The accompanying summary data on total cholesterol level (mmol/l) was obtained from a sample of Asian postmenopausal women who were vegans and another sample of such women who were omnivores. Diet Sample Size Sample Mean Sample SD Vegan 85 5.10 1.05 Omnivore 96 5.55 1.20 Calculate a 99% CI for the difference between population mean total cholesterol level for vegans and population mean total cholesterol level for omnivores. (Use μVegan − μOmnivore. Round your answers to three decimal places.)

Answer:

There is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.

Step-by-step explanation:

The question is incomplete as some information is not provided, please refer below the remaining information of the question.

Here are the data on grams of potato chips consumed (note that the study report gave the standard error of the mean rather than the standard deviation):

Group          n     bar{x}         s

Unrestrained  9     63.83      24.72

Restrained  11     34      39.8

a. The standard error of the difference between sample means  ( ± 0.0001 )  is:

b. The critical value  ( ± 0.001 )  from the t distribution for confidence interval 80 % using the conservative degrees of freedom is:

c. Give a 80 % confidence interval   ( ± 0.01 )  that describes the effect of restraint:

Answer:

a) The standard error of the difference between sample means:

S E  =  √ s 2 1 /  n 1  +  s 2 2 /n 2

    =  √ s e 2 1 +  s e 2 2

    =  √ 24.72 2 +  39.8 2

    =  46.85

b) The degrees of freedom:

D f  =  n 1  + n 2  − 2

=  9  +  11  −  2

= 18

The confidence level  =  0.80

The significance level,  α  =  0.20

The critical value from the t distribution for confidence interval 80%:

t c r i t i c a l  =  t α / 2 ,  d f  =  t 0.10 , 18  =  ±  1.33

c) The 80% confidence interval:

( μ 1 − μ 2 ) = ( ¯ x 1 − ¯ x 2 )  ±  t ⋅ S E

= ( 63.83 − 34 ) ±  1.33 × 46.85  

= 29.83 ± 62.31

− 32.48 < ( μ 1 − μ 2 ) <  92.14

As the interval contains the zero. So, it can be concluded that there is no effect of the good intentions i.e. to restrain their diet out of concern about their weight, have on their eating habits.

Good intentions regarding diet may impact eating habits among college students, but environmental influences often override intentions.

The effect of good intentions on eating habits, especially among those who are trying to restrain their diet due to concerns about weight, can vary greatly. Research shows that consumption patterns, such as an increase in eating fast food and sugary beverages, coupled with a decrease in physical activity, contribute to significant weight gain among college students. Implementing healthier eating and physical activity habits is crucial, but it is also important to acknowledge that neither changing diets nor exercise are effective on their own for preventing health issues. Both elements must be combined to reduce the risk of chronic diseases such as cardiovascular disease and cancers.

Good intentions may have some impact on eating habits, but the susceptibility to environmental influences, such as availability of unhealthy snacks, can override these intentions. Therefore, colleges and universities are actively pursuing comprehensive approaches to encourage both healthy eating and increased physical activity. For individuals concerned about their diet and weight, it is essential to engage in consistent healthy behaviors, rather than relying purely on intentions.

Answer:

Hence safely 9 watermelons can be placed in a single container.

Step-by-step explanation:

Given that the  weight (in pounds) of ``medium-size'' watermelons is normally distributed with mean 15 and variance 4.

X = weight in pounds of medium size watermelons is N(15, 2)

Let the water melons stored be n

Then the sample of n has a mean of (15) and std error = [tex]\frac{2}{\sqrt{n} }[/tex]

Capacity = 140

Hence we can say either 8 or 9

If n=9, we have weight = 15*9+1.96*2/3 = 136.40

Hence safely 9 watermelons can be placed in a single container.

Answer:

3

Step-by-step explanation:

View Image

Answer:

6 people ride in each car.

Step-by-step explanation:

42 people / 7 cars = 6 people/car

In this high school level mathematics question, we determine that 6 people ride in one car in the amusement park ride with 7 cars and 42 people. The calculation involves dividing the total number of people by the total number of cars.

Mathematics - In this question, we are determining how many people ride in one car in an amusement park ride where there are 7 cars and 42 people in total. Since the number of people is divided equally among the cars, we can calculate the number of people in one car by dividing the total number of people by the total number of cars.

Calculation:

Total People = 42

Total Cars = 7

Number of People in One Car = Total People / Total Cars = 42 / 7 = 6

Therefore, 6 people ride in one car in this amusement park ride.

Answer:

[tex]\chi^2_{\alpha/2}=43.195[/tex]

[tex]\chi^2_{1- \alpha/2}=14.573[/tex]

b. 14.573 and 43.195.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Solution to the problem

The confidence interval for the population variance is given by the following formula:

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]

We need on this case to calculate the critical values. First we need to calculate the degrees of freedom given by:

[tex]df=n-1=28-1=27[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table for the Chis square distribution with 27 degrees of freedom to find the critical values. We need a value that accumulates 0.025 of the area on the left tail and 0.025 of the area on the right tail.  

The excel commands would be: "=CHISQ.INV(0.025,27)" "=CHISQ.INV(0.975,27)". so for this case the critical values are:

[tex]\chi^2_{\alpha/2}=43.195[/tex]

[tex]\chi^2_{1- \alpha/2}=14.573[/tex]

b. 14.573 and 43.195.

Answer:

True, they form a proportion because their cross-multiplication products are equal.

Step-by-step explanation:

[tex]\frac{49}{21} =\frac{28}{12}[/tex] True or False?

If they are a proportion, the products you get from cross-multiplying are the same.

Multiply the left numerator by right denominator:

49 X 12 = 588

Multiply the left denominator by right numerator:

21 X 28 = 588

588 = 588

Therefore it is true, 49/21 and 28/12 form a proportion.

Answer:

Equation:

250 + 132.5y = x

Solution for solving for x:

x = 250 + 132.5y

Solution for solving for y:

y = (x - 250)/132.5

In this situation, the solutions mean that the equation is linear or first grade because the solutions can be calculated in any of the following four forms:

1. Adding the same number to both sides

2. Subtracting the same number from both sides

3. Multiplying both sides by the same number

4. Dividing both sides by the same number

Completing the question and statement correctly:

The equation  represents the relationship between the quantities in this situation, where x is the weight, in grams, of the filled box and y the number of shirts in the box.

Step-by-step explanation:

1. Let's review all the information provided to us to answer the question correctly:

Weight of an empty box = 250 grams

Weight of each T-shirt = 132.5 grams

2. Name two possible solutions to the equation that represent the relationship between the quantities in this situation, where x  is the weight, in grams, of the filled box and y the number of shirts in the box. What do the solutions mean in this situation?

x = weight, in grams, of the filled box

y = number of shirts in the box

Equation:

250 + 132.5y = x

Solution for solving for x:

x = 250 + 132.5y

Solution for solving for y:

y = (x - 250)/132.5

In this situation, the solutions mean that the equation is linear or first grade because the solutions can be calculated in any of the following four forms:

1. Adding the same number to both sides

2. Subtracting the same number from both sides

3. Multiplying both sides by the same number

4. Dividing both sides by the same number

Step-by-step explanation:

The two possible solution are 647.5 gm and 912.5 gm.

Let's use the following equation:

Weight of the filled box (W) = Weight of the empty box (250 grams) + (Number of T-shirts) * (Weight of each T-shirt, 132.5 grams)

So, the equation would be:

W = 250 + 132.5 * N

Where:

- W is the weight of the filled box in grams.

- N is the number of T-shirts in the box.

Now, let's find two possible solutions:

Solution 1:

Suppose the box is filled with 3 T-shirts.

W = 250 + 132.5 * 3

W = 250 + 397.5

W = 647.5 grams

Solution 2:

Suppose the box is filled with 5 T-shirts.

W = 250 + 132.5 * 5

W = 250 + 662.5

W = 912.5 grams

- Solution 1: If the box is filled with 3 T-shirts, it would weigh 647.5 grams in total.

This means that the combined weight of the 3 T-shirts added to the empty box's weight is 647.5 grams.

- Solution 2: If the box is filled with 5 T-shirts, it would weigh 912.5 grams in total.

This means that the combined weight of the 5 T-shirts added to the empty box's weight is 912.5 grams.

In general, the equation allows you to calculate the weight of the filled box based on the number of T-shirts you put in it. Each additional T-shirt adds 132.5 grams to the total weight, considering the empty box's weight of 250 grams.

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Answer:

0.278

Step-by-step explanation:

Given that Nuri joins a game for a car. The rule is that Nuri pick one key from box either A, B, or C. A box has two keys but only one can be used. B box has three keys but only one can be used. C box has two keys but none of them can be used.

Each box is equally likely to be selected.

In other words

[tex]P(A) = P(B) = P(C)=\frac{1}{3}[/tex]

If A is selected then probability of winning is using the correct key out of two keys i.e. 0.5

If B is selected then probability of winning is using the correct key out of three keys i.e. 0.333

If c is selected then probability of winning is using the correct key out of two keys i.e. 0.00

So the probability that Nuri can win the car

= [tex]\frac{1}{3} *0.5+\frac{1}{3} *0.333+\frac{1}{3} *0\\= 0.278[/tex]

Answer:

C. The parent population must be normally distributed

Step-by-step explanation:

The parent population must follow a normal distribution in order to use a t procedure

Answer:

Since p >0.05 at 5% level we find that there is no evidence    of a difference in the proportion of defective parts produced by the two suppliers

Step-by-step explanation:

Given that a manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective; an SRS of 100 parts from supplier 2 finds 10 defective.

Let p 1 p1 and p 2 p2 be the proportion of all parts from suppliers 1 and 2, respectively, that are defective.

[tex]H 0 : p 1 = p 2 \\H a : p 1 \neq  p 2 .[/tex]

(two tailed test )

Sample             I            II          total

N                     400      100        500

x                        20         10          30

p                       0.05    0.10      0.06

[tex]Std error =\sqrt{\bar p(1- \bar p)(\frac{1}{n_1} +\frac{1}{n_2} )}  \\=0.0266[/tex]

Z= -1.8831

p value = 0.0601

Since p >0.05 at 5% level we find that there is no evidence    of a difference in the proportion of defective parts produced by the two suppliers

Option 2 is correct. The  P-value of the test is 0.06.

The question requires testing if there is a significant difference between the proportions of defective parts from two suppliers. We test this using a two-proportion z-test and set up our hypotheses as H0: P₁ = P₂ and Ha: P₁ ≠ P₂.

Collect the Sample Data:

Sample 1: n₁ = 400, x₁ = 20

Sample 2: n₂ = 100, x₂ = 10

Calculate the Sample Proportions:

P₁ = x₁/n₁ = 20/ 400 = 0.05

P₂ = x₂/n₂ = 10/100 = 0.10

Calculate the Pooled Proportion:

[tex]\hat p[/tex] = (x₁ + x₂) / ( n₁ + n₂) = (20 + 10) / (400 + 100) = 30 / 500 = 0.06

Calculate the Standard Error (SE) of the Difference in Proportions:

[tex]SE &= \sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} \\ SE &= \sqrt{0.06 \times 0.94 \left( \frac{1}{400} + \frac{1}{100} \right)} \\ SE &= \sqrt{0.06 \times 0.94 \left( 0.0025 + 0.01 \right)} \\ SE &= \sqrt{0.06 \times 0.94 \times 0.0125} \\ SE &= \sqrt{0.000705} \\ SE &\approx 0.02655[/tex]

Calculate the Test Statistic z:

[tex]z &= \frac{\hat{p}_1 - \hat{p}_2}{SE} \\ z &= \frac{0.05 - 0.10}{0.02655} \\ z &\approx -1.886[/tex]

Find the P-value:

Since this is a two-tailed test, we need to find the probability of obtaining a test statistic as extreme as $z = -1.886$. The P-value is found using the standard normal distribution.

The P-value for $z = -1.886$ is approximately $0.06$.

Complete question:

A manufacturer receives parts from two suppliers. An SRS of 400 parts from supplier 1 finds 20 defective; an SRS of 100 parts from supplier 2 finds 10 defective. Let P₁ and P₂ be the proportion of all parts from suppliers 1 and 2, respectively, that are defective. The manufacturer wants to know if there is evidence of a difference in the proportion of defective parts produced by the two suppliers. To make this determination, you test the hypotheses H0 : P₁ = P₂ and Ha : P₁ ≠ P₂.

The P-value of your test is:

0.03

0.06

0.116

Answer:

B

Step-by-step explanation:

Absolute value is essentially just every number, but positive. (like -2's absolute value is 2). This is also a number's distance from zero. 2 is 2 away from zero, and -2 is 2 away from zero.

Absolute value returns the positive value of a numerical expression.

Absolute value is used to calculate distance on a coordinate plane because (a) Absolute value is the distance between 2 points on a number​ line, so it gives the distance between any 2 points.

The distance between points A and B on the coordinate plane is:

[tex]\mathbf{Distance = |A - B|}[/tex]

or

[tex]\mathbf{Distance = |B - A|}[/tex]

The above equations mean that:

Absolute values will calculate the distance between any two points, irrespective of whether the point is 0 or not.

Hence, the correct option is: (a)

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She will need to use 4/5 cups of flour

Step-by-step explanation:

First of all we have to calculate the 60% of the flour used in actual recipe then it will be added to the original amount,

So,

Given

Cups of flour used in actual recipe = 1/2

60% of 1/2:

[tex]= \frac{1}{2} * \frac{60}{100}\\= \fra{60}{200}\\=\frac{3}{10}[/tex]

So the total cups of flour for 60% more pancakes will be:

[tex]= \frac{1}{2} + \frac{3}{10}\\\\= \frac{5+3}{10}\\\\=\frac{8}{10}\\\\=\frac{4}{5}[/tex]

Hence,

She will need to use 4/5 cups of flour

Keywords: Percentage, percent

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Answer:

The minimum sample size is N=1537.

Step-by-step explanation:

We want to know the sample size to estimate the proportion of voters with a margin of error of 0.05, with a 95% of confidence.

The margin of error can be defined as:

[tex]e=UL-LL=(p+z*\sqrt{\frac{p(1-p)}{N}}) -(p-z*\sqrt{\frac{p(1-p)}{N}})=2z\sqrt{\frac{p(1-p)}{N}}[/tex]

We can calculate N from this

[tex]e=2z\sqrt{\frac{p(1-p)}{N}}\\\\\frac{p(1-p)}{N}=(\frac{e}{2z})^2\\\\N= (\frac{2z}{e})^2*p(1-p)[/tex]

The sample size will be calculated for p=0.5, which is the proportion that requires the larger sample size (maximum variance).

The z-value for a 95% CI is z=1.96.

The minimum sample size is then

[tex]N= (\frac{2z}{e})^2*p(1-p)=(\frac{2*1.96}{0.05})^2*0.5(1-0.5)=6146.56*0.5*0.5=1536.64[/tex]

The minimum sample size is N=1537.

Answer:

D) 9

Step-by-step explanation:

7 x 2 - 4 - 9x

14 - 4 - 9x

10 - 9x

You could easily see at first that the coefficient of x is 9 because the coefficient is the number that is before the variable, but you can also do the whole process.

Hope this was helpful :)

Answer:

The expected number of tickets written would be 6.5 per day.

Step-by-step explanation:

Given that the Sutton police  department must write, on average, 6 tickets a day to keep department revenues at budgeted levels.

Suppose the number of tickets written per day follows a Poisson distribution with a mean of 6.5 tickets per day.

This means population parameter i.e. expected value of tickets a day is 6.5

Option I is wrong, because 6.5 is the parameter of Poisson distribution for number of tickets

Option III is wrong because mean is the overall average and hence need not be balanced by exactly half.

Option IV is wrong because mean is the expected value and so we cannot say more than 6.5 tickets

Only correct option is option 2.

The expected number of tickets written would be 6.5 per day.

Answer: 0.0668

Step-by-step explanation:

Given : The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 ounces and a standard deviation of 0.30 ounce.

i.e. [tex]\mu=12.45[/tex] and [tex]\sigma=0.30[/tex]

Let x denotes the volume of soda a dispensing machine pours into a 12-ounce can.

Then, the proportion of the soda cans contain less than the advertised 12 ounces of soda will be :-

[tex]P(x<12)=P(\dfrac{x-\mu}{\sigma}<\dfrac{12-12.45}{0.30})\\\\=P(z<-1.5)\ \ [\because z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(z<1.5)\ \ [\because P(Z<-z)=1-P(Z<z)]\\\\=1-0.9332\ [\text{By z-table}]\\\\=0.0668[/tex]

Hence, the proportion of the soda cans contain less than the advertised 12 ounces of soda = 0.0668

Final answer:

The proportion of 12-ounce soda cans that contain less than the advertised amount of soda is approximately 6.68%, which is found by calculating the z-score of 12 ounces and looking up the corresponding proportion in the standard normal distribution table.

Explanation:

To find the proportion of soda cans that contain less than the advertised 12 ounces of soda, we need to calculate the z-score for 12 ounces using the mean and standard deviation of the soda volumes. The z-score tells us how many standard deviations away from the mean a certain value is.

The z-score formula is:

Z = (X - μ) / σ

Where X is the value (12 ounces), μ is the mean (12.45 ounces), and σ is the standard deviation (0.30 ounces).

Plugging in the values we get:

Z = (12 - 12.45) / 0.30
Z = -0.45 / 0.30
Z = -1.5

Once we have the z-score, we can use the standard normal distribution table to find the proportion of values that lie below this z-score. The table gives us the proportion of the distribution that is to the left of the z-score. A z-score of -1.5 corresponds to a proportion of approximately 0.0668.

Therefore, the proportion of soda cans containing less than 12 ounces is approximately 0.0668, or 6.68%.

Answer:

Step-by-step explanation:

given that an article in the Sacramento Bee (March 8, 1984, p. A1) reported on a study finding that "men who drank 500 ounces or more of beer a month (about 16 ounces a day) were three times more likely to develop cancer of the rectum than non-drinkers."

To support this some data collected should be given.  The data should be from a sample of a large size randomly drawn from 2 groups one men who drank 500 oz or more of beer and other group not drank that much.  These people medical history with cancer patients number also should have been given.

Final answer:

The Sacramento Bee article is missing key numerical details such as the baseline rate of rectal cancer in non-drinkers, the number of study participants, the total number of cancer cases, the study duration, and potential confounding factors. This information is crucial for interpreting the findings about the link between heavy alcohol consumption and increased cancer risk.

Explanation:

The report from the Sacramento Bee on a study finding that men who consumed a significant amount of beer had a higher risk of developing rectal cancer is missing several crucial pieces of numerical information. Specifically, it doesn't provide the baseline rate of rectal cancer in non-drinkers, which is necessary to understand the relative increase among the beer drinkers. Additionally, the exact number of participants in each group (drinkers vs. non-drinkers), the total number of rectal cancer cases reported, the duration of the study, and potential confounding factors that might influence the results were not disclosed. These details are essential to assess the study's validity and the significance of the findings regarding alcohol consumption and cancer risk.

Research has consistently demonstrated that excessive alcohol intake is linked with an increased risk of various cancers. Drinking too much alcohol is one lifestyle habit that can raise the risk of cancers of the mouth, esophagus, liver, breast, colon, and rectum. Notably, the National Cancer Institute has identified alcohol as a risk factor for these cancers, and multiple studies suggest that this risk increases with higher alcohol consumption. It is also established that moderate alcohol consumption could provide some cardiovascular benefits, but the trade-offs must be carefully weighed considering the increased risk of other health problems, such as cancers and hemorrhagic stroke.

Answer:

The sample proportion is 8.4%.

Step-by-step explanation:

The sample proportion f people in the United States who earn more than $75,000 each year is the number of people who reported earning more than $75,000 each year divided by the size of the sample.

The proportion is:

[tex]P = \frac{168}{2000} = 0.084[/tex]

The sample proportion is 8.4%.

The sample proportion of people in the United States who earn more than $75,000 each year is 0.084.

Number of people who reported earning more than $75,000: 168

Total number of people in the sample: 2,000

[tex]\[ \hat{p} = \frac{168}{2000} \][/tex]

[tex]\[ \hat{p} = 0.084 \][/tex]

 To three decimal places, the sample proportion is 0.084

To calculate the power of a hypothesis test for an IQ test with a true population mean of 105, we find the Z-score for the sample mean, use it to get the cumulative probability, and subtract that from 1. This results in the power of the test to be 0.0475.

[tex](15 / sqrt(25) = 3)[/tex]

Let's illustrate with numerical probabilities, assuming the Z-score of 1.67 corresponds to a cumulative probability of approximately 0.9525. The power of the test is then 1 - 0.9525 = 0.0475 or 4.75%. The power is relatively low, indicating a high risk of Type II error (failing to reject the null hypothesis when it should be rejected).

Standard Error: [tex]\(SE = \frac{σ}{\sqrt{n}} = 3\).[/tex] Z-score: [tex]\(Z = \frac{X - μ}{SE} \approx 3.33\).[/tex] Power: Find [tex]\(P(Z > 3.33)\).[/tex]

let's go through the process step by step:

Step 1: Standard Error of the Mean (SE)

The standard error of the mean (SE) is calculated using the formula:

[tex]\[ SE = \frac{σ}{\sqrt{n}} \][/tex]

Where:

- [tex]\( σ \)[/tex] is the population standard deviation (given as 15),

- [tex]\( n \)[/tex] is the sample size (given as 25).

Plugging in the values:

[tex]\[ SE = \frac{15}{\sqrt{25}} = \frac{15}{5} = 3 \][/tex]

So, the standard error of the mean is 3.

Step 2: Z-score Calculation

The Z-score measures how many standard deviations a data point is from the mean. It's calculated using the formula:

[tex]\[ Z = \frac{X - μ}{SE} \][/tex]

Where:

- [tex]\( X \)[/tex] is the sample mean,

-[tex]\( μ \)[/tex] is the population mean under the null hypothesis (given as 100),

- [tex]\( SE \)[/tex] is the standard error of the mean (calculated as 3).

We want to find the Z-score for a sample mean of 110:

[tex]\[ Z = \frac{110 - 100}{3} = \frac{10}{3} \]\[ Z \approx 3.33 \][/tex]

Step 3: Finding the Power

The power of a statistical test is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true.

In this case, the alternative hypothesis is that the true population mean is 105.

We need to find the probability of getting a Z-score greater than 3.33 when the true population mean is 105. This probability represents the likelihood of correctly rejecting the null hypothesis.

[tex]\[ P(Z > 3.33) = 1 - P(Z \leq 3.33) \][/tex]

We look up the probability [tex]\( P(Z \leq 3.33) \)[/tex] in a standard normal distribution table or calculate it using a calculator. Then, we subtract this probability from 1 to find [tex]\( P(Z > 3.33) \)[/tex]. This value gives us the power of the test.

So, in detail:

1. Calculate the standard error of the mean (SE) using the formula and given values.

2. Calculate the Z-score using the formula and the SE calculated in step 1.

3. Find the probability [tex]\( P(Z > 3.33) \)[/tex]  by subtracting [tex]\( P(Z \leq 3.33) \)[/tex] from 1.

Answer:

The degree measure of the slice representing the mint patties rounded to the nearest degree is 69°.

Step-by-step explanation:

Consider the provided information.

A two-pound bag of assorted candy contained 100 caramels,

83 mint patties,

93 chocolate squares,

80 nut clusters,

79 peanut butter taffy pieces.

Now find the degree measure of the slice representing the mint patties rounded to the nearest degree as shown:

[tex]\frac{83}{100+83+93+80+79} \times 360^0[/tex]

[tex]\frac{83}{435} \times 360^0[/tex]

[tex]68.69\approx69^0[/tex]

Hence, the degree measure of the slice representing the mint patties rounded to the nearest degree is 69°.

Answer:

Step-by-step explanation:

a)  While the mean is below 4 the standard deviation tells us that there is a pretty high chance for the value to be above 4.  One standard deviation away is 1.79 - 5.65.  We are concerned with things over 4 so we'll look at the upper half, which is 3.72 - 5.65.  Being one standard deviation to the right means there is a 34.1% chance of the readings being in this range.  And there is even chances of it being slightly higher, though that is comparatively low.  But even as low as 10% is usually considered too high a chance to risk.  If you don't understand how the standard deviation got me those percents let me know.

b) alternative hypothesis is always the option where we want to prove it.  So we want to prove the concentration is 4 or above.  So the null is less than 4 and the alternative is greater than or equal to 4.  Do you know the correct symbols?  if not I can get those written out.  As for the p value we need the confidence level for the question, do you have that?

The two hypothesis:

And the reasoning of the inspector is incorrect because it ignores the large standard deviation.

We know that the mean radon concentration must be smaller than 4.0 pC/L.

In this particular house, the mean is 3.72 pC/L with a really large standard deviation of 1.93 pC/L.

And the inspector says that the radon abatement is not necessary, as the mean is smaller than 4.0 pC/L.

Now, as you can see, the standard deviation is really large. This means that over a given period of time, the mean concentration per liter may be larger than 4.0 pC (and then decreases). But this would imply that the exposure over large periods of times could be really large. This is why the reasoning is incorrect.

b) The null hypothesis is what we want to prove. In this case, is that the mean concentration is smaller than 4.0 pC/L.

The alternative hypothesis is the other option, in this case, that the concentration is equal or larger than 4.0 pC/L.

using the correct notation and defining C as the concentration we can write:

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In this context, cos(θ)=0.9 represents the horizontal distance of the skier relative to the ski trail's center point, specifically, it's 0.9 times the length of the trail's radius to the east. Correct answer is - The skier is 0.88 radius lengths to the East of the center of the ski trail.

In this question, you're being asked about the meaning of cos(θ) = 0.9 in a real-world context. When a skier travels along a circular ski trail, they form an angle theta (θ) with the center of the circle. In trigonometry, cosine gives us the horizontal or x-coordinate in relation to the radius of the circle.

In this case, it means that the skier is 0.9 times the radius of the ski trail to the east of the center of the ski trail. As the radius of the ski trail is 1.25 km, this puts the skier 0.9 x 1.25 = 1.125 km to the east. So, the correct answer would be 'The skier is 0.88 radius lengths to the East of the center of the ski trail'.

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Answer:

Sample size should be atleast 60025

Step-by-step explanation:

Given that a school board wishes to know the current mean reading level of 6th graders throughout a very large school district. Past experience shows that the standard deviation of such reading scores is about 2.5

For 95% confidence we have significance level = 5%

Margin of error = 0.04

Margin of error should be less than 0.04

i.e. Z critical value for 95% *std error <0.04

[tex]1.96*\frac{2.5}{\sqrt{n} } <0.02\\\\n>60025[/tex]

Sample size should be atleast 60025

The school board needs to sample 151 6th graders to be 95% confident that the sample mean reading level is within 0.4 of the true mean.

The school board is interested in determining the required sample size needed to estimate the mean reading level of 6th graders. Given the standard deviation of 2.5 and the margin of error of 0.4 for a 95% confidence interval, we must use the formula for sample size in estimating a mean. The formula is:

n = (z*s/E)^2

where n is the sample size, z is the z-score corresponding to the desired confidence level, s is the population standard deviation, and E is the margin of error. For a 95% confidence level, the z-score is 1.96.

Therefore, the sample size n would be:

n = (1.96*2.5/0.4)^2

n = (4.9/0.4)^2

n = (12.25)^2

n = 150.0625

Since we cannot have a fraction of a sample, we round up to the nearest whole number. Thus, the school board needs to sample 151 6th graders to be 95% confident that the sample mean reading level is within 0.4 of the true mean.

Answer:

0.5 or 50%

Step-by-step explanation:

For any given value of 'x' representing the time between arrivals of two customers. If 0 < x <120, then the cumulative distribution function is:

[tex]\frac{x-0}{120-0}=\frac{x}{120}[/tex]

Therefore, the probability that the time between the arrivals of two customers will be more than 60 seconds is determined by:

[tex]P(X>60) = 1 -\frac{60}{120}\\P(X>60) = 0.5[/tex]

The probability is 0.5 or 50%.

Answer:

A. It's the amount of change in y when x increases by one unit

Step-by-step explanation:

For this case we have the following linear model adjusted:

[tex]\hat y = 36.5 +0.48 x[/tex]

Where y is the dependent variable, x the independent variable, 36.5 represent the intercept and 0.48 the slope.

We can analyze one by one the options to select the most appropiate.

A. It's the amount of change in y when x increases by one unit

True, for this case the slope is defined as:

[tex]m =0.48 \frac{\Delta y}{\Delta x}[/tex]

And is defined as the amount of change in y when x increase 1 unit.

B. It's the value of y when x = 0

False the value of y when x=0 is y= 36.5+0.48(0) = 36.5

C. It's the value of x when y = 0

False, when y=0 we have this:

[tex] 0=36.5 +0.48 x[/tex]

[tex]x= -\frac{36.5}{0.48}=-76.04[/tex]

D. It's the mean of the distribution for the response variable (y) around the explanatory variable (x)

False, the value 0.48 represent the slope obtained from an estimation of least squares and not represent the mean for the response variable y.

E. It's the amount of change in x when y increases by one unit.

False, is defined inverse as: the amount of change in y when x increase 1 unit.

The value .48 in the given linear equation represents the rate of change in y for each unit increase in x. It signifies the slope of the line, implying that the value of y increases by .48 for every unit increase in x.

In the equation y = 36.5 + .48x, the value .48 represents the slope of the line in the context of linear equation. The slope indicates the rate of change in y with respect to x. In this case,

.48 means that for each unit increase in the value of x (independent variable), the value of y (dependent variable) will increase by .48 units.

Therefore, the correct answer is A: It's the amount of change in y when x increases by one unit. This is a fundamental component of understanding linear equations, slopes and how they represent the relationship between two variables.

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Answer:

[tex]t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974[/tex]    

[tex]p_v =2*P(t_{39}<-1.8974)=0.0652[/tex]

If we compare the p value with a significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

Step-by-step explanation:

Assuming the following problem: "In Hamilton County, Ohio the mean number of days needed to sell a home is 86 days (Cincinnati Multiple Listing Service, April, 2012). Data for the sale 40 of homes in a nearby county showed a sample mean of 80 days with a sample standard deviation of 20 days. Conduct a hypotheses test to determine whether the mean number of days until a home is sold is different than the Hamilton county mean of 86 days in the nearby county. "

1) Data given and notation    

[tex]\bar X=80[/tex] represent the sample mean

[tex]s=20[/tex] represent the sample standard deviation

[tex]n=40[/tex] sample size    

[tex]\mu_o =86[/tex] represent the value that we want to test  

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)    

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

2) State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean is different from 86, the system of hypothesis would be:    

Null hypothesis:[tex]\mu = 86[/tex]    

Alternative hypothesis:[tex]\mu \neq 86[/tex]    

We don't know the population deviation, so for this case we can use the 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".

3) Calculate the statistic    

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

[tex]t=\frac{80-86}{\frac{20}{\sqrt{40}}}=-1.8974[/tex]    

4) Calculate the P-value    

First we need to find the degrees of freedom given by:

[tex]df=n-1=40-1=39[/tex]

Since is a two tailed test the p value would be:    

[tex]p_v =2*P(t_{39}<-1.8974)=0.0652[/tex]

In Excel we can use the following formula to find the p value "=2*T.DIST(-1.8974,39,TRUE)"  

5) Conclusion    

If we compare the p value with a significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we FAIL to reject the null hypothesis, so there is not enough evidence to conclude that the mean for the data is significantly different from 86 days.    

Answer:

Step-by-step explanation:

Given that X and Y are independent random variables with the following distributions:

x -1 10 1 2 Total

p 0.3 0.1 0.5 0.1 1

xp -0.3 1 0.5 0.2 1.4

x^2p 0.3 10 0.5 0.4 11.2

Mean of X = 1.4

Var(x) = 11.2-1.4^2 =  9.24    

y 2 3 5  

p 0.6 0.3 0.1  1

yp 1.2 0.9 0.5 0 2.6

y^2p 2.4 2.7 2.5 0 7.6

Mean of Y = 2.6

Var(Y) = 11.2-1.4^2 =  0.84

3) W=3+2x

Mean of w =3+2*Mean of x = 7.2

Var (w) = 0+2^2 Var(x)= 36.96

Answer:

A sample size of at least 737 specimens 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{1-0.99}{2} = 0.005[/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]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/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 = 3, \sigma = 31.62[/tex]

So:

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

[tex]3 = 2.575*\frac{31.62}{\sqrt{n}}[/tex]

[tex]3\sqrt{n} = 81.4215[/tex]

[tex]\sqrt{n} = 27.1405[/tex]

[tex]n = 736.60[/tex]

A sample size of at least 737 specimens is required.

Probability of graduating is 73% = 0.73

Probability of not graduating = 1-73% = 0.27

Number of students (n) = 15

Number of graduates (x) = 8

Use the binomial probability formula:

P(x) = (n/x) *p^x * (1-p)^n-x

P(8) = 15/8 * 0.73^8 * 0.27^7

P(8) = 0.0543

Rounded to nearest hundredth = 0.05

Final answer:

To find the probability, use the binomial formula: [tex]\( P(X = 8) = \binom{15}{8} \times (0.73)^8 \times (0.27)^7 \).[/tex] Calculate to get approximately 0.195.

Explanation:

To calculate the probability of exactly 8 out of 15 freshmen graduating, we can use the binomial probability formula:

[tex]\[ P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} \][/tex]

Where:

( n = 15 ) (total number of students)

( k = 8 ) (number of successes, i.e., number of students graduating)

( p = 0.73 ) (probability of success, i.e., the probability that a student graduates)

Using these values, we can plug them into the formula:

[tex]\[ P(X = 8) = \binom{15}{8} \times (0.73)^8 \times (1-0.73)^{15-8} \][/tex]

[tex]\[ P(X = 8) = \binom{15}{8} \times (0.73)^8 \times (0.27)^7 \][/tex]

Using a calculator or statistical software to compute the binomial coefficient, we find:

[tex]\[ \binom{15}{8} = 6435 \][/tex]

Plugging this into the formula:

[tex]\[ P(X = 8) = 6435 \times (0.73)^8 \times (0.27)^7 \][/tex]

[tex]\[ P(X = 8) \approx 0.195 \][/tex]

So, the probability that exactly 8 out of 15 freshmen graduate is approximately 0.195, rounded to the nearest hundredth.

Answer:

As the distance between the y-intercepts of f(x) and g(x) is 3.

So, the y-intercept of f(x) is 3 units below g(x), as shown in attached figure a.

Step-by-step explanation:

Let us consider the equation of straight line in the form of slope-intercept form such as:

[tex]y = mx + b[/tex]      

where m is the slope of the line and b is the y-intercept.

Determining the y-intercept of f(x) = x - 1

As the given function f(x) = x-1 is in the form of [tex]y = mx + b[/tex].

So, the y-intercept will be -1.

y-intercept can also be computed by putting x = 0, as at y-intercept the value of x is zero.

f(x) = x - 1

y = 0 - 1 ⇒ y = -1

Determining the y-intercept of g(x) = x + 2

As the given function g(x) = x + 2 is in the form of [tex]y = mx + b[/tex].

So, the y-intercept will be 2.

y-intercept can also be computed by putting x = 0, as at y-intercept the value of x is zero.

g(x) = x + 2

y = 0 + 2 ⇒ y = 2

Determining the distance between the y - intercepts of f(x) and g(x)

So, the distance between the y - intercepts of f(x) and g(x): 2 - (-1) = 3

So, the y-intercept of f(x) is 3 units below g(x), as shown in attached figure a.

Keywords: y-intercept, graph, function

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