A certain transportation system of buses and commuter trains is heavily utilized so that it is not practical to check every traveler's ticket. Rather, only a small, randomly selected group of travelers on any given trip will be asked to show their tickets. Suppose that in a random sample of 621 train travelers is selected and 69 of them admitted they did not buy a ticket. Find the upper bound of a 95% confidence interval for the true proportion of all train travelers who do not buy tickets

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

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

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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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

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.

Answer:

The expected number of vehicles owned to two decimal places is: 1.85.

Step-by-step explanation:

The table to the question is attached.

[tex]E(X) =[/tex]∑[tex]xp(x)[/tex]

Where:

E(X) = expected number of vehicles owned

∑ = Summation

x = number of vehicle owned

p(x) = probability of the vehicle owned

[tex]E(X) = (0 * 0.1) + (1 * 0.35) + (2 * 0.25) + (3 * 0.2) + (4 * 0.1)\\E(X) = 0 + 0.35 + 0.50 + 0.60 + 0.4\\E(X) = 1.85[/tex]

The expected number of vehicles owned is 1.85.

The expected number of vehicles owned, based on probability of ownership of 0 to 3 vehicles, is calculated by multiplying each possible number of vehicles by their corresponding probabilities and then summing up all the products. The calculated expected number is approximately 1.7 vehicles.

To find the expected number of vehicles owned, we first need to multiply each possible number of vehicles someone could own by the probability of them owning that many vehicles. Then, sum up all of these products.

For instance, if they could own up to 3 cars and the probability for owning 0, 1, 2, or 3 cars is 0.1, 0.3, 0.4, and 0.2 respectively:

For 0 cars: 0 * 0.1 = 0

For 1 car: 1 * 0.3 = 0.3

For 2 cars: 2 * 0.4 = 0.8

For 3 cars: 3 * 0.2 = 0.6    

Adding these together gives the expected number of cars:
0 + 0.3 + 0.8 + 0.6 = 1.7 (rounded to two decimal places).

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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:

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.

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.

To calculate the 95% confidence interval for the population proportion, we follow these steps:

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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.

Answer:

The procedure emphasizes the idea of the summation of one physical quantity. In this case, X.

Step-by-step explanation:

1. When we add fractions like these we do it simply by rewriting a new one, the summation of the numerators over the same denominator:

[tex]\frac{2}{3}X+\frac{4}{3})X=\frac{6}{3}X= 2X[/tex]

The procedure emphasizes the idea of the summation of one physical quantity, in this case, X.

2) This physical quantity x could be miles, oranges, gallons, etc.

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.

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.

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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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.

$ 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

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.

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!

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

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

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.

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]

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.

Answer:

The point estimate for population mean is 8.129 Mpa.

Step-by-step explanation:

We are given the following in the question:

Data on flexural strength(MPa) for concrete beams of a certain type:

11.8, 7.7, 6.5, 6.8, 9.7, 6.8, 7.3, 7.9, 9.7, 8.7, 8.1, 8.5, 6.3, 7.0, 7.3, 7.4, 5.3, 9.0, 8.1, 11.3, 6.3, 7.2, 7.7, 7.8, 11.6, 10.7, 7.0

a) Point estimate of the mean value of strength for the conceptual population of all beams manufactured

We use the sample mean, [tex]\bar{x}[/tex] as the point estimate for population mean.

Formula:

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]\bar{x} = \dfrac{\sum x_i}{n} = \dfrac{219.5}{27} = 8.129[/tex]

Thus, the point estimate for population mean is 8.129 Mpa.

To estimate the mean flexural strength, the sum of strengths (219.5 MPa) is divided by the total number of beams measured (26), which yields a mean value of 8.442 MPa when rounded to three decimal places. The estimator used is the sample mean.

To calculate a point estimate of the mean value for flexural strength (MPa) for a conceptual population of concrete beams, we use the sum of all measured strengths and divide by the number of measurements. The sum of the flexural strengths is provided as Σxi = 219.5 MPa.

Given the dataset:

The number of measurements is the number of data points, which is 26. To find the mean:

mean = Sum of strengths / Number of measurements

mean = 219.5 MPa / 26

mean = 8.442 MPa (rounded to three decimal places)

The estimator used here is the sample mean (×).

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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.

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]

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.

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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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)"

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

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]

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.