Answer:
a) Null hypothesis:[tex]\mu \geq 10[/tex]
Alternative hypothesis:[tex]\mu < 10[/tex]
b) [tex]p_v =P(t_{17}<-2.3)=0.017[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
c) [tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]
And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.
d) [tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X[/tex] represent the sample mean
[tex]s[/tex] represent the sample deviation
[tex]n=18[/tex] sample size
[tex]\mu_o =10[/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 check if the true mean is at least 10 hours, the system of hypothesis would be:
Part a
Null hypothesis:[tex]\mu \geq 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)
Part b
For this case we have t=-2.3 , [tex]\alpha=0.05[/tex]
First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]
Now since we are conducting a left tailed test the p value is given by:
[tex]p_v =P(t_{17}<-2.3)=0.017[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
Part c
For this case we have t=-1.8 , [tex]\alpha=0.01[/tex]
First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]
Now since we are conducting a left tailed test the p value is given by:
[tex]p_v =P(t_{17}<-1.8)=0.0448[/tex]
And on this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis.
Part d
For this case we have t=-3.6 , [tex]\alpha=0.05[/tex]
First we need to find the degrees of freedom [tex]df=n-1=18-1=17[/tex]
Now since we are conducting a left tailed test the p value is given by:
[tex]p_v =P(t_{17}<-3.6)=0.0011[/tex]
And on this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis.
The student is learning about conducting a one-sample t-test in a hypothesis testing problem related to the lifetime of a pen. The null and alternative hypothesis are defined, and conclusions are drawn based on various test statistics and significance levels.
Explanation:This situation relates to hypothesis testing in statistics, in particular the usage of a one-sample t-test.
The null hypothesis (H0) and alternative hypothesis (Ha) to be tested would be:
H0: μ ≥ 10 hours (The average lifetime of the pen is at least 10 hours)Ha: μ < 10 hours (The average lifetime of the pen is less than 10 hours)The t value represents the test statistic for conducting the hypothesis test, and the 'α' and 'u' are the significance levels.
With t = -23 and α = 0.05, this falls within the rejection region, as the t value is less than the critical value for a one-tailed test at significance level α = 0.05. Hence, we reject the null hypothesis. There is evidence to suggest that the pen's lifetime is less than 10 hours.
Similarly, with t = -1.8 and u = 0.01, the null hypothesis would be retained, as this value is within the acceptance region.The conclusion would be that there is not sufficient evidence to suggest that the pen's lifetime is less than 10 hours.
Finally, with t = -3.6, the null hypothesis is rejected, suggesting that the pen's lifetime is less than 10 hours. However, this is only applicable if the significance level is above the calculated p-value for t = -3.6.
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The concept of aging as a result of cellular duplication errors is based on the fact that the body's ability to make new cells that are exact copies of the original ones: a. stops at about age 80. b. continues unchanged throughout life. c. becomes less able to repair duplication errors. d. stops at about age 18.
Answer:
Becomes less able to repair duplication errors
Step-by-step explanation:
This is premised on the fact that aging has been connected with the deterioration of DNA maintenance and repair machinery, which tends to lose its ability to replicate new cell as a person age with time.
Ten experts rated a newly developed chocolate chip cookie on a scale of 1 to 50. Their ratings were:
34, 35, 41, 28, 26, 29, 32, 36, 38, and 40.
1. What is the mean deviation of the ratings?
Select one:
a. 8.00
b. 4.12
c. 12.67
d. 0.75
Answer:
Option B.
Step-by-step explanation:
The given data set is
34, 35, 41, 28, 26, 29, 32, 36, 38, 40
We need to find the mean deviation of the given data.
Number of observations, n = 10
Mean of the data is
[tex]Mean=\dfrac{\sum x}{n}[/tex]
[tex]Mean=\dfrac{34+35+41+28+26+29+32+36+38+40}{10}[/tex]
[tex]Mean=\dfrac{339}{10}[/tex]
[tex]Mean=33.9[/tex]
Formula for mean deviation is
[tex]\text{Mean deviation}=\dfrac{\sum |x-mean|}{n}[/tex]
[tex]\sum |x-mean|=|34-33.9|+|35-33.9|+|41-33.9|+|28-33.9|+|26-33.9|+|29-33.9|+ |32-33.9|+|36-33.9|+|38-33.9|+|40-33.9|=41.2[/tex]
[tex]\text{Mean deviation}=\dfrac{41.2}{10}[/tex]
[tex]\text{Mean deviation}=4.12[/tex]
The mean deviation of the ratings is 4.12.
Therefore, the correct option is B.
Answer:
b. 4.12
Step-by-step explanation:
We have been given that 10 experts rated a newly developed chocolate chip cookie on a scale of 1 to 50. Their ratings were:
34, 35, 41, 28, 26, 29, 32, 36, 38, and 40.
First of all, we will find the mean of the ratings.
[tex]\text{Mean of ratings}=\frac{34+35+41+28+26+29+32+36+38+40}{10}[/tex]
[tex]\text{Mean of ratings}=\frac{339}{10}[/tex]
[tex]\text{Mean of ratings}=33.9[/tex]
Let us find absolute deviation of each point from mean.
[tex]|34-33.9|=0.1[/tex]
[tex]|35-33.9|=1.1[/tex]
[tex]|41-33.9|=7.1[/tex]
[tex]|28-33.9|=5.9[/tex]
[tex]|26-33.9|=7.9[/tex]
[tex]|29-33.9|=4.9[/tex]
[tex]|32-33.9|=1.9[/tex]
[tex]|36-33.9|=2.1[/tex]
[tex]|38-33.9|=4.1[/tex]
[tex]|40-33.9|=6.1[/tex]
Now we will use mean deviation formula.
[tex]\text{Absolute mean deviation}=\frac{\Sigma |x-\mu|}{N}[/tex], where,
[tex]\mu=\text{Mean}[/tex] and N = Number of data points.
[tex]MD=\frac{0.1+1.1+7.1+5.9+7.9+4.9+1.9+2.1+4.1+6.1}{10}[/tex]
[tex]MD=\frac{41.2}{10}[/tex]
[tex]MD=4.12[/tex]
Therefore, the mean deviation of the ratings is 4.12 and option 'b' is the correct choice.
For the Data Set below, calculate the Variance to the nearest hundredth decimal place. (Do not use a coma in your answer) 175 349 234 512 638 549 500 611
Answer:
The variance of the data is 29966.3.
Step-by-step explanation:
The given data set is
175, 349, 234, 512, 638, 549, 500, 611
We need to find the variance to the nearest hundredth decimal place.
Mean of the data
[tex]Mean=\dfrac{\sum x}{n}[/tex]
where, n is number of observation.
[tex]Mean=\dfrac{3568}{8}=446[/tex]
The mean of the data is 446.
[tex]Variance=\dfrac{\sum (x-mean)^2}{n-1}[/tex]
[tex]Variance=\dfrac{(175-446)^2+(349-446)^2+(234-446)^2+(512-446)^2+(638-446)^2+(549-446)^2+(500-446)^2+(611-446)^2}{8-1}[/tex]
[tex]Variance=\dfrac{209764}{7}[/tex]
[tex]Variance=29966.2857[/tex]
[tex]Variance\approx 29966.3[/tex]
Therefore, the variance of the data is 29966.3.
Final answer:
The variance of the given data set is calculated by finding the mean, squaring the differences from the mean, summing these squares, and dividing by the count minus one. It results in a variance of 12790.54 when rounded to the nearest hundredth decimal place.
Explanation:
To calculate the variance of the data set, follow these steps:
First, find the mean (average) of the data set by adding all the numbers together and dividing by the total count.
Next, subtract the mean from each data point and square the result to get the squared differences.
Then, add up all of the squared differences.
Finally, divide the sum of the squared differences by the total number of data points minus one to get the variance (since this is a sample variance).
Data Set: 175, 349, 234, 512, 638, 549, 500, 611
Mean = (175 + 349 + 234 + 512 + 638 + 549 + 500 + 611) / 8 = 3793 / 8 = 474.125
Squared differences = (175 - 474.125)^2 + (349 - 474.125)^2 + (234 - 474.125)^2 + (512 - 474.125)^2 + (638 - 474.125)^2 + (549 - 474.125)^2 + (500 - 474.125)^2 + (611 - 474.125)^2
Sum of squared differences = 89533.78125
Variance = 89533.78125 / (8 - 1) = 12790.54
Therefore, the variance of the data set, to the nearest hundredth decimal place, is 12790.54.
A symbol used to name one or more parts of a whole or a set, or a location on a number line is a?
Answer:
Fraction
Step-by-step explanation:
Fraction is a symbol that represents a part of a whole. It consists of a numerator and a denominator. The numerator is the number above the fraction bar (also known as "Vinculum), while the denominator is the number below the fraction bar. The denominator is the total number of equal parts in a whole.
Examples of Fraction: [tex]\frac{1}{2}[/tex], [tex]\frac{2}{7}[/tex] and [tex]\frac{5}{8}[/tex].
In the first example, [tex]\frac{1}{2}[/tex], 1 is the numerator, while 2, is the denominator.
Additional Information
When the numerator is smaller than the denominator, the fraction is called "proper fraction." On the contrary, when the numerator is bigger than the denominator, the fraction is called "improper fraction."
A bag contains 8 red marbles, 3 blue marbles and 6 green marbles. If three marbles are drawn out of the bag, what is the probability, to the nearest 1000th, that all three marbles drawn will be red?
Answer:
0.082
Step-by-step explanation:
There are a total of 17 marbles, 8 of which are red.
The probability that the first marble is red is 8/17.
The probability that the second marble is red is 7/16.
The probability that the third marble is red is 6/15.
Therefore, the probability that all three marbles are red is:
P = 8/17 × 7/16 × 6/15
P = 7/85
P = 0.082
When individuals in a sample of 150 were asked whether or notthey supported capital punishment, the following information wasobtained.
Doyousupport Numberof
capitalpunishment? individuals
Yes 40
No 60
No Opinion 50
We are interested in determining whether or not the opinionsof the individuals (as to Yes, No and No Opinion) are uniformlydistributed.
The expected frequency for each group is?
a. 0.333
b. 0.50
c. 1/3
d. 50
Answer:
They are not uniformly distributed.
The expected frequency of each group is 50
Step-by-step explanation:
In probability distributions, uniform distribution refers to a probability distribution for which all of the values that a random variable can take on occur with equal probability.
In other words, for n number of events, the probability of occurrence 1,2,3,4......n is 1/n
There are 3 possible occurrence in the question above
1. Yes
2. No
3. No Opinion.
For the above events to have a uniform distribution, then they must have a probability of ⅓ each.
The expected frequency of each would then be ⅓ of n where n = 150
⅓ of 150 = 50
You perform a X2 goodness-of-fit test to see if the number of birthdays occurring each month matches the expected number (assuming each month is equally likely to be the birth month for any given individual). You get 20.5 as your X2 value. What is the P-value for this test?
Answer:
[tex]p_v = P(\chi^2_{11} >20.5)=0.0389[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(20.5,11,TRUE)"
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
We need to conduct a chi square test in order to check the following hypothesis:
H0: Each month is equally likely to be the birth month for any given individual
H1: Each month is NOT equally likely to be the birth month for any given individual
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
After calculate the statistic we got [tex]\chi^2 = 20.5[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=categories-1=12-1=11[/tex]
And we have categories =12 since we have 12 months in a year
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{11} >20.5)=0.0389[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(20.5,11,TRUE)"
The lumen output was determined for each of I = 3 different brands of lightbulbs having the same wattage, with J = 8 bulbs of each brand tested. The sums of squares were computed as SSE = 4776.3 and SSTr = 599.5. State the hypotheses of interest (including word definitions of parameters).
Answer:
Step-by-step explanation:
Hello!
The study variable is
X: Lumen of a bulb of the i brand. i=3
There are 3 populations of bulbs, Brand 1, Brand 2 and brand 3.
The objective is to test if the population means are equal.
The study parameters are:
μ₁: population mean lumen of the population of light bulbs of brand 1.
μ₂: population mean lumen of the population of light bulbs of brand 2.
μ₃: population mean lumen of the population of light bulbs of brand 3.
The hypothesis is:
H₀:μ₁= μ₂= μ₃= μ
H₁: At least one of the population means is different.
To test this hypothesis, considering the given information, I'll use an ANOVA test, then the statistic is defined as:
[tex]F= \frac{MSTr}{MSerror}[/tex]~[tex]F_{(I-1)(J-1)}[/tex]
Rejection region
This region is always one-tailed (right), the statistic is constructed as the mean square of the treatments divided by the mean square of the error, if the number of F is big, this means that the treatments have more effect over the populations. If the value of F is small, this means that there is no difference between the variability caused by the treatments and the one caused by the residues.
Since there is no significance level specified, I'll use α: 0.05
[tex]F_{(I-1);(J-1); 1 - \alpha } = F_{2; 7; 0.95} = 19.35[/tex]
You will reject the null hypothesis when F[tex]_{H_0}[/tex] ≥ 19.35
To calculate the statistic value you need to calculate the Mean Square of Treatments and the Mean Square of errors:
MSTr= SSTr/DfTr = 599.5/2= 299.75
MSerror= SSerror/Dferror= 4776.3/5= 955.26
F[tex]_{H_0}[/tex]= [tex]\frac{299.75}{955.26}[/tex]= 0.31
At this level the decision is to not reject the null hypothesis.
I hope it helps!
As reported on the package of seeds, the mean time until maturity of a certain variety of tomato plant is 70 days. It is also known that the maturity of this variety of tomato follows a normal distributed with standard deviation σ = 2.4. A researcher believes that it will indeed take more time in a given condition. To test his belief, he selects a sample of 36 plants of this variety under the given condition and measure the time until maturity. The sample mean is found to be 72 days. The value of the test statistic and p-value for testing H0: μ= 70, Ha: μ>70 are:
Group of answer choices
A. 1.00
B. 0.00
C. 0.786
Water is being pumped continuously from a pool at a rate proportional to the amount of water left in the pool. Initially there was 15,000 gallons of water in the pool; six minutes later there was 13,800 gallons.
At what rate was the amount of water in the pool decreasing when there were 14,000 gallons remaining and when will there be 5,000 gallons remaining?
Please show all steps.
Answer:
194.6 gpm at 14,000 gallons69.5 gpm at 5,000 gallons.Step-by-step explanation:
When a value is decreasing at a rate proportional to that value, it can be modeled by the formula
a = a0·e^(-kt)
where k is the constant of proportionality.
Alternatively, we can write the exponential function describing the pool volume* as ...
a = 15000·(138/150)^(t/6) = 15000·((138/150)^(1/6))^t
Comparing these, we see that ...
e^(-kt) = (138/150)^(t/6)
or ...
k = -ln(138/150)/6 ≈ 0.0138969
__
So, when 14000 gallons remain, the rate of decrease is ...
14000·0.0138969 ≈ 194.6 . . . gallons per minute
When 5000 gallons remain, the rate of decrease is ...
5000·0.0138969 ≈ 69.5 . . . gallons per minute
_____
* The generic form of this is ...
(initial value) · (multiplier per interval)^(number of intervals)
Here, the multiplier over a 6-minute period is 13800/15000 = 138/150, and the number of 6-minute intervals is t/6 when t is in minutes.
_____
Effectively, we make use of the fact that for ...
a = a0·e^(-kt)
the derivative is ...
da/dt = -k(a0·e^(-kt)) = -k·a
That is, k is the constant of proportionality mentioned in the first sentence of the problem statement.
The mean time taken to design a house plan by 38 architects was found to be 22 hours with a standard deviation of 3.70 hours. a. Construct a 95% confidence interval for the population mean μ. Round your answers to two decimal places.
To construct a 95% confidence interval for the population mean of the time taken to design a house plan, use the formula which states 21.08 to 22.92 hours.
Explanation:To construct a 95% confidence interval for the population mean, we can use the formula:
Confidence Interval = mean ± (critical value) * (standard deviation/sqrt(sample size))
Given that the mean time taken to design a house plan is 22 hours, the standard deviation is 3.70 hours, and the sample size is 38, we can calculate the confidence interval:
Confidence Interval = 22 ± (1.96) * (3.70/sqrt(38))
Calculating this gives us a confidence interval of approximately 21.08 to 22.92 hours.
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Juan roller skates at the constant speed of 8 miles per hour. How far can he travel in 2.4 hours.
Answer: Juan can travel 19 miles in 2.4 hours at a speed of 8 miles per hour
Step-by-step explanation:
Juan roller skates at the constant speed of 8 miles per hour. Distance travelled is expressed as
Distance = speed × time
Therefore, the distance that Juan can travel in 2.4 hours is
Distance = 2.4 × 8 = 19.2 miles
Approximating to the nearest whole number, it becomes 19 miles
A random sample of 100 high school students was surveyed regarding their favorite subject. The following counts were obtained: Favorite Subject Number of Students English Math Science 30 Art/Music The researcher conducted a test to determine whether the proportion of students was equal for all four subjects. What is the value of the test statistic? O b. 25 OOOO d. -4 How many degrees of freedom does the chi-square test statistic for a goodness of fit have when there are 10 categories? a. 74 OOOO d. 62
Answer:
a) [tex]\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6[/tex]
b) [tex]df=Categories-1=10-1=9[/tex]
Step-by-step explanation:
We assume the following info:
Favorite Subject Number of students
English 25
Math 30
Science 30
Art/Music 15
Total 100
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Part a
The system of hypothesis on this case are:
H0: There is no difference with the distribution proposed
H1: There is a difference with the distribution proposed
The level os significance assumed for this case is [tex]\alpha=0.05[/tex]
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The table given represent the observed values, we just need to calculate the expected values are 25 for each category.
And the calculations are given by:
[tex]E_{English} =25[/tex]
[tex]E_{Math} =25[/tex]
[tex]E_{Science} =25[/tex]
[tex]E_{Music} =25[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=Categories-1=4-1=3[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{3} >6)=0.112[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(6,3,TRUE)"
Part b
For this case we have this formula:
[tex]df=Categories-1=10-1=9[/tex]
Find the area of the following regions, expressing your results in terms of the positive integer n ≥ 2. The region bounded by f(x)=x and g(x)=x^1/n, for x≥0
Answer:
The area of the searched region is [tex]A= a+b+ \frac{2n}{n+1}- \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}-2[/tex]
Step-by-step explanation:
If you want to find the area of a region bounded by functions f(x) and G(x) between two limits (a,b), you have to do a double integral. you must first know which of the functions is greater than the other for the entire domain.
In this case, for 0<x<1, f(x)<g(x)
while for 1<x, g(x)<f(x).
Therefore if our domain is all real numbers superior to 0 (where the limit 0<a<1 and 1<b), we have to do 2 integrals:
A=A(a<x<1)+A(1<x<b)
[tex]A(a<x<1)=\int\limits^1_a {\int\limits^{x^{\frac{1}{n}} }_{x} } {} \, dy } \, dx = \int\limits^1_a {x^{\frac{1}{n} } -x \, dx = a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1}[/tex]
[tex]A(1<x<b)=\int\limits^b_1 {\int\limits^{x}_{x^{\frac{1}{n} } } {} \, dy } \, dx = \int\limits^b_1 {x-x^{\frac{1}{n} } \, dx =b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1}[/tex]
[tex]A=a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1} + b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1} = a+b+ \frac{2n}{n+1} - \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1} -2[/tex]
Sophia buys a certain brand of cereal that costs $5 per box. Yani changes to a super-saving brand of the same size. The equation shows the price, y, as a function of the number of boxes, x, for the new brand.
y = 4.35x
Part A: How many more dollars is the price of a box Sophia's original brand of cereal than the price of a box of the super-saving cereal? Show your work.
Part B: How much money does she save each month with the change in cereal brand if he buys 5 cereal boxes each month? Show your work.
$ 0.65 more dollars is the price of a box Sophia's original brand of cereal than the price of a box of the super-saving cereal
Amount saved each month with the change in cereal brand if he buys 5 cereal boxes each month is $ 3.25
Solution:
Given that Sophia buys a certain brand of cereal that costs $5 per box
The equation shows the price, y, as a function of the number of boxes, x, for the new brand:
y = 4.35x
Part A:
New brand, y = 4.35x where y is the price and x is the number of boxes
Original brand, y = 5x since given that cereal that costs $5 per box
If Sophia old cereal preference was $5, and the equation shows that the new cereal preference is $4.35, if I subtract the amount of the new one from the old,
we get , 5 - 4.35 = 0.65
Therefore, $ 0.65 more dollars is the price of a box Sophia's original brand of cereal than the price of a box of the super-saving cereal
Part B:
Given that if he buys 5 cereal boxes, let us calculate price for old and new brand
New brand, y = 4.35x
New brand, y = 4.35(5) = 21.75
Original brand, y = 5x = 5(5) = 25
Amount saved = $ 25 - $ 21.75 = $ 3.25
Thus amount saved each month with the change in cereal brand if he buys 5 cereal boxes each month is $ 3.25
A professor is trying to determine if her students guessed on a certain multiple choice question. She expects that if the students guessed, the distribution of answers would be uniform for that question. She compares the observed distribution of answers with the uniform distribution. The professor conducts a chi-square Goodness-of-Fit hypothesis test at the 5% significance level.
a) The null and alternative hypotheses are: H0: The student answers have the uniform distribution. Ha: The student answers do not have the uniform distribution.
b) χ20=13.167.
c) χ20.05=7.815.
d) What conclusions can be made?
Select all that apply:
A. We should reject H0.
B. We should not reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
D. At the 5% significance level, there is not enough evidence to conclude that the students were not guessing.
Answer:
A. We should reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
We need to conduct a chi square test in order to check the following hypothesis:
H0: The student answers have the uniform distribution.
H1: The student answers do not have the uniform distribution.
The level os significance assumed for this case is [tex]\alpha=0.05[/tex]
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]
On this case we assume that the calculated statistic is given by:
Statistic calculated
[tex]\chi^2_{calc}=13.167[/tex]
P value
Assuming the we have 2 rows and 4 columns on the contingency table.
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(rows-1)(cols-1)=(2-1)(4-1)=3[/tex]
We can calculate the critical value with this formula in excel:" =CHISQ.INV(0.95,3)" On this case we got that the critical value is:
[tex]\chi^2_{crit}=7.815[/tex]
Since our calculated value is higher than the cirtical value we have enough evidence to reject the null hypothesis at the significance level of 5%.
And we can also calculate the p value given by:
[tex]p_v = P(\chi^2_{3} >13.167)=0.0043[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(13.167,3,TRUE)"
Since the p value is lower than the significance level we reject the null hypothesis at 5% of significance.
A. We should reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
The reason why we select option C is because if we reject the null hypothesis of uniform distribution then we are rejecting the claim that the students are guessing.
If n is a positive integer, which of following statement is individually sufficient to prove whether 289 is a factor of n?a. The greatest common divisor of n and 344 is 86. b. Least common multiple of n and 272 is 4624. c. The least common multiple of n and 289 is 289n.
Answer:
The statement b) is individually sufficient to prove than 289 is a factor of n
Step-by-step explanation:
The least common multiple of n and 272 is the smallest number that is a multiple of n and a multiple of 272. Therefore:
272 x X = 4624 ⇒ X = 17 but 272 = 17 · 16 and 289 = 17 · 17
Therefore 17·17 must be a factor of n. That means 289 is a factor of n
Suppose that n(U) = 200, n(A) = 105, n(B) = 110, and n( A ∩ B ) = 30. Find n( A c ∪ B ). a) 80 b) 110 c) 125 d) 95 e) 75 f) None of the above.
Answer:
Option c - [tex]n(A^c\cup B)=125[/tex]
Step-by-step explanation:
Given : Suppose that n(U) = 200, n(A) = 105, n(B) = 110, and n(A∩B) = 30.
To find : The value of [tex]n(A^c\cup B)[/tex] ?
Solution :
n(U) = 200, n(A) = 105, n(B) = 110, and n(A∩B) = 30
We know that,
[tex]n(A^c)=n(U)-n(A)[/tex]
[tex]n(A^c)=200-105[/tex]
[tex]n(A^c)=95[/tex]
and [tex]n(A^c \cap B)=n(B)-n(A\cap B)[/tex]
[tex]n(A^c \cap B)=110-30[/tex]
[tex]n(A^c \cap B)=80[/tex]
Now, [tex]n(A^c\cup B)=n(A^c)+n(B)-n(A^c \cap B)[/tex]
[tex]n(A^c\cup B)=95+110-80[/tex]
[tex]n(A^c\cup B)=125[/tex]
Therefore, option c is correct.
The value of the union set given as n(A^c U B) is; C: 125
What is the union of the set?
We are given;
n(U) = 200, n(A) = 105, n(B) = 110, and n(A ∩ B) = 30.
In sets, we know that complement of set A is;
n(A^c) = n(U) - n(A)
Thus; n(A^c) = 200 - 105
n(A^c) = 95
Also, we know that;
n(A^c ∩ B) = n(B) - n(A ∩ B)
n(A^c ∩ B) = 110 - 30
n(A^c ∩ B) = 80
Thus;
n(A^c U B) = n(A^c) + n(B) - n(A^c ∩ B)
n(A^c U B) = 95 + 110 - 80
n(A^c U B) = 125
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Use the given data to find the best predicted value of the response variable. Ten pairs of data yield requals0.003 and the regression equation ModifyingAbove y with caret equals2plus3x. Also, y overbarequals5.0. What is the best predicted value of y for xequals2?
The best predicted value of 'y' when 'x' is 2, using the linear regression equation ŷ = 2 + 3x, is 8. However, the correlation coefficient of 0.003 indicates this prediction may not be accurate due to the weak linear relationship between the variables.
Explanation:The question is about predicting a value using a given linear regression equation. Given the regression equation ŷ = 2 + 3x, to predict 'y' when x = 2, we just replace 'x' with '2' in the regression equation. The equation becomes ŷ = 2 + 3*2 = 2 + 6 = 8. Therefore, the best predicted value of 'y' when 'x' is 2 is 8.
Note that the provided correlation coefficient (r) of 0.003 indicates a very weak linear relationship between the variables, hence this prediction might not be very reliable.
We use the regression line equation to make the prediction, this line of best fit has been calculated using the data provided. These predictions are most reliable when there is a strong correlation between the variables used.
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Consider a manufacturing process with a quality inspection station. In the past, 15% of parts are defective. As soon as one defective part is found, the process is stopped. If 8 parts have been inspected without finding a defective part, what is the probability that at least 11 total parts will be inspected before the process is stopped?
Answer:
0.614125
Step-by-step explanation:
Given that a manufacturing process with a quality inspection station has on an average 15% of parts are defective.
As soon as one defective part is found, the process is stopped.
We find that number of defectives would be binomial because each part randomly selected has a constant probability of 0.15 being defective
Probability that at least 11 total parts will be inspected before the process is stopped/8 parts have been inspected without finding a defective part
=[tex]P(x\geq 11)/P(x=8)\\[/tex]
= Probability of 9th, 10th, 11th should not be defective
= [tex](1-0.15)^3\\= 0.614125[/tex]
Great Lakes Health Care common stock offers an expected total return of 9.2 percent. The last annual dividend was $2.10 a share. Dividends increase at a constant 2.6 percent per year. What is the dividend yield?
A. 3.75 percent
B. 4.20 percent
C. 4.55 percent
D. 5.25 percent
E. 6.60 percent
Answer:
E. 6.60 percent
Step-by-step explanation:
We have been given that Great Lakes Health Care common stock offers an expected total return of 9.2 percent. The last annual dividend was $2.10 a share. Dividends increase at a constant 2.6 percent per year.
We will use total return formula to answer our given problem.
[tex]\text{Total return}=\text{Dividend yield}+\text{Growth rate}[/tex]
Upon substituting our given values in above formula, we will get:
[tex]9.2\%=\text{Dividend yield}+2.6\%[/tex]
[tex]\text{Dividend yield}=9.2\%-2.6\%[/tex]
[tex]\text{Dividend yield}=6.6\%[/tex]
Therefore, the dividend yield would be 6.60% and option E is the correct choice.
The price of a new computer is p dollars. The computer is on sale for 30% off. Which expression shows the savings that are being offered on the computer?
A. p - 0.3p B. 0.7 × p C. 0.3 × p D. p ÷ 30
Option C
Expression that shows the savings that are being offered on the computer is 0.3p
Solution:
Given that price of a new computer is p dollars
The computer is on sale for 30% offer
To find: Expression that shows the savings that are being offered on the computer
Computer is on sale for 30% offer which means 30 % offer on original price "p"
Original price = "p" dollars
offer price / saved price = 30 % of "p"
[tex]\text{ saved price } = 30 \% \times p\\\\\text{ saved price } = \frac{30}{100} \times p\\\\\text{ saved price } = 0.3p[/tex]
Thus the required expression is 0.3p
Thus option C is correct.
An advertisement for a word-processing class claims that students who complete the class will, on average, be able to type 60 words per minute (wpm) with a standard deviation of 6 wpm. At the end of the class, 49 students are tested and their mean score is 58.5 wpm.
A. Is this evidence that the true mean is different from 60 wpm? Give a complete answer, using a significance level of .05. (12 points)
B. Suppose the person conducting the research had believed, before collecting data, that the graduates weren't as good as claimed. Would this belief have changed the analysis you performed in part (A)? (12 points)
C. Construct a 99% confidence interval for the true mean of the population. What conclusion can you draw from this interval, and do you have evidence to reject the claim that the average graduate can type 60 wpm? (6 points)
Answer:
Reject at 5%, accept at 1% the null hypothesis
Step-by-step explanation:
Set up hypotheses as
[tex]H_0: \bar x = 60\\H_a: \bar x < 60[/tex]
(Left tailed test)
Population std dev = 6
Sample std error = [tex]\frac{6}{\sqrt{49} } \\=0.8555[/tex]
Mean difference = -1.5
Since sigma is known we can use Z test
Z = mean diff/std error = -1.7533
p value = 0.039
a) Since p value <0.05 we reject H0. There is evidence that the true mean is different from 60 wpm
b) Yes, because his sample would have been biased since he may want to prove his belief so slow or inefficient persons he would have selected in the sample.
c) For 99% confidence interval critical value = 2.58
Confidence interval for population mean = 58.5±2.58*std error
=(56.2928, 60.7072)
Since this contains 60, the hypothesized mean, we accept null hypothesis.
we do not have evidence to reject the claim that the average graduate can type 60 wpm at 1% level of significance.
A researcher developing scanners to search for hidden weapons at airports has concluded that a new scanner is significantly better than the current scanner. He made his decision based on a test using alpha equals 0.025 . Would he have made the same decision at alpha equals 0.10 question mark How about alpha equals 0.005 question mark Explain.
Step-by-step explanation:
Since the decision is made on the test based on the use of alpha equals 0.025, the p-value of the test would have been higher than the level of significance provided that is 0.025 since the test is not important.
p > 0.025
Now if we know that p > 0.025, this would not necessarily mean that p > 0.1 also, therefore we do not know with the given information that he would have made the same decision for 0.1 level of significance, ( we are not sure about his decision in that case ).
Now for the level of significance of 0.005, we would be sure that p > 0.005 as it is greater than 0.025, therefore the test is not significant at this level of significance as well. Therefore he would have made the same decision for 0.005 level of significance.
Suppose that ten bats were used in the experiment. For each trail, the zoo keeper pointed to one of two "feeders". Suppose that the bats went to the correct feeder (the one that the zoo keeper pointed at) 8 times. Find the 95% confidence interval for the population proportion of times that the bats would follow the point. (0.62, 1.0) (0.477, 0.951) (0.321, 0.831)
Answer: (0.477, 0.951)
Step-by-step explanation:
Given : Number of observations : n = 10
Number of successes : x = 8
Let p be the population proportion of times that the bats would follow the point.
Because the number of observation is not enough large , so we use plus four confidence interval for p.
Plus four estimate of p=[tex]\hat{p}=\dfrac{\text{No. of successes}+2}{\text{No. of observations}+4}[/tex]
[tex]\hat{p}=\dfrac{8+2}{10+4}=\dfrac{10}{14}\approx0.714[/tex]
We know that , the critical value for 95% confidence level : z* = 1.96 [By using z-table]
Now, the required confidence interval will be :
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}}[/tex] , where N= 14
[tex]0.714\pm (1.96)\sqrt{\dfrac{0.714(1-0.714)}{14}}[/tex]
[tex]0.714\pm (1.96)\sqrt{0.014586}[/tex]
[tex]0.714\pm (1.96)(0.120772513429)[/tex]
[tex]\approx0.714\pm0.237=(0.714-0.237,\ 0.714+0.237)[/tex]
[tex](0.477,\ 0.951)[/tex]
Hence, the 95% confidence interval for the population proportion of times that the bats would follow the point = (0.477, 0.951)
The 95% confidence interval for the proportion of the times that bats would follow the point is (0.552, 1.0). The result was adjusted because proportions cannot exceed 1.
Explanation:To calculate the 95% confidence interval for the population proportion, we follow these steps:
First, we calculate the sample proportion (p-hat) as the number of successes (bats going to the correct feeder) divided by the total number of observations. In this case, p-hat = 8 ÷ 10 = 0.8.Next, our goal is to construct the confidence interval using the formula p-hat ± Z * sqrt[p-hat(1 - p-hat) / n], where Z is the Z-value in the standard normal distribution corresponding to the desired confidence level (1.96 for 95% confidence level), n is the number of observations, and p-hat is the calculated sample proportion.Substituting all values into the formula, we get 0.8 ± 1.96 * sqrt[0.8(0.2) / 10] = 0.8 ± 1.96 * 0.126 = (0.552, 1.048).However, this interval contains value bigger than 1, which is not possible because proportion cannot exceed 1.Hence, we adjust our interval to (0.552, 1.0).Learn more about Confidence Interval here:https://brainly.com/question/34700241
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Tyler has a baseball bat that weighs 28 ounces. Find this weight in kilograms and in grams. (Note 1 kilogram=35 ounces)
Answer:0.8 kilograms
800 grams
Step-by-step explanation:
The weight of Tyler's baseball bat is 28 ounces. We would convert the weight in ounces to kilogram and grams.
Let x represent the number of kilograms that is equal to 28 ounces. Therefore
1 kilogram = 35 ounces
x kilogram = 28 ounces
Cross multiplying, it becomes
35 × x = 28 × 1
35x = 28
x = 28/35 = 0.8 kilograms
We would convert 0.8 kilograms to grams
Let y represent the number of grams that is equal to 0.8 kilograms. Therefore,
1000 grams = 1 kilogram
y grams = 0.8 kilograms
Cross multiplying,
y × 1 = 0.8 × 1000
y = 800 grams
Answer:
0.2
Step-by-step explanation:
A psychologist wants to see if a certain company has fair hiring practices in an industry where 60% of the workers are men and 40% are women. She finds that the company has 55 women and 52 men. Test to see if these numbers are different from the industry numbers, and if so, how are they different? Use alpha -.05 and four steps. A) what is the null hypothesis? B) what is the alternative hypothesis? C) what is the critical value of the test statistic? D) what is the value of the test statistic? E) Reject or accept the null? And why?
The hypothesis test examines if the company's hiring distribution differs from industry standards. The null hypothesis represents no difference, while the alternative suggests a discrepancy.
The critical value for the test statistic at a 0.05 significance level is ±1.96 for a two-tailed test, and we either reject or fail to reject the null based on the comparison of the calculated Chi-square statistic to the critical value.
To determine if there is a significant difference between the hiring practices of a certain company and the industry standard, we use a hypothesis test for proportions.
A. Null Hypothesis (H₀)
The null hypothesis H0: P_(men) = 0.60 and P_(women) = 0.40, where P represents the proportion of men and women in the company, respectively.
B. Alternative Hypothesis (Ha)
The alternative hypothesis Ha: P_(men) ≠ 0.60 and P_(women) ≠ 0.40.
C. Critical Value of Test Statistic
The critical value for a two-tailed test at alpha = 0.05 is z = ±1.96.
D. Value of the Test Statistic
To calculate the test statistic, we use the formula for a test of proportions:
Calculate the expected counts based on industry proportions: expected men = 107 * 0.60 = 64.2, expected women = 107 * 0.40 = 42.8.
Compute the Chi-square test statistic: Χ2 = ((52-64.2)2/64.2) + ((55-42.8)2/42.8).
The resulting Χ₂ statistic can then be compared against the critical Χ₂ value with 1 degree of freedom at alpha = 0.05, which is 3.841.
E. Reject or Accept the Null Hypothesis
If the calculated Χ₂ is greater than 3.841, we reject the null hypothesis; if not, we fail to reject the null hypothesis. Without the actual calculation of the Χ₂, we cannot definitively conclude the action on the null hypothesis in this context.
Assume that you have a sample of n 1 equals 8n1=8, with the sample mean Upper X overbar 1 equals 42X1=42, and a sample standard deviation of Upper S 1 equals 4S1=4, and you have an independent sample of n 2 equals 15n2=15 from another population with a sample mean of Upper X overbar 2 equals 34X2=34 and a sample standard deviation of Upper S 2 equals 5S2=5. What assumptions about the two populations are necessary in order to perform thepooled-variance t test for the hypothesis Upper H 0 : μ 1 equals μ 2H0: μ1=μ2 against the alternative Upper H 1 : μ 1 >μ 2H1: μ1>μ2 and make a statistical decision?
Answer:
Check the explanation below
Step-by-step explanation:
Hello!
To make a pooled variance t-test you have to make the following assumptions:
The study variables X₁ and X₂ must be independent.
Both variables should have a normal distribution, X₁~N(μ₁; σ₁²) and X₂~N(μ₂; σ₂²)
The population variances should be equal but unknown, σ₁² = σ₂² = ?.
You have the information of two samples:
Sample 1
n₁=8
sample mean X[bar]₁= 42
sample standard deviation S₁=4
Sample 2
n₂=15
sample mean X[bar]₂= 34
sample standard deviation S₂= 5
For the hypothesis:
H₀: μ₁ = μ₂
H₁: μ₁ > μ₂
The statistic is:
t= (X[bar]₁ - X[bar]₂) - (μ₁ - μ₂) ~[tex]t_{n_1 + n_2 - 2}[/tex]
Sa[tex]\sqrt{\frac{1}{n_1} + \frac{1}{n_2} }[/tex]
Sa²= [tex]\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}[/tex]
Sa²= 22
Sa= 4.69
[tex]t_{H0}[/tex]= 3.8962 ≅ 3.9
The critical region is one-tailed, for example for α: 0.05
[tex]t_{n_1 + n_2 - 2; 1 - \alpha } = t_{21; 0.95} = 1.721[/tex]
Since [tex]t_{H0}[/tex] > 1.721, then the decision is to reject the null hypothesis.
I hope it helps!
A leprechaun places a magic penny under a girl's pillow. The next night there are 2 magic pennies under her pillow. The following morning she finds four pennies. Apparently, while she sleeps each penny turns into two magic pennies. The total number of pennies seen under the pillow each day is the grand total; that is, the pennies from each of the previous days are not being stored away until more pennies magically appear. How many days would elapse before she has a total of more than $2 billion ? (Proceed by trial and error.)
nothing days
(Type a whole number.) PLEASE HURRY
Answer: 31
Step-by-step explanation:
2^x=2 000 000 000
log2^x=log2 000 000 000
xlog2 = log 2 000 000 000
x= log (2000 000 000)/log 2
x= 30.897352854
round to 31
gotchu bro
Which of the following is used to determine the significance of predictions made by a best fitting linear equation?A. correlational analysisB. analysis of varianceC. analysis of regressionD. method of least squares
Answer:
D. method of least squares
Step-by-step explanation:
The Least Squares Method (LSM) is a mathematical method used to solve various problems, based on minimizing the sum of the squared deviations of some functions from the desired variables. It can be used to “solve” over-determined systems of equations (when the number of equations exceeds the number of unknowns), to find a solution in the case of ordinary (not redefined) linear or nonlinear systems of equations, to approximate the point values of a function. OLS is one of the basic regression analysis methods for estimating the unknown parameters of regression models from sample data.
Correlation analysis is a statistical method used to assess the strength of the relationship between two quantitative variables. A high correlation means that two or more variables have a strong relationship with each other, while a weak correlation means that the variables are hardly related. In other words, it is a process of studying the strength of this relationship with available statistics.
Analysis of Variance (or ANOVA) is a collection of statistical models used to analyze group averages and related processes (such as intra- and inter-group variation) in statistical science. When using Variance Analysis, the observed variance of a specified variable is divided into the variance component that can be based on different sources of change. In its simplest form, "Analysis of Variance" is a inferential statistical test to test whether the averages of several groups are equal or not, and this test generalizes the t-test test for two-groups to multiple-groups. If multiple two-sample-t-tests are desired for multivariate analysis, it is clear that this results in increased probability of type I error. Therefore, the variance analysis would be more useful to compare the statistical significance of three or more means (for groups or for variables) with the test.
Regression analysis is an analysis method used to measure the relationship between two or more variables. If analysis is performed using a single variable, it is called univariate regression, and if more than one variable is used, it is called multivariate regression analysis. With the regression analysis, the existence of the relationship between the variables, if there is a relationship between the strength of the information can be obtained. The logic here is that the variable to the left of the equation is affected by the variables to the right. The variables on the right are not affected by other variables. Not being influenced here means that when we put these variables into a linear equation in mathematical sense, it has an effect. Multiple linearity, sequential dependency problems are not meant.