For a school field trip the students had two options for lunch, a turkey or egg salad sandwich, so it is impossible for a student have both lunches. If the probability that a student chooses a turkey sandwich is 0.10, and the probability that a student chooses an egg salad sandwich is 0.67, what is the probability that a student chooses a turkey or egg salad sanwich?

Answer:

The correct option is C) 0.3

Step-by-step explanation:

Consider the provided information.

A small entrepreneurial can develop two product, product A and product B.

If they develop Product A:

There are 50% chance of selling it to the large company with annual purchases of about 20,000 units.

If the large company won't purchase it, then they think they have an 80% chance of placing it with a smaller company, with sales of 15,000 units.

If they develop Product B.

They have a 40% chance of selling it to the large company, resulting in annual sales of about 17,000 units. If the large company doesn't buy it, they have a 50% chance of selling it to the small company with sales of 20,000 units.

If they develop product B than the probability of purchasing by small company is 60% or 0.6

If large company doesn't buy it, then they have 50%or 0.5 chance of selling it to small company.

Hence, the total probability is: [tex]0.6\times0.5=0.3[/tex]

Therefore, the correct option is C) 0.3

Final answer:

The probability that Product B will be purchased by the smaller company is 30%, which corresponds to answer choice C.

Explanation:

The probability that Product B will be purchased by the smaller company is calculated by considering that Product B will only be offered to the smaller company if the large company does not purchase it. So, first, we must calculate the probability that the large company does not purchase Product B, which is 1 - 0.4 (since there is a 40% chance that the large company will purchase it). This results in a 60% chance that the large company will not purchase Product B. The probability that Product B will then be purchased by the smaller company is 50% of that 60%, which is calculated as 0.6 × 0.5 = 0.3 or 30%.

Answer:

A sample size of at least 328 students is required.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]

Now, find the width M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

[tex]M = 10, \sigma = 110[/tex]

So:

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]10 = 1.645*\frac{110}{\sqrt{n}}[/tex]

[tex]10\sqrt{n} = 180.95[/tex]

[tex]\sqrt{n} = 18.095[/tex]

[tex]n = 327.43[/tex]

A sample size of at least 328 students is required.

Answer:

A sample size of at least 328 students is required.

Step-by-step explanation:

We have that to find our  level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the width M as such

In which  is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

So:

A sample size of at least 328 students is required.

Answer:

The probability that a student earns a grade of A is 1/7.

Step-by-step explanation:

Let E be an event and S be the sample space. The probability of E, denoted by P(E) could be computed as:

P(E) = n(E) / n(S)

As the total number of students = n(S) = 35

Students getting the grade A = n(E) = 5

So, the probability that a student earns a grade of A:

                      P(E) = n(E) / n(S)

                              = 5/35

                              = 1/7

Hence, the probability that a student earns a grade of A is 1/7.

Keywords: probability, sample space, event

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Answer:

.14

Step-by-step explanation:

Answer: [tex](98.04,\ 98.36)[/tex]

Step-by-step explanation:

Given : Sample size of healthy adults: n= 106

Degree of freedom = df =n-1 = 105

Sample mean body temperature  : [tex]\overline{x}=98.2[/tex]

Sample standard deviation : [tex]s= 0.62[/tex]

Significance level ; [tex]\alpha= 1-0.99=0.01[/tex]

∵ population standard deviation is unknown , so we use t- critical value.

Confidence interval for the population mean :

[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]

Using t-distribution table , we have

Critical value for df = 105 and [tex]\alpha=0.01[/tex]= [tex]t_{\alpha/2, df}=t_{0.005 , 105}=2.623[/tex]

A 99% confidence interval estimate of the mean body temperature of all healthy humans will be :

[tex]98.2\pm (2.623)\dfrac{0.62}{\sqrt{106}}[/tex]

[tex]98.2\pm (2.623)(0.0602197234662)[/tex]

[tex]98.2\pm (0.157956334652)[/tex]

[tex]\approx98.2\pm 0.16[/tex]

[tex](98.2-0.16,\ 98.2+0.16)[/tex]

[tex](98.04,\ 98.36)[/tex]

Hence, a 99% confidence interval estimate of the mean body temperature of all healthy humans = [tex](98.04,\ 98.36)[/tex]

Answer: [tex]\dfrac{7}{8}[/tex]

Step-by-step explanation:

Given : When a coin is tossed twice , the total outcomes will become

HHH , HHT , HTH , THH , TTH , THT , HTT ,TTT

Total outcomes = 8

Favorable outcomes to obtain 1 and 3 inclusive = HHT , HTH , THH , TTH , THT , HTT ,TTT

Number of favorable outcomes = 7

Probability that the number of tails obtained will be between 1 and 3 inclusive = [tex]\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]

[tex]=\dfrac{7}{8}[/tex]

Hence, the probability that the number of tails obtained will be between 1 and 3 inclusive [tex]=\dfrac{7}{8}[/tex]

Answer:

(B) 0.2843

Step-by-step explanation:

Let d be the difference in proportions from Country X and Country Y who were first married between the ages of 18 and 32.

Then hypotheses are

[tex]H_{0}[/tex]: d=0.15

[tex]H_{a}[/tex]: d<0.15

Test statistic can be found using the equation

[tex]z=\frac{p1-p2-0.15}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}[/tex] where

Then [tex]z=\frac{0.36-0.26-0.15}{\sqrt{{0.31*0.69*(\frac{1}{60} +\frac{1}{50}) }}}[/tex] ≈ 0.5646

p-value of test statistic is ≈ 0.2843

p-value states the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is at least 0.15

To find the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is greater than 0.15, we need to calculate the sampling distribution of the difference in proportions. The answer is (E) 0.7157.

To find the probability that the difference in proportions between a random sample of 60 adults from Country X and a random sample of 50 adults from Country Y who were first married between the ages of 18 and 32 is greater than 0.15, we need to calculate the sampling distribution of the difference in proportions.

The standard error of the difference in proportions can be calculated as:

SE = sqrt((p1 * (1 - p1)) / n1 + (p2 * (1 - p2)) / n2)

where p1 and p2 are the proportions of first married adults from Country X and Country Y respectively, and n1 and n2 are the sample sizes for Country X and Country Y.

Let's plug in the values:

SE = sqrt((0.36 * (1 - 0.36)) / 60 + (0.26 * (1 - 0.26)) / 50)

SE = sqrt(0.0024 / 60 + 0.0016 / 50)

SE = sqrt(0.00004 + 0.00003)

SE = sqrt(0.00007)

SE ≈ 0.00836660027

Now, we can use the standard normal distribution to find the probability that the difference in proportions is greater than 0.15 by calculating the z-score and looking it up in the z-table or using a calculator.

Z = (0.15 - 0) / 0.00836660027

Z ≈ 17.914253

The probability that the difference in proportions is greater than 0.15 is very close to 1, which means it is highly likely that the difference in proportions will be greater than 0.15 in a random sample. Therefore, the answer is (E) 0.7157.

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Answer:

There is enough evidence to make the conclusion that the population mean amount of time to assemble the Meat Man barbecue is not equal to 10 minutes (P-value=0.009).

Step-by-step explanation:

We have to perform an hypothesis test on the mean.

The null and alternative hypothesis are:

[tex]H_0: \mu=10\\\\H_1: \mu \neq 10[/tex]

The significance level is [tex]\alpha=0.05[/tex].

The test statistic t can be calculated as:

[tex]t=\frac{M-\mu}{s/\sqrt{N} } =\frac{11.2-10}{3.1/\sqrt{50} }=2.737[/tex]

The degrees of freedom are:

[tex]df=N-1=50-1=49[/tex]

The P-value (two-tailed test) for t=2.737 and df=49 is P=0.00862.

This P-value (0.009) is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough evidence to make the conclusion that the population mean amount of time to assemble the Meat Man barbecue is not equal to 10 minutes.

Final answer:

A hypothesis test can be conducted to test the claim made in the commercial for the new Meat Man Barbecue. The test results indicate that there is evidence to support that the population mean assembly time is not equal to 10 minutes.

Explanation:

To test the claim made in the commercial for the new Meat Man Barbecue, a hypothesis test can be conducted. The null hypothesis, H0, states that the population mean assembly time is 10 minutes, while the alternative hypothesis, Ha, states that it is not equal to 10 minutes. The test statistic to use in this case is the t-test, as we are comparing the sample mean to a known value. The p-value for this test is 0.009, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis and conclude that there is evidence to support that the population mean assembly time is not equal to 10 minutes.

Answer:

P-value = 0.0261

We conclude that the machine is under-filling the bags.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 433 gram

Sample mean, [tex]\bar{x}[/tex] = 427 grams

Sample size, n = 26

Alpha, α = 0.05

Sample standard deviation, σ = 15 grams

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 433\text{ grams}\\H_A: \mu < 433\text{ grams}[/tex]

We use one-tailed(left) t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{427 - 433}{\frac{15}{\sqrt{26}} } = -2.0396[/tex]

Now, we calculate the p-value using the standard table.

P-value = 0.0261

Since the p-value is lower than the significance level, we fail to accept the null hypothesis and reject it.We accept the alternate hypothesis.

We conclude that the machine is under-filling the bags.

Answer:

a) [tex]t=\frac{524.6-550}{\frac{27.682}{\sqrt{5}}}=-2.052[/tex]    

[tex]p_v =P(t_{(4)}<-2.052)=0.055[/tex]    

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly different from 550.

b) The 95% confidence interval would be given (490.184;559.016).  

We are confident at 95% that the true mean is between (490.184;559.016).

Step-by-step explanation:

Part a

Data given and notation  

560, 525, 496, 543, 499

The mean and sample deviation can be calculated from the following formulas:

[tex]\bar X =\frac{\sum_{i=1}^n x_i}{n}[/tex]

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}[/tex]

[tex]\bar X=524.6[/tex] represent the sample mean  

[tex]s=27.682[/tex] represent the sample standard deviation  

[tex]n=5[/tex] sample size  

[tex]\mu_o =550[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/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 population mean is equal to 550, the system of hypothesis are:

Null hypothesis:[tex]\mu = 550[/tex]  

Alternative hypothesis:[tex]\mu \neq 550[/tex]  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)  

t test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]t=\frac{524.6-550}{\frac{27.682}{\sqrt{5}}}=-2.052[/tex]  

P-value  

We need to calculate the degrees of freedom first given by:  

[tex]df=n-1=5-1=4[/tex]  

Since is a one-side left tailed test the p value would given by:  

[tex]p_v =P(t_{(4)}<-2.052)=0.055[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.  

We can say that at 5% of significance the true mean is not significantly different from 550.

Part b

The confidence interval for a proportion is given by this formula  

[tex]\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]  

The degrees of freedom are 4 from part a

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]t_{\alpha/2}=2.78[/tex]  

And replacing into the confidence interval formula we got:  

[tex]524.6 - 2.78 \frac{27.682}{\sqrt{5}}=490.184[/tex]  

[tex]524.6 + 2.78 \frac{27.682}{\sqrt{5}}=559.016[/tex]  

And the 95% confidence interval would be given (490.184;559.016).  

We are confident at 95% that the true mean is between (490.184;559.016).

Answer:

Yes, one of the properties of determinants is that they are real numbers (including zero) not matrix. This if the entries of the matrix are real. Determinants can be both positive or negative numbers.

Step-by-step explanation:

The determinant of a matrix is a number

What is a Matrix?

In mathematics, a matrix is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns, which is used to represent a mathematical object or a property of such an object

The determinant of inverse matrix can never be zero.

The determinant of an identity matrix is always 1.

The determinant of a zero matrix is always 0.

Given data ,

Let the matrix be represented as

[tex]A = \left[\begin{array}{ccc}a_{11} &a_{12} \\ a_{21} &a_{22} \\\end{array}\right][/tex]

Let the determinant of the matrix be d

Now , the determinant of the matrix A is calculated as

d = ( a₁₁ ) x ( a₂₂ ) - ( a₁₂ ) x ( a₂₁ )

So , the determinant will be a single number and will not be another matrix

The value of d will be scaling factor for the transformation of a matrix

Hence , the determinant of a matrix is a number

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Answer:

See proof below.

Step-by-step explanation:

Remember that a collection of subsets B of a set X is a base for a topology of X if the following conditions hold

In this case, X=R and B={[a,b]: a is rational and b is irrational}. Let's prove the statements above

Answer:

[tex]P(\bar X <m)=0.844[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the mass of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]  

And the z score is defined as:

[tex]z=\frac{X-\mu}{\sigma}[/tex]

We know that for some amount  m we have this:

[tex]P(X>m)=0.4[/tex]   (a)

[tex]P(X<m)=0.6[/tex]   (b)

We can use the z table or excel in order to find a quantile that satisfy the two conditions. The excel code would be:

"=NORM.INV(0.6,0,1)" and using condition (b) we have that Z=0.253

So we have this:

[tex]0.253=\frac{m-\mu}{\sigma}[/tex]

If we solve for m from the last equation we got:

[tex]m=0.253\sigma +\mu [/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

And we want this probability:

[tex]P(\bar X<m)[/tex]

We can apply the z score formula for this case given by:

[tex]z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{m-\mu}{\frac{\sigma}{\sqrt{16}}})[/tex]

[tex]P(Z<\frac{4(m-\mu)}{\sigma}[/tex]

If we use the expression obtained for m we got:

[tex]P(Z<\frac{4(0.253\sigma +\mu-\mu)}{\sigma}[/tex]

[tex]P(Z<\frac{1.012 \sigma}{\sigma})=P(Z<1.012)=0.844[/tex]

Answer:

8/27 ≈ 29.6%

Step-by-step explanation:

Two of the six faces are B, which means four of the six faces are not B.

The probability of rolling not B three times is:

P = (4/6)^3

P = (2/3)^3

P = 8/27

P ≈ 29.6%

Answer:

8/27 = 0.296 = 29.6%

Step-by-step explanation:

Given that the die faces are as follows:

A A A   B B    C

i.e :

P( rolls A) = 3/6

P (rolls B) = 2/6

P (rolls C) = 1/6

for any single roll,

P (rolls not B) = P(rolls A) + P(rolls C)

P (rolls not B) = 3/6 + 1/6 = 4/6 = 2/3

for 3 consecutive rolls

P ( B does not appear) = P(rolls not B) * P(rolls not B) * P(rolls not B)

= P(rolls not B) ³

= (2/3)³

= 8/27 = 0.296 = 29.6%

Answer:

B. Goodness-of-Fit Test

Step-by-step explanation:

Given that a professor expects that the grades on a recent exam are normally distributed with a mean of 80 and a standard deviation of 10.

She records the actual distribution of grades and wants to compare it to the normal distribution

For this expected values are to be got assuming normal and the observed and expected are used to calculate chi square.

Depending on the chi square statistic conclusion is made

Here the test to be done is

B) Goodness of fit test.

A is wrong because test of independence is done when there are more than one categorical variable in the rows.

C is wrong because here homogeneity is not tested

Only B is right.

Answer:

C.ModifyingAbove p with caret p=0.224. The symbol ModifyingAbove p with caret p is used to represent a sample proportion.

Step-by-step explanation:

Data given and notation

X= 459 represent the number of people that answer yes

1241 people answer no and 352 had no opinion

n= 459+1241+352=2052 represent the sampplpe size selected

Solution to the problem

What is the sample proportion of yes​ responses, and what notation is used to represent​ it?

If we want to find the sample proportion of yes we need to apply the following formula:

[tex]\hat p =\frac{X}{n}[/tex]

Where X represent the number of people with the characteristic desired in the sample and n the sample size selected.

For this case we have:

[tex]\hat p =\frac{X}{n}=\frac{459}{2052}=0.224[/tex]

Let's analyze the possible options given:

A.ModifyingAbove p with caret p=0.270. The symbol ModifyingAbove p with caret p is used to represent a sample proportion.

False the sample proportion is 0.224 not 0.270

B.p=0.224. The symbol p is used to represent a sample proportion.

The sample proportion is 0.224 not the population proportion for this case this statement is False.

C.ModifyingAbove p with caret p=0.224. The symbol ModifyingAbove p with caret p is used to represent a sample proportion.

Correct the [tex]\hat p= 0.224[/tex] represent th sample proportion and it's the correct way to express it.

D.p=0.270. The symbol p is used to represent a sample proportion.

False the sample proportion is 0.224 not 0.270

To find the sample proportion of yes responses, divide the number of yes responses by the total responses. The sample proportion is 0.224, represented as p'. The correct answer is option C.

To determine the sample proportion of yes responses in the Harris poll, we need to use the formula for sample proportion:

p' = x/n, where x represents the number of successes (yes responses) and n represents the sample size (total responses).

Step-by-Step:

The sample proportion of yes responses is 0.224. In statistical notation, this is represented as p'.

The correct answer is: C. p' = 0.224. The symbol p' is used to represent a sample proportion.

Answer:

P-value =  0.4846

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4.73

Sample mean, [tex]\bar{x}[/tex] = 4.35

Sample size, n = 51

Alpha, α = 0.05

Sample standard deviation, s = 3.88

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 4.73\\H_A: \mu \neq 4.73[/tex]

We use Two-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{4.35 - 4.73}{\frac{3.88}{\sqrt{51}} } = -0.699[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]

We can calculate the p-value with the help of standard normal table.

P-value =  0.4846

Since the p-value is higher than the significance level, we fail to reject the null hypothesis and accept it.

We conclude that this college has same drinking habit as the college students in general.

Answer:

The obtained value of the appropriate statistic is ____.

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06[/tex]  

d. 3.06

[tex]p_v =2*P(t_{(4)}>3.06) =0.0376[/tex]  

d. reject H0; the class appeared to improve study habits

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations we can use it.  

Let put some notation  

x=before method , y = after method  

x: 10 12 14 16 12

y: 15 14 17 17 20

The system of hypothesis for this case are:  

Null hypothesis: [tex]\mu_y- \mu_x = 0[/tex]  

Alternative hypothesis: [tex]\mu_y -\mu_x \neq 0[/tex]  

The first step is define the difference [tex]d_i=y_i-x_i[/tex], that is given so we have:

d: 5,2,3,1,8

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=3.8[/tex]  

The third step would be calculate the standard deviation for the differences, and we got:  

[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =2.775[/tex]  

The fourth step is calculate the statistic given by :  

[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06[/tex]  

The next step is calculate the degrees of freedom given by:  

[tex]df=n-1=5-1=4[/tex]  

Now we can calculate the p value, since we have a two tailed test the p value is given by:  

[tex]p_v =2*P(t_{(4)}>3.06) =0.0376[/tex]  

The p value is less than the significance level given [tex]\alpha=0.05[/tex], so then we can conclude that we reject the null hypothesis.

d. reject H0; the class appeared to improve study habits

Answer:

Step-by-step explanation:

At the significance level given, the conclusion is B. Reject the null hypothesis.

The first thing to do is to calculate the test statistic. This will be 1.98 from the z table.

In this case, the rejection region will be when the number is more than 1.28. Here, 1.98 is more than 1.28.

Therefore, the best thing to do is to simply reject the null hypothesis.

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Answer: (0.881, 0.919)

Step-by-step explanation:

Let p be the population proportion of respondents who say the Internet has been a good thing for them personally.

Confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex] , where z* = Critical value ,

[tex]\hat{p}[/tex] = Sample proportion

n = sample size.

As per given , we have

n= 957

[tex]\hat{p}=0.90[/tex]

By z-table , Critical value for 95% confidence interval : z*= 1.96

Then, the 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally. will be :

[tex]0.90\pm 1.96\sqrt{\dfrac{0.90(1-0.90)}{957}}[/tex]

[tex]0.90\pm 1.96\sqrt{0.0000940438871473}[/tex]

[tex]0.90\pm 1.96(0.00969762275753)[/tex]

[tex]\approx0.90\pm 0.019\\\\=(0.90-0.019,\ 0.90+0.019)[/tex]

[tex](0.881,\ 0.919)[/tex]

Hence, the required 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them personally = (0.881, 0.919)

To develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them, we can use the formula: CI = p ± Z * sqrt((p * (1 - p)) / n). Plugging in the values, we can calculate a 95% confidence interval of [0.8808, 0.9192].

To develop a 95% confidence interval for the proportion of respondents who say the Internet has been a good thing for them, we can use the formula:

CI = p ± Z * sqrt((p * (1 - p)) / n)

Where:

Plugging in the values, we can calculate:

CI = 0.90 ± 1.96 * sqrt((0.90 * (1 - 0.90)) / 957)

This gives us a 95% confidence interval for the proportion, which is [0.8808, 0.9192]. So we can be 95% confident that the true proportion of respondents who say the Internet has been a good thing for them is between 0.8808 and 0.9192.

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Answer:

The initial population at the beginning of the 10 years was 129.

Step-by-step explanation:

The population of the village may be modeled by the following function.

[tex]P(t) = P_{0}e^{rt}[/tex]

In which P is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the growth rate, in decimal.

In this problem, we have that:

[tex]P(10) = 158, r = 0.02[/tex].

So

[tex]158 = P_{0}e^{0.02*10}[/tex]

[tex]P_{0} = 158*e^{-0.2}[/tex]

[tex]P_{0} = 129[/tex]

The initial population at the beginning of the 10 years was 129.

The population of the village at the beginning of the 10 years was approximately 130. This is determined by using the formula for exponential growth that is continuously compounded and solving for the initial population.

We can use the formula for exponential growth that is continuously compounded, which is P=Pe^rt, where P is the final population, P is the initial population, r is the growth rate, and t is the time.

In this case, we know that P=158, r=0.02 (2% expressed as a decimal), and t=10. We need to find P.

Substituting the given values into the formula, we get 158=Pe^(0.02*10).

We can calculate e^(0.02*10) using a calculator to get approximately 1.22. Therefore, the equation becomes 158=P*1.22.

To find P, we can divide both sides of the equation by 1.22 resulting in P ≈ 158/1.22 ≈ 129.5. Since we are asked to round up to the next whole number, the original population was approximately 130.

Answer:

So the value of score that separates the bottom 99% of data from the top 1% is 33.316.  

And rounded to 1 decimal place would be 33.3

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the scores for the ACT of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(21.5,5.2)[/tex]  

Where [tex]\mu=21.5[/tex] and [tex]\sigma=5.2[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.01[/tex]   (a)

[tex]P(X<a)=0.99[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.99 of the area on the left and 0.01 of the area on the right it's z=2.33. On this case P(Z<2.33)=0.99 and P(Z>2.33)=0.01

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.99[/tex]  

[tex]P(Z<\frac{a-\mu}{\sigma})=0.99[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]Z=2.33<\frac{a-21.2}{5.2}[/tex]

And if we solve for a we got

[tex]a=21.2 +2.33*5.2=33.316[/tex]

So the value of score that separates the bottom 99% of data from the top 1% is 33.316.  

And rounded to 1 decimal place would be 33.3

Answer:

[tex]p_v =2*P(Z>2.4)=0.016[/tex]  

And we can use the following excel code:

"=2*(1-NORM.DIST(2.4;0;1;TRUE))"

Step-by-step explanation:

1) Data given and notation

n represent the random sample taken

X represent the business students who have personal computers (PC's) at home

[tex]\hat p[/tex] estimated proportion of business students who have personal computers (PC's) at home

[tex]p_o=0.25[/tex] is the value that we want to test

[tex]\alpha[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.25:  

Null hypothesis:[tex]p=0.25[/tex]  

Alternative hypothesis:[tex]p \neq 0.25[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

For this case the value of the statistic is given by z=2.4 and that's given.

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(Z>2.4)=0.016[/tex]  

And we can use the following excel code:

"=2*(1-NORM.DIST(2.4;0;1;TRUE))"

Answer:

The average cost function is [tex]\bar C(x)=-\frac{1200}{x}+34[/tex].

The cost function is [tex]C(x)=-1200+34x[/tex].

The fixed costs are -$1200.

Step-by-step explanation:

If [tex]y = C(x)[/tex] is the total cost of producing x items, then the average cost, or cost per unit, is [tex]\bar C(x)=\frac{C(x)}{x}[/tex] and the marginal average cost is [tex]\bar C'(x)=\frac{d}{dx}\bar C(x)[/tex].

We know that the marginal average cost is given by

[tex]\bar C'(x)=\frac{1200}{x^2}[/tex]

[tex]\bar C'(x)=\frac{1200}{x^2}\\\\\bar C(x)=\int {\frac{1200}{x^2}} \, dx[/tex]

[tex]\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\\\1200\cdot \int \frac{1}{x^2}dx\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\1200\cdot \frac{x^{-2+1}}{-2+1}\\\\-\frac{1200}{x}\\\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\\\-\frac{1200}{x}+C[/tex]

Because [tex]\bar C(100)=22[/tex] C is

[tex]22=-\frac{1200}{100}+C\\C=34[/tex]

And the average cost function is

[tex]\bar C(x)=-\frac{1200}{x}+34[/tex]

[tex]x\cdot \bar C(x)=C(x)\\\\C(x)=(-\frac{1200}{x}+34)\cdot x\\\\C(x)=-1200+34x[/tex]

Cost functions are express as

[tex]C=(fixed \:costs)+(variable \:costs)\\C=a+bx[/tex]

According with this definition and the cost function that we have the fixed costs are -$1200.

To find the average cost function, use the given formula and substitute the value of x. To find the cost function, integrate the average cost function. The fixed costs are $22.

To find the average cost function, we will substitute the value of C^overbar(x) into the formula. The average cost function is given by C^overbar(x) = 1,200/x^2.

To find the cost function, we will integrate the average cost function. The cost function m C(x) is the integral of C^overbar(x) with respect to x. F

inally, to find the fixed costs, we will substitute the value of C^overbar(100) into the average cost function.

The average cost function is C^overbar(x) = 1,200/x^2.

The cost function m C(x) = 1,200/x.

The fixed costs are $22.

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Final answer:

To calculate the probability that Kim wins the match, we sum the probabilities of winning in 2 sets, 3 sets, and 4 sets. The probability that Kim wins the match is 0.6553. The probability that Kim wins the match in exactly 2 sets is 0.2339. The probability that 3 sets are played is 0.1728.

Explanation:

To calculate the probability that Kim wins the match, we need to consider the different possible outcomes. Kim can win the match in two sets, which happens with a probability of [tex](0.64)^2 = 0.4096[/tex] . Kim can also win the match in three sets, which happens with a probability of [tex](0.64)^2 * (1-0.64) * (1-0.64) = 0.1728[/tex]. And finally, Kim can win the match in four sets, which happens with a probability of [tex](0.64)^2 * (1-0.64)^2 = 0.0729.[/tex] Therefore, the probability that Kim wins the match is the sum of these probabilities: 0.4096 + 0.1728 + 0.0729 = 0.6553.

To calculate the probability that Kim wins the match in exactly 2 sets, we use the probability of winning a set twice and the probability of losing a set once:[tex](0.64)^2 * (1-0.64) = 0.2339.[/tex]

To calculate the probability that 3 sets are played, we need to consider the different possible outcomes. Kim can win the match in three sets, which happens with a probability of [tex](0.64)^2[/tex]* (1-0.64) * (1-0.64) = 0.1728. Therefore, the probability that 3 sets are played is 0.1728.

Answer: (0.089, 0.125)

Step-by-step explanation:

Confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where n= sample size.

[tex]\hat{p}[/tex] = Sample proportion.

z*= Critical z-value.

Let p be the population proportion of people the company contacts who may buy something.

As per given , sample size : n= 1140

Number of recipients ordered = 122

Then, [tex]\hat{p}=\dfrac{122}{1140}\approx0.107[/tex]

Critical value for 95% confidence interval = z*= 1.96 (By z-table)

So , the 95% confidence interval for the percentage of people the company contacts who may buy something:

[tex]0.107\pm (1.96)\sqrt{\dfrac{0.107(1-0.107)}{1140}}[/tex]

[tex]=0.107\pm (1.96)\sqrt{0.000083817}[/tex]

[tex]=0.107\pm (1.96)(0.00915516)[/tex]

[tex]=0.107\pm 0.018[/tex]

[tex]=(0.107-0.018,\ 0.107+0.018)=(0.089,\ 0.125)[/tex]

Hence, the 95% confidence interval for the percentage of people the company contacts who may buy something = (0.089, 0.125)

Answer:

[tex]\chi^2 = \frac{(141-141)^2}{141}+\frac{(291-282)^2}{282}+\frac{(132-141)^2}{141}=0.862[/tex]

[tex]p_v = P(\chi^2_{2} >0.861)=0.6502[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(0.861,2,TRUE)"

Since the p value is higher than the significance level [tex]0.6502>0.05[/tex] we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences for the proportions assumed.

Step-by-step explanation:

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".

Solution to the problem

Assume the following dataset:

White = 141, Pink = 291, Red= 132

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference with the proportions assumed [tex]p_{white}=1/4, p_{pink}=1/2, p_{red}=1/4[/tex]

H1: There is difference with the proportions assumed [tex]p_{white}=1/4, p_{pink}=1/2, p_{red}=1/4[/tex]

The level of 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 = p_i *Total[/tex]

And the calculations are given by:

[tex]E_{White} =\fra{1}{4} * 564=141[/tex]

[tex]E_{Pink} =\frac{1}{2} *564=282[/tex]

[tex]E_{Red} =\frac{1}{4}*564=141[/tex]

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(141-141)^2}{141}+\frac{(291-282)^2}{282}+\frac{(132-141)^2}{141}=0.862[/tex]

Now we can calculate the degrees of freedom for the statistic given by:

[tex]df=categories-1=3-1=2[/tex]

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{2} >0.861)=0.6502[/tex]

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(0.861,2,TRUE)"

Since the p value is higher than the significance level [tex]0.6502>0.05[/tex] we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we don't have significant differences for the proportions assumed.

The tangent to [tex]r(\theta)=e^\theta[/tex] has slope [tex]\frac{\mathrm dy}{\mathrm dx}[/tex], where

[tex]\begin{cases}x(\theta)=r(\theta)\cos\theta\\y(\theta)=r(\theta)=\sin\theta\end{cases}[/tex]

By the chain rule, we have

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}[/tex]

and by the product rule,

[tex]\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r(\theta)\sin\theta[/tex]

[tex]\dfrac{\mathrm dy}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r(\theta)\cos\theta[/tex]

so that with [tex]\frac{\mathrm dr}{\mathrm d\theta}=e^\theta[/tex], we get

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}=-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}[/tex]

The tangent line is horizontal when the slope is 0; this happens for

[tex]-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}=0\implies\sin(2\theta)=-1\implies2\theta=-\dfrac\pi2+2n\pi\implies\theta=-\dfrac\pi4+n\pi[/tex]

where [tex]n[/tex] is any integer. In the interval [tex]0\le\theta\le2\pi[/tex], this happens for [tex]n=1,2[/tex], or

[tex]\theta=\dfrac{3\pi}4\text{ and }\theta=\dfrac{7\pi}4[/tex]

i.e at the points

[tex](r,\theta)=\left(e^{3\pi/4},\dfrac{3\pi}4\right)[/tex]

and

[tex](r,\theta)=\left(e^{7\pi/4},\dfrac{7\pi}4\right)[/tex]

Answer:

We accept the null hypothesis

Step-by-step explanation:

Given

The valve was tested on 120 engines

Mean pressure = 5 lbs/square inch.

Variance = 1

Standard Deviation, σ = √1 = 1

The valve was designed to produce a mean pressure of 5.1 lbs/square inch

So, μ = 5.1

Null hypothesis: H₀ : μ = 5.1

Alternative Hypothesis: H₁ : μ≠ 5.1

Since n > 30 and population standard deviation is given

So, We will use z test

Formula : z = (x - μ)/( σ /√n) ---_-_--- Substitute the values

z = [tex]\frac{5 - 5.1}{1/\sqrt{120} }[/tex]

z = −1.1

The p - value of -1.1 (refer to z table) is 0.13576

Since it is a two tailed test So, p = 2(1-  0.13576) = 1.7285

α = 0.02

p value > α

So, we accept the null hypothesis

Hence There is no sufficient evidence at the 0.02 level that the valve does not performs below to the specifications

Answer: x = 4 , y = -3

Step-by-step explanation:

Going by the Cramer's rule , we first determine the determinant by dealing with only  the coefficients of x and y in the 2 x 2 matrix.

2x  -  3y = 17

5x  + 4y = 8

2        -3

5         4,   going by the rule now, we now have

(2 x 4) - (5 x -3)

8 + 15

= 23.

Now to find the value of x , replace the constants with the coefficient of x and divide by he determinants.

 17        -3

 8          4

---------------  

2           -3

5            4

( 17 x 4 ) - ( 8 x -3 )

---------------------------

         23

     =  68 + 24

         ------------

              23

    =       92/23

    =           4.

 x =   4

Now  to find y, just repeat the process by replacing the coefficient of y with the constants.

2        17

5        8

-----------

2        -3

5         4

( 2 x 8 ) - ( 5 x 17 )

-----------------------

            23

       16 - 85

       ---------

          23

  =  -69/23

  =   -3

y =  -3.

check

substitute for the values in any of the equations above.

2(4)  - 3(-3)

8  + 9

= 17

Answer:

a. 95​% confidence interval estimate for the population mean amount of paint included in a​ 1-gallon can is 0.998±0.0055

b.  No, because a​ 1-gallon paint can containing exactly​ 1-gallon of paint lies within the 95​% confidence interval.

c. Yes.  The population amount of paint per can is assumed normally distributed, because confidence interval calculations assume normal distribution of the parameter.

d. 90% confidence interval is 0.998±0.0046. ​The answer in b. didn't change; 1-gallon paint can containing exactly​ 1-gallon of paint lies within the 90​% confidence interval. The manager doesn't have a right to complain to the manufacturer.

Step-by-step explanation:

Confidence Interval can be calculated using M±ME where

M is the sample mean amount of paint per 1​-gallon can (0.998 gallon)

ME is the margin of error from the mean

And margin of error (ME) can be calculated using the equation

ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where

Then ME=[tex]\frac{1.96*0.02}{\sqrt{50} }[/tex]≈0.0055

95% confidence interval is 0.998±0.0055

90% confidence interval can be calculated similary, only z statistic is 1.64.

ME=[tex]\frac{1.64*0.02}{\sqrt{50} }[/tex]  ≈0.0046

90% confidence interval is 0.998±0.0046