated t-statistic exceeds the critical value for a one-tailed test at the 0.01 level, we would reject the null hypothesis and conclude that there is a significant difference, supporting the claim that single-earner couples spend more time watching television together.
Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test of mu = 100 vs not = 100 Variable N Mean StDev SE Mean 95% CI (Lower) 95% CI (Upper) T X 19 98.77 4.77 Enter your answer; SE Mean Enter your answer; 95% CI (Lower) Enter your answer; 95% CI (Upper) Enter your answer; T (a) How many degrees of freedom are there on the t-statistic? Enter your answer in accordance to the item a) of the question statement (b) What is your conclusion if ? Choose your answer in accordance to the item b) of the question statement(c) What is your conclusion if the hypothesis is versus ? Choose your answer in accordance to the item c) of the question statement
Final answer:
The degrees of freedom for the t-statistic are 18. The conclusion cannot be determined without the p-value. If the null hypothesis is true, there is not enough evidence to support the alternative hypothesis.
Explanation:
(a) The degrees of freedom for the t-statistic are found by subtracting 1 from the sample size. In this case, the sample size is 19 so the degrees of freedom would be 19 - 1 = 18.
(b) If the p-value is less than the alpha level (typically 0.05), we reject the null hypothesis. In this case, the p-value is not provided, so we cannot determine the conclusion.
(c) If the null hypothesis is true, we would not reject it and conclude that there is not enough evidence to support the alternative hypothesis.
WILL GIVE LARGE REWARD!
Given: Sector BAC with r=8
Radius of inscribed circle O is 2
Find the area of the sector BAC
Answer:
21.75 square units
Step-by-step explanation:
Draw a radial line from O to the point where AB intersects the circle. We'll call this point P.
Draw another radial line from O to the point where arc BC intersects the circle. We'll call this point Q.
OQ is equal to the radius of the circle, 2. And AQ is equal to the radius of the sector, 8. Therefore, the length of AO is 6.
Next, OP is equal to the radius of the circle, 2. Since AB is tangent to the circle, it is perpendicular to OP.
So we have a right triangle with a hypotenuse of 6 and a short leg of 2. Finding the angle ∠PAO:
sin ∠PAO = 2/6
∠PAO = asin(1/3)
That means the angle ∠BAC is double that:
∠BAC = 2 asin(1/3)
∠BAC ≈ 38.94°
Therefore, the area of the sector is:
A = (θ/360) πr²
A = (38.94/360) π(8)²
A ≈ 21.75
The area of the sector is approximately 21.75 square units.
A l-meter steel rod is bent into a rectangle so that its length exceeds quadruple its width
by 5 cm. What are the dimensions of the newly formed rectangle?
Answer:
The dimensions of the newly formed rectangle is length is 41 cm and width is 9 cm.
Step-by-step explanation:
Given,
A 1 meter steel rod is bent into a rectangle.
That means the perimeter of the rectangle is 1 meter.
So, Perimeter = [tex]1\ m=100\ cm[/tex]
Let the width of the rectangle be 'x'.
Now According to question, length exceeds quadruple its width by 5 cm.
Hence framing the above sentence in equation form, we get;
[tex]Length=4x+5[/tex]
Now we use the formula of perimeter of rectangle,
[tex]Perimeter=2(Length+width)[/tex]
On substituting the values, we get;
[tex]2(4x+5+x)=100\\\\(5x+5)=\frac{100}{2}\\\\5x+5=50\\\\5x=50-5\\\\5x=45\\\\x=\frac{45}{5}=9[/tex]
Now, we substitute the value of x to get the value of length;
[tex]Length=4x+5=4\times9+5=36+5=41\ cm[/tex]
Thus, Width=9 cm Length=41 cm
Hence the dimensions of the newly formed rectangle is length is 41 cm and width is 9 cm.
Salmon Weights: Assume that the weights of spawning Chinook salmon in the Columbia river are normally distributed. You randomly catch and weigh 17 such salmon. The mean weight from your sample is 19.2pounds with a standard deviation of 4.4 pounds. You want to construct a 90% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River.
(a) What is the point estimate for the mean weight of all spawning Chinook salmon in the Columbia River?
pounds
(b) Construct the 90% confidence interval for the mean weight of all spawning Chinook salmon in the Columbia River. Round your answers to 1 decimal place.
< ? <
(c) Are you 90% confident that the mean weight of all spawning Chinook salmon in the Columbia River is greater than 18 pounds and why?
No, because 18 is above the lower limit of the confidence interval.
Yes, because 18 is below the lower limit of the confidence interval.
No, because 18 is below the lower limit of the confidence interval.
Yes, because 18 is above the lower limit of the confidence interval.
(d) Recognizing the sample size is less than 30, why could we use the above method to find the confidence interval?
Because the sample size is greater than 10.
Because we do not know the distribution of the parent population.
Because the parent population is assumed to be normally distributed.
Because the sample size is less than 100.
Answer:
a) [tex]\bar X= 19.2[/tex] represent the sample mean. And that represent the best estimator for the population mean since [tex]\hat \mu =\bar X=19.2[/tex] b) The 90% confidence interval is given by (17.3;21.1)
c) No, because 18 is above the lower limit of the confidence interval.
d) Because the parent population is assumed to be normally distributed.
The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.
Step-by-step explanation:
1) Notation and definitions
n=17 represent the sample size
Part a
[tex]\bar X= 19.2[/tex] represent the sample mean. And that represent the best estimator for the population mean since [tex]\hat \mu =\bar X=19.2[/tex] Part b
[tex]s=4.4[/tex] represent the sample standard deviation
m represent the margin of error
Confidence =90% or 0.90
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
2) Calculate the critical value tc
In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. The degrees of freedom are given by:
[tex]df=n-1=17-1=16[/tex]
We can find the critical values in excel using the following formulas:
"=T.INV(0.05,16)" for [tex]t_{\alpha/2}=-1.75[/tex]
"=T.INV(1-0.05,16)" for [tex]t_{1-\alpha/2}=1.75[/tex]
The critical value [tex]tc=\pm 1.75[/tex]
3) Calculate the margin of error (m)
The margin of error for the sample mean is given by this formula:
[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]
[tex]m=1.75 \frac{4.4}{\sqrt{17}}=1.868[/tex]
4) Calculate the confidence interval
The interval for the mean is given by this formula:
[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]
And calculating the limits we got:
[tex]19.2 - 1.75 \frac{4.4}{\sqrt{17}}=17.332[/tex]
[tex]19.2 + 1.75 \frac{4.4}{\sqrt{17}}=21.068[/tex]
The 90% confidence interval is given by (17.332;21.068) and rounded would be: (17.3;21.1)
Part c
No, because 18 is above the lower limit of the confidence interval.
Part d
Because the parent population is assumed to be normally distributed.
The reason of this is because the t distribution is an special case of the normal distribution when the degrees of freedom increase.
Find the critical value in the t-distribution for a two-tailed hypothesis test in which α = .05 and df = 15 Round your answer to the thousandth decimal place.
Answer:[tex]\pm2.131[/tex]
Step-by-step explanation:
The critical value in the t-distribution for a two-tailed hypothesis test is the t-value in the students's t-distribution table corresponding to the [tex]\alpha/2[/tex] and df , where [tex]\alpha=[/tex] significance level and df = Degree of freedom.
We are given , [tex]\alpha=0.05[/tex]
Then, [tex]\alpha/2=0.025[/tex]
df = 15
Now , from the students's t-distribution table
The critical value in the t-distribution for a two-tailed hypothesis test is [tex]t=\pm2.131449\approx\pm2.131[/tex]
Hence, the required t-value = [tex]\pm2.131[/tex]
Each lap around pavia park is 1 7/8 miles. Ellen rode her bike for 3 1/2 laps before leaving the park. How many total miles did ellen ride her bike in pavia park?
Answer:
Step-by-step explanation:
The distance of each lap around pavia park is 1 7/8 miles. Converting
1 7/8 miles to improper fraction, it becomes 15/8 miles.
Ellen rode her bike for 3 1/2 laps before leaving the park. Converting
3 1/2 laps to improper fraction, it becomes 7/2 laps.
The total number of miles that Ellen rode her bike in pavia park would be the product of the distance of each lap and the number of laps that he covered. It becomes
15/8 × 7/2 = 105/16 = 6.5626 miles