Answer:
Part 1: The statistic
[tex]t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)
And the degrees of freedom are given by [tex]df=n_1 +n_2 -2=35+35-2=68[/tex]
Replacing we got
[tex]t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31[/tex]
Part 2: P value
Since is a right tailed test the p value would be:
[tex]p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002[/tex]
Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense
Step-by-step explanation:
Data given
[tex]\bar X_{1}=5135[/tex] represent the mean for four year college
[tex]\bar X_{2}=4436[/tex] represent the mean for two year college
[tex]s_{1}=783[/tex] represent the sample standard deviation for four year college
[tex]s_{2}=553[/tex] represent the sample standard deviation two year college
[tex]n_{1}=35[/tex] sample size for the group four year college
[tex]n_{2}=35[/tex] sample size for the group two year college
[tex]\alpha=0.01[/tex] Significance level provided
t would represent the statistic (variable of interest)
System of hypothesis
We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:
Null hypothesis:[tex]\mu_{1}-\mu_{2}\leq 0[/tex]
Alternative hypothesis:[tex]\mu_{1} - \mu_{2}> 0[/tex]
We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.
Part 1: The statistic
[tex]t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)
And the degrees of freedom are given by [tex]df=n_1 +n_2 -2=35+35-2=68[/tex]
Replacing we got
[tex]t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31[/tex]
Part 2: P value
Since is a right tailed test the p value would be:
[tex]p_v =P(t_{68}>4.31)=0.000022[/tex]
Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense
Solve the equation, 4 - (x + 1) = 6.
Answer:
X=-3
Step-by-step explanation
First of all, remove the parenthesis by distributing the value of the negative sign so it would be 4-x-1=6 , next find the like terms which is 4-1 is 3. so 3-x=6.
transpose the 3 to the other side so it would be -x=6-3
-x=3 divide both sides by -1. X=-3
A number line showing 2/3 divided by 1/12
Answer:
how do you thik,,, n vm
Step-by-step explanation:
4.37 is what% of 460
Answer:
0.95%
Step-by-step explanation:
Simply divide 100 with 460 then multiply 4.37
100/460 = 0.2173913043
Multiply by 4.37 and you get 0.95
One day, Donnie observes that the wind is blowing at 6 miles per hour. A unladen swallow nesting near Donnie’s house flies three quarters of a mile down the road (in the direction of the wind), turns around, and returns exactly 4 minutes later. What is the airspeed of the unladen swallow? (Here, ‘airspeed’ is the speed that the swallow can fly in still air.) socratic.org
Answer: 11.24 miler per hour
Step-by-step explanation:
Which two events are most likely to be independent? A) Being a senior; going to homeroom. B) Registering to vote; being left-handed. C) Having a car accident; having a junior license. D) Doing the Statistics homework; getting an A on the test.
Answer:
It is B) Registering to vote; being left-handed.
Step-by-step explanation:
These two events would be the only situation where the occurrence of one has no effect on the other, and each one of the other examples contain an event that affects the other probability of the other event.
Registering to vote; being left-handed are independent events so option (B) will be correct.
What are dependent and independent events?If the probability of an event affects the other then that will call a dependent event.
For example, getting two red balls from a bag without replacement.
While independent events are events, where the probability of the first event doesn't affect another.
For example tossing a coin.
Given the two events in all options.
In option (A) if you are a senior then it is possible that you can go to homeroom.
In option (B) if you are left-handed then also you can get a mark on the right finger it doesn't matter so that will be an independent event.
Hence "Registering to vote; being left-handed are independent events".
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A scientist claims that 7%7% of viruses are airborne. If the scientist is accurate, what is the probability that the proportion of airborne viruses in a sample of 418418 viruses would be greater than 6%6%? Round your answer to four decimal places.
The probability that the proportion of airborne viruses in the sample would be greater than 6% is approximately 0.5596.
Explanation:To find the probability that the proportion of airborne viruses in a sample of 418 viruses would be greater than 6%, we can use the normal distribution. First, let's find the mean and standard deviation of the proportion of airborne viruses. The mean is 7% (0.07) and the standard deviation can be calculated using the formula: sqrt((p(1-p))/n), where p is the proportion and n is the sample size.
Using the given information, the standard deviation is sqrt((0.07(1-0.07))/418) ≈ 0.0064.
Next, we convert the 6% threshold to a z-score using the formula (x - mean) / standard deviation. So, the z-score is (0.06 - 0.07) / 0.0064 ≈ -0.156.
To find the probability that the proportion is greater than 6%, we look up the z-score (-0.156) in the standard normal distribution table and subtract the corresponding probability from 1. The probability is approximately 0.5596.
10 boys share 7 cereal bars equally what fraction of the cereal bars does each boy get? show the work.
Answer:
1 43/100
Step-by-step explanation:
10 boys divide 7 cereal bars equally or
10 / 7 = 1.428571428571429
Or 1.43 rounded Or 1 43/100
A freight train rumbles by as bob watches. Each freight container on the train is shaped like a rectangular prism 17 meters long,3 meters wide, and 3 meters tall. What is the volume of a freight container on that train
Answer:
The answer is 153
Step-by-step explanation:
You multiply 17 by 3 by 3.
The volume of a rectangular prism-shaped freight container is calculated by multiplying its length, width, and height. For a container that is 17 meters long, 3 meters wide, and 3 meters high, the volume is 153 cubic meters.
Explanation:The question is asking for the volume of the rectangular prism-shaped freight container. In mathematics, the volume of a rectangular prism is found by multiplying its length, width, and height. For the freight container with dimensions of 17 meters long, 3 meters wide, and 3 meters high, we can find the volume by multiplying all these values together.
So, the volume V = length x width x height = 17m x 3m x 3m = 153 cubic meters. Therefore, each freight container on the train has a volume of 153 cubic meters.
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Kesha threw her baton up in the air from the marching band platform during practice. The equation h(t) = −16t² + 54t + 40 gives the height of the baton, in feet, t seconds after it is thrown from the platform. What is the height of the platform? At what speed was the baton thrown? If she doesn't catch it, when will it hit the ground?
Answer:
a) 40 feet
b) 54 ft/min
c) 4 mins
Step-by-step explanation:
Solution:-
- Kesha models the height ( h ) of the baton from the ground level but thrown from a platform of height hi.
- The function h ( t ) is modeled to follow a quadratic - parabolic path mathematically expressed as:
h ( t ) = −16t² + 54t + 40
Which gives the height of the baton from ground at time t mins.
- The initial point is of the height of the platform which is at a height of ( hi ) from the ground level.
- So the initial condition is expressed by time = 0 mins, the height of the baton h ( t ) would be:
h ( 0 ) = hi = -16*(0)^2 + 54*0 + 40
h ( 0 ) = hi = 0 + 0 + 40 = 40 feet
Answer: The height of the platform hi is 40 feet.
- The speed ( v ) during the parabolic path of the baton also varies with time t.
- The function of speed ( v ) with respect to time ( t ) can be determined by taking the derivative of displacement of baton from ground with respect to time t mins.
v ( t ) = dh / dt
v ( t )= d ( −16t² + 54t + 40 ) / dt
v ( t )= -2*(16)*t + 54
v ( t )= -32t + 54
- The velocity with which Kesha threw the baton is represented by tim t = 0 mins.
Hence,
v ( 0 ) = vi = -32*( 0 ) + 54
v ( 0 ) = vi = 54 ft / min
Answer: Kesha threw te baton with an initial speed of vo = 54 ft/min
- The baton reaches is maximum height h_max and comes down when all the kinetic energy is converted to potential energy. The baton starts to come down and cross the platform height hi = 40 feet and hits the ground.
- The height of the ball at ground is zero. Hence,
h ( t ) = 0
0 = −16t² + 54t + 40
0 = -8t^2 + 27t + 20
- Use the quadratic formula to solve the quadratic equation:
[tex]t = \frac{27+/-\sqrt{27^2 - 4*8*(-20)} }{2*8}\\\\t = \frac{27+/-\sqrt{1369} }{16}\\\\t = \frac{27+/-37 }{16}\\\\t = \frac{27 + 37}{16} \\\\t = 4[/tex]
Answer: The time taken for the baton to hit the ground is t = 4 mins
A credit card company wants to test the hypothesis that its account holders spend an average of $100 per month at gasoline stations. They take a sample of 1000 accounts and find an average spend of $115 with a standard deviation of $41. Conduct this hypothesis test with a .01 level of significance. What is the test statistic?
Answer:
test statistics = 11.57
Step-by-step explanation:
test statics = [tex]\frac{average spend from the sample - average spend from theory}{Standard deviation / \sqrt{no of samples} }[/tex]
= [tex]\frac{115-100}{41/\sqrt{1000} }[/tex] = 11.57
the sum 27 and a number g
Answer:
27 + g is how you write if u want a solution tell me what G is = to
Step-by-step explanation:
Rhada has a 6-pound bag of day . Her craft project requires 6 ounces of clay for each batch of 3 ornaments,
Answer:
Rhada can make 48 ornaments if she uses all of the clay.
Step-by-step explanation:
The complete question is:
Rhada has a 6-pound bag of day . Her craft project requires 6 ounces of clay for each batch of 3 ornaments. If she uses all of the clay, how many ornaments can Rhada make?
Solution:
The amount of clay Rhada has, X = 6-pound.
Convert the weight of clay bag into ounces as follows:
1 pound = 16 ounces
6 pound = 16 × 6
= 96 ounces.
So, Rhada has 96 ounces of clay.
It is provided that her craft project requires 6 ounces of clay for each batch of ornaments.
Compute the number of batches that can be made by 96 ounces of clay as follows:
Number of batches = Total weight of clay ÷ Amount required for each batch
[tex]=\frac{96}{6}\\=16[/tex]
Thus, Rhada can make 16 batches of ornaments.
Now, it is also provided that each batch has 3 ornaments.
Compute the number of ornaments in 16 batches as follows:
Number of ornaments = Number of batches × No. of ornament in 1 batch
[tex]=16\times 3\\=48[/tex]
Thus, Rhada can make 48 ornaments if she uses all of the clay.
The package of a particular brand of rubber band says that the bands can hold a weight of 7 lbs. Suppose that we suspect this might be an overstatement of the breaking weight. So we decide to take a random sample of 36 of these rubber bands and record the weight required to break each of them. The mean breaking weight of our sample of 36 rubber bands is 6.6 lbs. Assume that the standard deviation of the breaking weight for the entire population of these rubber bands is 2 lbs. True or false
Answer:
The statement is True.
Step-by-step explanation:
In this case we need to determine whether the rubber bands in a package of a particular brand of rubber band can hold a weight of 7 lbs or less.
A one-sample test can be used to perform the analysis.
The hypothesis can be defined as follows:
H₀: The mean weight the rubber bands can hold is 7 lbs, i.e. μ = 7.
Hₐ: The mean weight the rubber bands can hold is less than 7 lbs, i.e. μ < 7.
The information provided is:
[tex]n=36\\\bar x=6.6\ \text{lbs}\\\sigma=2\ \text{lbs}[/tex]
As the population standard deviation is provided, we will use a z-test for single mean.
Compute the test statistic value as follows:
[tex]z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{6.6-7}{2/\sqrt{36}}=-1.20[/tex]
The test statistic value is -1.20.
Decision rule:
If the p-value of the test is less than the significance level then the null hypothesis will be rejected.
Compute the p-value for the two-tailed test as follows:
[tex]p-value=P(Z<-1.20)=0.1151[/tex]
*Use a z-table for the probability.
The p-value of the test is 0.1151.
The p-value of the test is very large for all the commonly used significance level. The null hypothesis will not be rejected.
Thus, it can be concluded that the mean weight the rubber bands can hold is 7 lbs.
Hence, the statement is True.
The number P, in hundreds of bacteria in a sample, can be modeled by the Equation P= T^4+ 5T^3 + 5T^2 + 6t where t is measured in weeks. explain how to determine the number of weeks which the population would be greater than 10,000
Answer:
The values of T which satisfies the inequality are: [tex](8.8, \infty)[/tex]
Step-by-step explanation:
In the equation [tex]P= T^4+ 5T^3 + 5T^2 + 6T[/tex]
Represent P by 10,000Write an inequality where the expression is greater than 10000: [tex]T^4+ 5T^3 + 5T^2 + 6T>10000[/tex]Get 0 on one side of the inequality.[tex]T^4+ 5T^3 + 5T^2 + 6T-10000>0[/tex]
Graph the polynomial function. We have real x-intercepts of 8.84 and -11.38.Determine intervals where the graph is above the x-axis. Since negative values of x in this situation are irrelevant, the values of T which satisfies the inequality are: [tex](8.8, \infty)[/tex]We now test a value from the set of solution to see if it is valid.Let T=9[tex]T^4+ 5T^3 + 5T^2 + 6T>10000\\9^4+ 5(9)^3 + 5(9)^2 + 6(9)>10000\\10665>10000[/tex]
Since 10665 is grater than 10000, the result is reasonable.
A process that fills packages is stopped whenever a package is detected whose weight falls outside the specification. Assume that each package has probability 0.01 of falling outside the specification and that the weights of the packages are independent. Find the mean number of packages that will be filled before the process is stopped.
The mean number of packages that will be filled before the process is stopped is 100
Step-by-step explanation:
Step 1
Given that each package has probability 0.01 of falling outside the specification (Probability of Failure)
The probability of Success is (100-probability of failure)=(100-0.01)=0.99
It is important to note that the weight of the packages are independent
Using the data given in the question we get the following:
P(fail) =Probability of Failure
P(success)=Probability of Success
P(fail) = 0.01 and P(success) = 0.99
Step 2
The mean of the Geometric distribution is:
P = 0.01;
μx = 1/p = 1/0.01 = 100
Step 3
Thus, we can say that the mean number of packages that will be filled before the process is stopped is 100
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
Recall that a percent is a ratio of a number to 100. For example 39% means 39 out of 100.
1.Last year, there were 20 students on the track team. Eight of them competed in the long
jump. What percent of the team members competed in the long jump?
Answer:
1.6% should be right.
Step-by-step explanation:
A teacher is experimenting with computer-based instruction. In which situation could the teacher use a hypothesis test for a population mean?
A) She gives each student a pretest. Then she teaches a lesson using a computer program. Afterwards, she gives each student a posttest. The teacher wants to see if the difference in scores will show an improvement.
B) She randomly divides the class into two groups. One groups receives computer-based instruction. The other group receives traditional instruction without computers. After instruction, each student has to solve a single problem. The teachers wants to compare the proportion of each group who can solve the problem.
C) The teacher uses a combination of traditional methods and computer-based instruction. She asks students which they liked better. She wants to determine if the majority prefer the computer-based instruction.
A teacher can use a hypothesis test for a population mean in situation A, where pretest scores are compared to posttest scores to determine if there is a significant improvement.
Explanation:A hypothesis test for a population mean can be used in situation A. The teacher can use the pretest scores as the population mean and then compare it to the posttest scores to see if there is a statistically significant improvement. The hypothesis test will determine if the difference in scores is due to chance or if it is a result of the computer-based instruction.
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5) The records of 100 SNHU students show the following courses taken: 53 students took History 41 students took Marketing 48 students took Writing 18 students took History and Marketing 21 students took Marketing and Writing 7 students took all 3 courses 9 students took none of these courses Answer the following questions. Show how you obtained your solution. a) How many students took Marketing and Writing, but not History? b) How many students took only History?.
Answer:
a) How many students took Marketing and Writing, but not History?
14 Students.
b) How many students took only History?
23 Students.
Step-by-step explanation:
We were given the following values in the question
Total number of students = 100
53 students took History = n(H)
41 students took Marketing = n(M)
48 students took Writing = n(W)
18 students took History and Marketing = n(H n M)
21 students took Marketing and Writing = n ( M n W)
7 students took all 3 courses = n ( H n M n W)
9 students took none of these courses
a) How many students took Marketing and Writing, but not History?
This is calculated as
= n(M n W) - n ( H n M n W)
= 21 - 7
= 14
Therefore, only 14 Students took Marketing and Writing but not History.
b) How many students took only History?
Step 1
Subtract the Total number of students from the number of students who did not take any course
100 - 9 = 91 students
Step 2
The number of students who took History and Writing (H n W) was not given, so we find it.
Therefore,
n(H) + n(M) + n(W) -n( H n M) - n( M n W) - n( H n W) + n( H n M n W) = 91
53 + 41 + 48 - 18 - 21 - n( H n W) + 7 = 91
n( H n W) = 149 - 39 -91
n( H n W) = 149 - 130
n( H n W) = 19
Step 3
To find the number of students taking History only, we calculate it as
Number of Students that took History only = n( H) - n(H n W) - n( H n M) + n( H n M n W)
= 53 - 19 - 18 + 7
= 53 + 7 -19 - 18
= 60 - 37
= 23
Therefore, the number of students that took only History is 23 Students.
An adult African elephant weighs 8.25 tons. How much do 27 elephants weigh in tons? Answer without units.
Answer:
222.75 tons
Step-by-step explanation:
1 elephant = 8.25 tons
x 27 x 27
27 elephants = 222.75 tons
I hope this helps.
Final answer:
To find the weight of 27 elephants, multiply the weight of one elephant by 27. The total weight of 27 elephants, each weighing 8.25 tons, is 222.75 tons.
Explanation:
To calculate the total weight of 27 elephants each weighing 8.25 tons, we simply multiply the weight of one elephant by the number of elephants:
8.25 tons/elephant × 27 elephants = 222.75 tons
It's clear we need to multiply when converting from a single elephant's weight to the combined weight of multiple elephants. We use the fact that 1 ton = 2,000 pounds for conversion between units if needed, but here we keep the weight in tons as the question requested.
write a quadratic function f whose zeros are 6 and 1
Answer:
x^2-7x+6
Step-by-step explanation:
(x-6)(x-1)
x^2-1x-6x+6
x^2-7x+6
The quadratic function whose zeros are 6 and 1 is: f(x) = x² - 7x + 6.
How do we find the quadratic function with zero's at x = 6 and x = 1?To write a quadratic function f(x) with zeros at x=6 and x=1, we can start by representing it in factored form:
f(x) = a(x−6)(x−1)
Where a is a constant. Depending on the desired leading coefficient, you can choose any value for a. For simplicity, let's choose a=1.
Therefore, the quadratic function with zeros at 6 and 1 is:
f(x) = (x−6)(x−1)
f(x) = x² − 7x + 6
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b. Interpret the Rsquared value. Does the multiple regression equation help us predict the total golf score much better than we could without knowing that equation? A. No, because Rsquared is close to 1. B. Yes, because Rsquared is not close to 1. C. No, because Rsquared is not close to 1. D. Yes, because Rsquared is close to 1.
Answer:
Depending upon the golf score predictors we can conclude the answer from given Conditional statements
If Someone want Total golf score better then, following .condition
Yes, Because R squared is close to 1 . ....(If Regression values are prefect to fit with model then only)
(But the data is not sufficient to say 100%, it depends on score given)
Step-by-step explanation:
Given :
Explanatory Question on r-squared in regression.
To Find :
Does regression equation help us to know total golf score much better.
Solution;
As in Question there is no golf score given so we cant see "yes" or "no".
But we can be conditional here ,
1) If golf score to be accurate then , multiple regression predict points should fit the model prefect ,hence the R squared value "must be close to 1" .
2) If gold score is not accurate then multiple regression suggest that model does not explain any variables, no linear relationship.
then r squared value "must not be close to 1".
Depending upon the golf score predictors we can conclude the answer from given Conditional statements.
that Yes, Because R squared is close to 1 .
What is the whole number for 6/1?
Answer:
6
Step-by-step explanation:
Answer:
6
Step-by-step explanation:
Because 1/1 = 1 whole and you add 5 more 1/1 and you get 6/1 so their is 6 wholes
A manufacturer claims that the thickness of the spearmint gum it produces is 7.5 one-hundredths of an inch. A quality control specialist regularly checks this claim. From his experience, the specialist knows the distribution is right skewed and that the standard deviation is 0.4 one-hundredths of an inch. On one production run, he took a random sample of n = 100 pieces of gum and measured their thickness. Which hypothesis test would be most appropriate for this task?
The most appropriate hypothesis test for this task would be a one-sample t-test.
Explanation:For this task, the most appropriate hypothesis test to use would be a one-sample t-test. A one-sample t-test is used to determine whether the mean of a sample is significantly different from a population mean.
In this case, the quality control specialist wants to test whether the mean thickness of the gum produced by the manufacturer is significantly different from the claimed value of 7.5 one-hundredths of an inch. The specialist takes a random sample of n = 100 pieces of gum and measures their thickness.
The one-sample t-test is appropriate because the population distribution is assumed to be right skewed and the standard deviation is known. The t-test allows for the use of a smaller sample size and accommodates the assumption of a non-normal distribution.
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A study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, consider the hypothesis testing with H subscript 0 colon space p space equals space 0.64 space space space v s. space H subscript a colon space p space not equal to 0.64If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good as the national brand, what is the p-value
Answer:
[tex]z=\frac{0.52 -0.64}{\sqrt{\frac{0.64(1-0.64)}{100}}}=-2.5[/tex]
The p value for this case is given by:
[tex]p_v =2*P(z<-2.5)=0.0124[/tex]
Step-by-step explanation:
Data given and notatio
n=100 represent the random sample selected
X=52 represent the shoppers stating that the supermarket brand was as good as the national brand
[tex]\hat p=\frac{52}{100}=0.52[/tex] estimated proportion of stating that the supermarket brand was as good as the national brand
[tex]p_o=0.64[/tex] is the value that we want to test
z would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We need to conduct a hypothesis in order to test the claim that the true proportion of shoppers stating that the supermarket brand was as good as the national brand is 0.64 or not, then the system of hypothesis are.:
Null hypothesis:[tex]p=0.64[/tex]
Alternative hypothesis:[tex]p \neq 0.64[/tex]
The statistic is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
Calculate the statistic
The statistic is given by:
[tex]z=\frac{0.52 -0.64}{\sqrt{\frac{0.64(1-0.64)}{100}}}=-2.5[/tex]
Statistical decision
The p value for this case is given by:
[tex]p_v =2*P(z<-2.5)=0.0124[/tex]
The sodium content of a popular sports drink is listed as 205 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sample mean of 219.2 mg with a sample standard deviation of 18.0 mg. (a) State the hypotheses for a two-tailed test of the claimed sodium content. H0: μ ≥ 205 vs. H1: μ < 205 H0: μ ≤ 205 vs. H1: μ > 205 H0: μ = 205 vs. H1: μ ≠ 205
Answer:
[tex]t=\frac{219.2-205}{\frac{18}{\sqrt{20}}}=3.528[/tex]
[tex]p_v =2*P(t_{19}>3.528)=0.0022[/tex]
If we compare the p value and a significance level 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, and we can conclude that the lifetime is signficantly different from 205 hours.
Step-by-step explanation:
Data given and notation
[tex]\bar X=219.2[/tex] represent the sample mean
[tex]s=18[/tex] represent the sample standard deviation
[tex]n=20[/tex] sample size
[tex]\mu_o =205[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to apply a two tailed test.
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu = 205[/tex]
Alternative hypothesis :[tex]\mu \neq 205[/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{219.2-205}{\frac{18}{\sqrt{20}}}=3.528[/tex]
Give the appropriate conclusion for the test
The degreed of freedom are:
[tex] df = n-1= 19[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{19}>3.528)=0.0022[/tex]
Conclusion
If we compare the p value and a significance level 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, and we can conclude that the lifetime is signficantly different from 205 hours.
Answer:
H0: mu equals 205 vs. H1: mu not equals 205.
Step-by-step explanation:
A null hypothesis (H0) is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It expresses equality.
An alternate hypothesis (H1) is also a statement from the population parameter which negates the null hypothesis and is accepted if the null hypothesis is true. It expresses inequality.
A two-tailed test is one in which the alternate hypothesis is expressed using any of the inequality signs below:
not equal to, less than or equal to, greater than or equal to.
Find the greatest common factor of 4c and 10c".
Answer:
Step-by-step explanation:
Create 3 fraction whose product is -5/24
Step-by-step explanation:
1.)
[tex] - \frac{5}{3} \times \frac{1}{8} = - \frac{5}{24} \\ \\ [/tex]
2.)
[tex] - \frac{5}{2} \times \frac{1}{12} = - \frac{5}{24} \\ [/tex]
3.)
[tex] - \frac{1}{6} \times \frac{5}{4} = - \frac{5}{24} \\ [/tex]
Kevin installed a certain brand of automatic garage door opener that utilizes a transmitter control with four independent switches, each one set on or off. The receiver (wired to the door) must be set with the same pattern as the transmitter. If six neighbors with the same type of opener set their switches independently, what is the probability of at least one pair of neighbors using the same settings?
Kevin installed a certain brand of automatic garage door opener that utilizes a transmitter control with four independent switches, each one set on or off. The receiver (wired to the door) must be set with the same pattern as the transmitter. If six neighbors with the same type of opener set their switches independently.The probability of at least one pair of neighbors using the same settings is 0.65633
Step-by-step explanation:
Step 1
In the question it is given that
Automatic garage door opener utilizes a transmitter control with four independent switches
So .the number of Combinations possible with the Transmitters =
2*2*2*2= 16
Step 2
Probability of at least one pair of neighbors using the same settings = 1- Probability of All Neighbors using different settings.
= 1- 16*15*14*13*12*11/(16^6)
Step 3
Probability of at least one pair of neighbors using the same settings=
= 1- 0.343666
Step 4
So the probability of at least one pair of neighbors using the same settings
is 0.65633
Which two events are independent? A.) go sledding, snow B.) sleep,dream C.) count to 100, fix a bike D.) do chores, earn allowance
Answer:
A and C
Step-by-step explanation:
The meaning of an event being independent is that one event doesn't effect the probability of they other event happening. So B is wrong because whether yo sleep or not can affect whether you dream or not and D is wrong because you earning and allowance is based off whether you do your chores or not.
What is the approximate value of the expression below?
-3,033 ÷ (-56)
Answer:
54.1607142857
Step-by-step explanation:
just simplify it
The approximate value of -3,033 divided by -56 is -54 when rounded to two significant figures.
To approximate the value of -3,033 ÷ (-56), you can perform the division by changing both numbers into their absolute values and divide normally as the negative signs will cancel each other out. This results in the division of 3,033 by 56. Using a calculator, you would get 54.1607142857. However, since we need an approximate value with two significant figures, the result would be rounded to -54.