I really need help with this!

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:

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:

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

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:

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: 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

$ 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:

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:

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."

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:

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

Read more about Union of Sets at; https://brainly.com/question/1563195

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

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

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:

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.

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]

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.

Answer:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

[tex]t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416[/tex]

Step-by-step explanation:

Data given

American: 21,17,17,20,25,16,20,16 (Sample 1)

French: 24,18,20,28,18,29,17 (Sample 2)

When we have two independent samples from two normal distributions with equal variances we are assuming that  

[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]

And the statistic is given by this formula:

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]

Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:

[tex]S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]

This last one is an unbiased estimator of the common variance [tex]\sigma^2[/tex]

The system of hypothesis on this case are:

Null hypothesis: [tex]\mu_1 = \mu_2[/tex]

Alternative hypothesis: [tex]\mu_1 \neq \mu_2[/tex]

Or equivalently:

Null hypothesis: [tex]\mu_1 - \mu_2 = 0[/tex]

Alternative hypothesis: [tex]\mu_1 -\mu_2 \neq 0[/tex]

Our notation on this case :

[tex]n_1 =8[/tex] represent the sample size for group 1

[tex]n_2 =7[/tex] represent the sample size for group 2

[tex]\bar X_1 =19[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =22[/tex] represent the sample mean for the group 2

[tex]s_1=3.117[/tex] represent the sample standard deviation for group 1

[tex]s_2=5.0[/tex] represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

[tex]S^2_p =\frac{(8-1)(3.117)^2 +(7 -1)(5.0)^2}{8 +7 -2}=16.770[/tex]

And the deviation would be just the square root of the variance:

[tex]S_p=4.095[/tex]

And now we can calculate the statistic:

[tex]t=\frac{19 -22)-(0)}{4.095\sqrt{\frac{1}{8}+\frac{1}{7}}}=-1.416[/tex]

Now we can calculate the degrees of freedom given by:

[tex]df=8+7-2=13[/tex]

And now we can calculate the p value using the altenative hypothesis:

[tex]p_v =2*P(t_{13}<-1.416) =0.1803[/tex]

So with the p value obtained and using the significance level assumed [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance we don't have significant differences between the two means.  

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]

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.

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

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:

Answer: 0.6022

Consider the following calculation

Step-by-step explanation:

Let F shows the event that guardian father is biological father. So

P(F) = 0.75

By the complement rule,

P(F') = 1 - P(F) =1 - 0.75 = 0.25

Let B shows the event that child has blood type B. So we have

P(B|F) = 0.50, P(B' |F') = 0.0091

By the complement rule we have

P(B|F') = 1 - P(B' |F') = 0.9909

The probability that the guarding father is the child’s biological father given that child have blood type B is

P(BFPF) P(F|B) = PRI P(BF)P(F) + P(BF)P(F) 0.50 -0.75 0.50 -0.75 +0.9909 - 0.25

0.375 0.622725 = 0.6022

Answer: 0.6022

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:

70 is answer

Step-by-step explanation:

Given that a function in x is

[tex]f(x) = 5x^2[/tex]

we have to find f'(7)

we know by derivative rule derivative of a function is

[tex]f'(x) = lim_({h-->0}) \frac{f(x+h)-f(x)}{h}[/tex]

For finding out at 7 we replace x by 7

[tex]f'(7) = lim_({h-->0}) \frac{f(7+h)-f(7)}{h}[/tex]

=[tex]lim\frac{5(7+h)^2-5*7^2}{h} \\= lim \frac{10h*7+h^2}{h} \\= 70+h = 70[/tex]

So f'(7) = 70

answer is 70

Answer:

f'(7)=70

Step-by-step explanation:

We have the definition of the derivative as:

[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}[/tex]

Now we have a function [tex]f(x)=5x^2[/tex] and we want to approximate the first derivative around x=7, that is [tex]f'(7)[/tex].

We can replace this in the first formula as:

[tex]f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}= \lim_{h \to 0} \dfrac{5(x+h)^2-5x^2}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{5(x^2+2xh+h^2-x^2)}{h}\\\\f'(x)=\lim_{h \to 0}\dfrac{5(2xh+h^2)}{h}\\\\f'(x)=\lim_{h \to 0}5(2x+h)\\\\f'(x)=10x+lim_{h \to 0}h=10x+0=10x[/tex]

Then, the value for f'(7) is:

[tex]f'(7)=10\cdot 7=70[/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:

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!