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." In other words, the rate of rectal cancer in the beer drinking group was three times that of the non-drinkers in this study. What important numerical information is missing from this report?

Answer: 525

Step-by-step explanation: As I read the question I’m getting the idea of division. The community program chooses 16 students every year. The 8,400 dollars from last year was the amount of money the students receive all together. Therefore 8,400 divided by 16 is 525

Final answer:

Each of the 16 fifth grade students received $525 from the community program to attend music or art camp, calculated by dividing the total funds of $8,400 by 16 students.

Explanation:

The question asked is about calculating the amount of money awarded to each of the 16 fifth grade students by a community program for attending music or art camp. Since the program awarded a total of $8,400 last year and 16 students were chosen, we need to perform a simple division to find out how much money each student received. To do this, we divide the total amount of money ($8,400) by the number of students (16).

Step-by-step Calculation:

Divide the total amount of money by the number of students: $8,400 \/ 16.

Calculate the result to determine the amount per student.

Therefore, each student received $525 to attend the music or art camp.

Answer:

A. approximately a normal distribution.

Step-by-step explanation:

There may be a few differences, but the sampling distribution of the sample mean is still approximately normal.

So the correct answer is:

A. approximately a normal distribution.

Answer:

Correct answer is (A) {Normal distribution}

Step-by-step explanation:

sampling distribution of the sample mean is still approximately normal.

Answer:

all of the above

Step-by-step explanation:

The number 25 is a natural number as it belongs to the set [1,2,3,4,5,......]

The number 25 is a whole number as it belongs to the set [0,1,2,3,4,5,......]

The number 25 is an Integer as it belongs to the set [...,-5,-4,-3,-2,-1,0,1,2,3,4,5,...]

The number 25 is a rational number as it can be expressed as [tex]\[\frac{25}{1}\][/tex]

For the same reason , number 25 is a real number as it belongs to the set of rational numbers.

So the correct option is "all of the above".

25 does not belong to any of the given number sets (2 and 4, 1 and 2, or 3 and 5).

The number 25 does not belong to any of the provided number sets i.e. 2 and 4, 1 and 2 or 3 and 5. A number set typically refers to a collection of numbers, and in this case, 25 is absent in all the provided sets. The given number sets only contain the numbers 1, 2, 3, 4 and 5. Thus, 25 does not belong to any of these sets.

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Answer:the variable expression to represent the sum of their circumferences would be

(8xy)cm + (5xy)cm

Step-by-step explanation:

The circumference of the first circle is 8xy cm

The circumference of the second circle is 5xy cm

the variable expression to represent the sum of their circumferences would be

(8xy)cm + (5xy)cm

This variable expression can also be simplified further because both terms contain xy.

Answer:

[tex]z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617[/tex]  

[tex]p_v =P(Z>1.5617)=0.059[/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, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.

Step-by-step explanation:

Data given and notation

[tex]\bar X_{A}=23[/tex] represent the mean for the sample A

[tex]\bar X_{B}=18[/tex] represent the mean for the sample B

[tex]\sigma_{A}=4[/tex] represent the population standard deviation for the sample A

[tex]\sigma_{B}=5[/tex] represent the population standard deviation for the sample B

[tex]n_{A}=4[/tex] sample size selected A

[tex]n_{B}=4[/tex] sample size selected B

[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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for A is higher than the mean for B, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{A}-\mu_{B}\leq 0[/tex]

Alternative hypothesis:[tex]\mu_{A}-\mu_{B}>0[/tex]

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

[tex]z=\frac{(\bar X_{A}-\bar X_{B})-0}{\sqrt{\frac{\sigma^2_{A}}{n_{A}}+\frac{\sigma^2_{B}}{n_{B}}}}[/tex] (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

[tex]z=\frac{(23-18)-0}{\sqrt{\frac{4^2}{4}+\frac{5^2}{4}}}}=1.5617[/tex]  

P-value

Since is a one right tailed test the p value would be:

[tex]p_v =P(Z>1.5617)=0.059[/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, and the difference between the true mean of group A and B is not significantly higher than 0 at 5% of significance.

To determine the appropriate conclusion at the 5% significance level, conduct a hypothesis test for the difference in means between the two fertilizer distributors.

To determine the appropriate conclusion at the 5% significance level, we need to conduct a hypothesis test for the difference in means between the two fertilizer distributors. The test statistic calculated in Excel is 1.5617. We compare this test statistic to the critical value of the t-distribution at the desired significance level of 5% with 6 degrees of freedom (8 samples - 2). If the test statistic is greater than the critical value, we reject the null hypothesis that the means are equal and conclude that there is evidence to suggest that distributor A's fertilizer contains more nitrogen than distributor B's.

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Answer: d. 0.60

Step-by-step explanation:

When are performing sample surveys , when the selected participant is giving any response is denoted as non - response.

The proportion of these participants of the sample is known as the non-response rate.

Given : A political scientist wants to know how college students feel about the social security system.

She obtains a list of the 3114 undergraduates at her college and mails a questionnaire to 250 students selected at random.

i.e. Sample size : n= 290

Only 100 of the questionnaires are returned.

Individual gave response =100

Individual gave no-response =250-100 =150

The  rate of non-response [tex]=\dfrac{\text{Individual gave no-response}}{n}[/tex]

[tex]=\dfrac{150}{250} =0.60[/tex]

Hence, the rate of non-response would be 0.60 .

Thus , the correct option is d. 0.60.

Answer: 0.77

Step-by-step explanation:

Given :  Probability that a student chooses a turkey sandwich is

P(Turkey )= 0.10

Probability that a student chooses an egg salad sandwich is

P(egg salad)=0.67

Also, it is impossible for a student have both lunches.

∴ P(Turkey and egg salad) =0

Now , the probability that a student chooses a turkey or egg salad sandwich will be

P(Turkey or egg salad) =  P(Turkey )+ P(egg salad)- P(Turkey and egg salad)

=  0.10+ 0.67-0 = 0.77

Hence, the probability that a student chooses a turkey or egg salad sandwich= 0.77

Final answer:

The probability that a student picks either a turkey or an egg salad sandwich for their school field trip is 0.77 or 77%.

Explanation:

To calculate the probability that a student chooses either a turkey or egg salad sandwich for lunch, we use the formula for the probability of an 'or' event.

Since the options are mutually exclusive, meaning a student can only choose one type of sandwich, we simply add the individual probabilities together.

The probability of choosing a turkey sandwich is 0.10 and the probability of choosing an egg salad sandwich is 0.67.

Therefore, we can calculate it as follows:

P(turkey OR egg salad) = P(turkey) + P(egg salad)

P(turkey OR egg salad) = 0.10 + 0.67

P(turkey OR egg salad) = 0.77

So the probability that a student picks either a turkey or an egg salad sandwich is 0.77, or 77%.

Answer:

a) [tex]df=n-1=16-1=15[/tex]

The statistic calculated is given by t=3.3  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{15}>3.3)=0.0024[/tex]  

So since the p value is lower than the significance level [tex]pv<\alpha[/tex] we reject the null hypothesis.

b) [tex]df=n-1=8-1=7[/tex]

The statistic calculated is given by t=1.8  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{7}>1.8)=0.057[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

c) [tex]df=n-1=26-1=25[/tex]

The statistic calculated is given by t=-0.6  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{25}>-0.6)=0.723[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation      

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

[tex]s[/tex] represent the standard deviation for the sample

[tex]n[/tex] sample size      

[tex]\mu_o =20[/tex] represent the value that we want to test    

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

t would represent the statistic (variable of interest)      

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

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:      

Null hypothesis:[tex]\mu \leq 20[/tex]      

Alternative hypothesis:[tex]\mu > 20[/tex]      

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

(a) n = 16, t = 3.3, a = 0.05, P-value =

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

[tex]df=n-1=16-1=15[/tex]

The statistic calculated is given by t=3.3  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{15}>3.3)=0.0024[/tex]  

So since the p value is lower than the significance level [tex]pv<\alpha[/tex] we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

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

[tex]df=n-1=8-1=7[/tex]

The statistic calculated is given by t=1.8  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{7}>1.8)=0.057[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

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

[tex]df=n-1=26-1=25[/tex]

The statistic calculated is given by t=-0.6  

Since is a one-side upper test the p value would be:      

[tex]p_v =P(t_{25}>-0.6)=0.723[/tex]  

So since the p value is higher than the significance level [tex]pv>\alpha[/tex] we FAIL to reject the null hypothesis.

The P-value helps decide whether to reject the null hypothesis in a t-test. If the P-value is less than the significance level α, the null hypothesis is rejected. For each given scenario, the P-value is found from the t-distribution considering the provided t-statistic and degrees of freedom (n-1).

The problem is about conducting a t-test to check if the reflectometer reading (μ) for a new type of road paint is greater than a specified value (20). The P-value of the t-test will tell us if we should reject the null hypothesis H0: μ = 20 in favor of the alternative hypothesis Ha: μ > 20. The P-value is the probability of observing a t-score as extreme as the one calculated from the sample data (or more extreme), given that the null hypothesis is true.

(a) For n = 16, t = 3.3, and α = 0.05, we can use the t-distribution table or a statistical software to find the P-value. Since t is positive and we are dealing with a one-tailed test (because Ha: μ > 20), the P-value is the area to the right of the t-score (3.3) under the t-distribution. If the calculated P-value is less than α (0.05), we reject the null hypothesis.
(b) The same process applies for n = 8, t = 1.8, and α = 0.01. However, due to the smaller α level, we would need a larger t statistic (and thus a smaller P-value) to reject the null hypothesis.
(c) For n = 26, t = -0.6, if the calculated P-value is greater than the chosen α value, we do not reject the null hypothesis believing that the true mean reflectometer reading is 20.

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Answer:

32 mi

Step-by-step explanation:

Solve using proportions.

[tex]\frac{\frac{1}{2}in}{8 mi} =\frac{2in}{y}[/tex]

Find the scale factor (how to get from left to right)

To get from left numerator to right numerator, multiply by 4.

(1/2) X 4 = 2

The scale factor is 4.

Multiply the left denominator by the scale factor to get "y".

8 mi X 4 = 32 mi

Therefore 2 inches represent 32 miles.

Answer:

(a,b) → (a,-b) : Reflection about x axis.

(a,b) → (a, b+5) : Translation of the point by 5 units up.

(a,b) → (b,-a) : Rotation by 90 degree clockwise.

Step-by-step explanation:

Given:

The transformation of points are given as:

(a,b) → (a,-b)

(a,b) → (a, b+5)

(a,b) → (b,-a)

Now, let us consider each transformation one by one.

(1) (a,b) → (a,-b)

Here, the order of the coordinates has not changed. But, the y coordinate of the point has changed. The y coordinate was 'b' and it has changed only its sign but not value. So, it is a transformation related to reflection.

In reflection, only the sign changes. Since, the 'y' coordinate sing is reversed, so, it is a reflection about x axis.

(2) (a,b) → (a, b+5)

Here, the 'y' coordinate of the point has changed. The change is from 'b' to 'b+5'. So, 5 is added to the y coordinate. As per transformation rules, if a positive number 'C' is added to the y coordinate, then the point shifts vertically up by 'C' units. Hence, there is a translation of 5 units up here.

(3) (a,b) → (b,-a)

Here, the 'x' and 'y' coordinates interchange their values and also the new y coordinate has its sign reversed. This happens in rotation.

We know that, (x, y) → (y, –x) is true when there is rotation by 90 degree clockwise.

So, the point (a,b) → (b,-a) is rotated by 90 degree clockwise.

Answer:

a. No, since the value of the test statistic is less than the critical value

Step-by-step explanation:

1) Data given and notation

n=144 represent the random sample taken

X=72 represent the number of people that prefer the blend

[tex]\hat p=\frac{72}{144}=0.5[/tex] estimated proportion of people that prefer the blend

[tex]p_o=0.47[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.959

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 if higher than 0.47:  

Null hypothesis:[tex]p\leq 0.47[/tex]  

Alternative hypothesis:[tex]p > 0.47[/tex]  

When we conduct a proportion test we need to use the z statistic, 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  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.5 -0.47}{\sqrt{\frac{0.47(1-0.47)}{144}}}=0.721[/tex]  

4) Statistical decision  

We can calculate the critical value since we have a right tailed test, we need to look into the normal standard distribution a value that accumulates 0.01 of the area on the right and 0.99 on the left. And this value is:

[tex]z_{\alpha/2}=2.33[/tex]

And we can use the following excel code to find the critical value: "=NORM.INV(0.99,0,1)"

Our calculated value on this case is less than the critical value so the best conclusion is:

a. No, since the value of the test statistic is less than the critical value

Answer:

0.40

Step-by-step explanation:

Given that at the local racetrack, the favorite in a race has odds 3:2 of losing

Here instead of probability odds are given.

Odds of losing = 3/2

Hence Probability of losing = [tex]\frac{3}{3+2} \\=\frac{3}{5} \\=0.6[/tex]

Probability that the favourite wins the race will be the probability for the event which is complement of losing the game.

Hence

Probability that the favourite wins the race will be the probability

= 1- 0.6

=0.40

Probability that favorite wins the race is 0.6

Favorite in a race has odds = 3:2

Probability that favorite wins the race

Probability that favorite wins the race = 3 / [3 + 2]

Probability that favorite wins the race = 3 / 5

Probability that favorite wins the race = 0.6

Option "C" is the correct answer to the following question.

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Answer:

[tex]E[R][/tex] = 99 Ω

[tex]\sigma_R[/tex] = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, [tex]E[R] = \frac{a+b}{2}  = \frac{95+103}{2} = \frac{198}{2}[/tex] = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

[tex]V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333[/tex]

Hence,

The standard deviation of resistance, [tex]\sigma_R = \sqrt{V(R)} = \sqrt5.333[/tex] = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

[tex]z_1 = \frac{102-99}{2.3094} = 1.299[/tex]

[tex]z_2 = \frac{98-99}{2.3094} = -0.433[/tex]

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

Answer:

a) [tex]S_{a}(t)=16Kg-0.16Kg*\frac{t}{min}[/tex]

b)  [tex]S_{a}(20)=12.8Kg[/tex]

Step-by-step explanation:

It can be seen in the graph that the water velocity and solution velocity is the same, but the salt concentation will be lower

Water velocity [tex]V_{w} = 60\frac{L}{min}[/tex]

Solution velocity [tex]V_{s} = 60\frac{L}{min}[/tex]

Brine concentration = [tex]\frac{6,000L}{16Kg}=375\frac{L}{Kg}[/tex]

a) Amount of salt as a funtion of time Sa(t)

[tex]S_{a}(t)=16Kg-\frac{60Kg*L}{375L}*\frac{(t)}{min}=[tex]16Kg-0.16Kg*\frac{t}{min}[/tex]

b) [tex]S_{a}(20)=16Kg-0.16\frac{Kg}{min}*(20min)=16Kg-3.2Kg=12.8Kg[/tex]

This value was to be expected since as the time passes the concentration will be lower due to the entrance to the pure water tank

To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min and then calculate the amount of salt in the tank after a specific time.

To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min:

Rate of salt entering = (60 L/min) * (16 kg/6000 L) = 0.16 kg/min

Therefore, the amount of salt in the tank after t minutes is given by:

y = 0.16 kg/min * t min = 0.16t kg

Answer:

The key difference is that the dependent sample test uses usually the same individuals to obtain the info for two different moments. And the independent sample test uses two different groups in order to compare a parameter on specific, usually the mean.

Step-by-step explanation:

A paired t-test is used to compare two population means when we 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 in order to see an improvement or not for a method we can use it ( Without treatment and With specific treatment).  

An independent sample test is used when we want to compare "two sample means to determine whether the population means are significantly different". For example if we want to compare the scores for male and female in a test.  

The key difference is that the dependent sample test uses usually the same individuals to obtain the info for two different moments. And the independent sample test uses two different groups in order to compare a parameter on specific, usually the mean.

When conducting hypothesis testing, the choice between using a related sample or an independent sample depends on the nature of the data being analyzed. A related sample is used when the two samples are dependent or paired, while an independent sample is used when the two samples are not related and can be considered as separate groups.

When conducting hypothesis testing, the choice between using a related sample or an independent sample depends on the nature of the data being analyzed. A related sample is used when the two samples are dependent or paired, meaning that there is a one-to-one correspondence between the data points in the two samples. An independent sample is used when the two samples are not related and can be considered as separate groups.

For example, in a related sample scenario, you might compare the blood pressure of the same group of individuals before and after a treatment. The paired samples would be the pre-treatment and post-treatment measurements of each individual. In an independent sample scenario, you might compare the test scores of two different groups of students who were taught using different teaching methods.

Answer:

The indicated confidence interval for the difference between the two population means is  (-1.5159, 7.5159)

Step-by-step explanation:

Let the drying times of type A be the first population and the drying times of type B be the second population. Then

We have small sample sizes [tex]n_{1} = 11[/tex] and [tex]n_{2} = 9[/tex], besides [tex]\bar{x}_{1} = 71.5[/tex], [tex]s_{1} = 3.4[/tex] , [tex]\bar{x}_{2} = 68.5[/tex] and [tex]s_{2} = 3.6[/tex]. Therefore, the pooled

estimate is given by  

[tex]s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(11-1)(3.4)^{2}+(9-1)(3.6)^{2}}{11+9-2} = 12.1822[/tex]

The 99% confidence interval for the true mean difference between the mean drying time of type A and the mean drying time of type B is given by

[tex](\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{11}+\frac{1}{9}}[/tex], i.e.,

[tex](71.5-68.5)\pm t_{0.005}(3.4903)\sqrt{\frac{1}{11}+\frac{1}{9}}[/tex]

where [tex]t_{0.005}[/tex] is the 0.5th quantile of the t distribution with (11+9-2) = 18 degrees of freedom. So

[tex]3\pm(-2.8784)(3.4903)(0.4495)[/tex], i.e.,

the indicated confidence interval for the difference between the two population means is  (-1.5159, 7.5159)

Answer: Option 'a' is correct.

Step-by-step explanation:

Since we have given that

P(A) = 0.8

P(B|A) = 0.4

Since A and B are independent events.

Since P(B|A) is given, we will use the formula for "conditional probability":

So, [tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\0.4=\dfrac{P(A\cap B)}{0.8}\\\\0.4\times 0.8=P(A\cap B)}\\\\0.32=P(A\cap B)[/tex]

Hence, it is true.

Therefore, Option 'a' is correct.

Answer:

a)[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]

[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]

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

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

b)

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

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".

Assume the following dataset:

Quality management        Excellent      Good     Fair    Total

Excellent                                40                35         25       100

Good                                      25                35         10         70

Fair                                         5                   10          15        30

Total                                       70                 80         50       200

Part a

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

H0: There is independence between the two categorical variables

H1: There is association between the two categorical variables

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 = \frac{total col * total row}{grand total}[/tex]

And the calculations are given by:

[tex]E_{1} =\frac{70*100}{200}=35[/tex]

[tex]E_{2} =\frac{80*100}{200}=40[/tex]

[tex]E_{3} =\frac{50*100}{200}=25[/tex]

[tex]E_{4} =\frac{70*70}{200}=24.5[/tex]

[tex]E_{5} =\frac{80*70}{200}=28[/tex]

[tex]E_{6} =\frac{50*70}{200}=17.5[/tex]

[tex]E_{7} =\frac{70*30}{200}=10.5[/tex]

[tex]E_{8} =\frac{80*30}{200}=12[/tex]

[tex]E_{9} =\frac{50*30}{200}=7.5[/tex]

And the expected values are given by:

Quality management        Excellent      Good     Fair       Total

Excellent                                35              40          25         100

Good                                      24.5           28          17.5        85

Fair                                         10.5            12           7.5         30

Total                                       70                 80         65        215

And now we can calculate the statistic:

[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]

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

[tex]df=(rows-1)(cols-1)=(3-1)(3-1)=4[/tex]

And we can calculate the p value given by:

[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]

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

"=1-CHISQ.DIST(17.03,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.

Part b

We can find the probabilities that Quality of Management and the Reputation of the Company would be the same like this:

Let's define some notation first.

E= Quality Management excellent     Ex=Reputation of company excellent

G= Quality Management good     Gx=Reputation of company good

F= Quality Management fait     Ex=Reputation of company fair

P(EΛ Ex) =40/215=0.186

P(GΛ Gx) =35/215=0.163

P(FΛ Fx) =15/215=0.0697

If we have dependence then the conditional probabilities would be higher values.

P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714

P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5

P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357

P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375

P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2

P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714

P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125

P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3

And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.

Answer: B. (0.44,0.94)

Step-by-step explanation:

Given : Number of observations : n = 9

Number of successes  : x = 7

Let p be the population proportion of times that the bats would follow the point.

Since the sample size is small , 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]=\dfrac{7+2}{9+4}\approx0.69[/tex]

By z-table , the critical value for 95% confidence level : z* = 1.96

Then, the 95% confidence interval for the population proportion of times that the bats would follow the point. will be :

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{N}}[/tex] , where N= 13

[tex]0.69\pm (1.96)\sqrt{\dfrac{0.69(1-0.69)}{13}}[/tex]

[tex]0.69\pm (1.96)\sqrt{0.0163862084615}[/tex]

[tex]0.69\pm (1.96)(0.128008626512)[/tex]

[tex]\approx0.69\pm 0.25=(0.69-0.25,\ 0.69+0.25)[/tex]

[tex](0.44,\ 0.94)[/tex]

Hence, the  95% confidence interval for the population proportion of times that the bats would follow the point = [tex](0.44,\ 0.94)[/tex]

Thus the correct answer is B. (0.44,0.94)

To find the 95% confidence interval for the population proportion, use the formula CI = p ± z * √((p(1-p))/n), where p is the sample proportion, z is the z-score, and n is the sample size. Substituting values, the 95% confidence interval is approximately (0.685, 0.869).

To find the 95% confidence interval for the population proportion, we can use the formula:


CI = p ± z * √((p(1-p))/n)


where p is the sample proportion, z is the z-score for the desired confidence level, and n is the sample size.


In this case, the sample proportion is 7/9 and n is 9. Since we want a 95% confidence interval, the z-score is approximately 1.96.

Substituting these values into the formula:


CI = (7/9) ± 1.96 * √(((7/9)(2/9))/9)


CI = 0.777 ± 1.96 * √(0.123/9)


CI ≈ 0.777 ± 1.96 * 0.047


CI ≈ (0.777 - 0.092, 0.777 + 0.092)


CI ≈ (0.685, 0.869)

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Answer:

7. 25% of the merchants who purchase goods from Asia also purchase from Europe.

Step-by-step explanation:

I am going to say that:

A is the percentage of merchants who purchase goods from Asia.

B is the percentage of merchants who purchase goods from Europe.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that a merchant purchases goods from Asia but not from Europe and [tex]A \cap B[/tex] is the probability that a merchant purchases goods from both Asia and Europe.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

Which of following statement is individually sufficient to calculate what percent of the merchants in the group purchase goods from Europe but not form Asia?

We already have B.

Knowing [tex]A \cap B[/tex], that is, the percentage of those who purchase from both Asia and Europe, we can find b.

So the correct answer is:

7. 25% of the merchants who purchase goods from Asia also purchase from Europe.

Answer:

Step-by-step explanation:

Given

span of bridge [tex]L=1270\ ft[/tex]

height of span [tex]h=157\ ft[/tex]

Equation of Parabola

[tex]y=0.00039x^2[/tex]

[tex]|x|<635[/tex] i.e.

[tex]-635<x<635[/tex]

[tex]\frac{dy}{dx}=2\times 0.00039[/tex]

length of Arc[tex]=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}[/tex]

[tex]=\int_{-635}^{635}\sqrt{1+(\frac{dy}{dx})^2}[/tex]

[tex]=\int_{-635}^{635}\sqrt{1+(0.00078x)^2}[/tex]

[tex]=2\times \int_{0}^{635}\sqrt{1+(0.00078x)^2}[/tex]

[tex]=2\times (660.08)[/tex]

[tex]=1320.16\ ft[/tex]

The approximate length of the cables is approximately 4534.24 feet.

To approximate the length of the cables that stretch between the tops of the two towers of the suspension bridge, we can use the integral calculus to find the length of the curve defined by the equation [tex]\(y = 0.00039x^2\).[/tex]

The formula for finding the length of a curve between two points [tex]\([a, b]\)[/tex] is given by the integral:

[tex]\[L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\][/tex]

Where:

- L is the length of the curve.

- a and b are the x-coordinates of the two points between which we want to find the length.

- f(x) is the function representing the curve.

- f'(x) is the derivative of the function.

In this case, we want to find the length of the cables between the two towers, which corresponds to the x-values from -635 to 635 since the width of the bridge is 1270 feet. The curve is defined by [tex]\(y = 0.00039x^2\)[/tex], so:

- a = -635

- b = 635

- [tex]\(f(x) = 0.00039x^2\)[/tex]

- [tex]\(f'(x) = 2 \cdot 0.00039x\)[/tex]

Now, let's calculate the length:

[tex]\[L = \int_{-635}^{635} \sqrt{1 + (2 \cdot 0.00039x)^2} \, dx\][/tex]

[tex]\[L = \int_{-635}^{635} \sqrt{1 + 0.0001536x^2} \, dx\][/tex]

Now, we can evaluate this integral:

[tex]\[L = \int_{-635}^{635} \sqrt{1 + 0.0001536x^2} \, dx \approx 4534.24\][/tex]

So, the approximate length of the cables that stretch between the tops of the two towers of the suspension bridge is approximately 4534.24 feet.

Learn more about Length of suspension bridge cables here:

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When the sample size is small (< 30) and the population standard deviation is unknown , then we use t-test.

The confidence interval for population mean will be :

[tex]\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}[/tex]   (1)

, where [tex]\overline{x}[/tex] = sample mean

t* = Critical value (based on degree of freedom and significance level).

s= sample standard deviation

n= sample size.

As per given we have

n= 9

Degree of freedom = n-1 = 8

[tex]\overline{x}=2.4[/tex]

s= 0.75

Significance level =[tex]\alpha=1-0.80=0.20[/tex]

Using students' t distribution table ,

Critical value : [tex]t^*=t_{\alpha/2,df}=t_{0.10,8}=1.3304[/tex]

We assume that the population is approximately normal.

Then, a 80% confidence interval for the mean waste recycled per person per day for the population of Florida will be :

[tex]2.4\pm (1.3304)\dfrac{0.75}{\sqrt{8}}[/tex]   (Substitute the values in (1))

[tex]2.4\pm (1.3304)\dfrac{0.75}{2.82842712475}[/tex]

[tex]2.4\pm (1.3304)(0.265165042945)[/tex]

[tex]2.4\pm 0.352775573134\approx2.4\pm0.353=(2.4-0.353,\ 2.4+0.353)=(2.047,\ 2.753)[/tex]

Hence, the 80% confidence interval for the mean waste recycled per person per day for the population of Florida. = (2.047, 2.753)

Answer:

0.624 is the probability that a randomly selected college student will take between 2 and 6 minutes to find a parking spot in the main parking lot.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  5 minutes

Standard Deviation, σ = 2 minutes

We are given that the distribution of time is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P(student will take between 2 and 6 minutes )

[tex]P(2 \leq x \leq 6) = P(\displaystyle\frac{2 - 5}{2} \leq z \leq \displaystyle\frac{6-5}{2}) = P(-1.5 \leq z \leq 0.5)\\\\= P(z \leq 0.5) - P(z < -1.5)\\= 0.691 - 0.067 = 0.624 = 62.4\%[/tex]

[tex]P(2 \leq x \leq 6) = 62.4\%[/tex]

0.624 is the probability that a randomly selected college student will take between 2 and 6 minutes to find a parking spot in the main parking lot.

Answer:

Step-by-step explanation:

If you pay cash, the total amount that you will pay for the television is $349

If you pay $75 down, the balance would be paid in 18 monthly payments of 22.50 which is the installment price of the TV. Total amount paid in 18 months would be

22.5 × 18 = $405

Total cost of the TV when you pay in installments would be

405 + 75 = $480

Difference between the installment price and the cash price would be

480 - 349 = $131

The percent by which the installment price would be greater than the cash price is

131/349 × 100 = 37.5%

The estimated cost to feed a family for a week with a household income of $12,000 would be $9,341.33.

To find out how much money would be needed to feed a family for a week whose household income is $12,000, we need to use the regression equation provided. The equation is Food/wk = 101.33 + 0.77HIncome. We substitute the value of HIncome with $12,000 and solve for Food/wk.

Food/wk = 101.33 + 0.77(12,000)

Food/wk = 101.33 + 9240

Food/wk = $9,341.33

Therefore, the estimated cost to feed a family for a week with a household income of $12,000 would be $9,341.33.

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The probability of a motorbike rider in this city experiencing carbon monoxide exposure exceeding 20 ppm is 0.3932, while the probability of exceeding 25 ppm is 0.1316, based on the normal distribution with a mean of 18.6 ppm and a standard deviation of 5.9 ppm.

Define the random variable X as the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in this city.

Since the distribution of X is normally distributed with a mean of 18.6 ppm and a standard deviation of 5.9 ppm, we can use the standard normal distribution to calculate probabilities.

To find the probability of X being greater than 20 ppm, we need to find the area to the right of 20 ppm under the standard normal curve.

We can calculate this area using a z-score, which is defined as the number of standard deviations a specific point is away from the mean. In this case, the z-score for 20 ppm is (20 ppm - 18.6 ppm) / 5.9 ppm = 0.271.

Using a standard normal table or calculator, we can find that the area to the right of 0.271 is 0.3932.

Therefore, the probability of someone riding a motorbike for 5 km on a highway in this city experiencing a carbon monoxide exposure of more than 20 ppm is 0.3932.

Follow the same steps as in part a, but use a z-score of (25 ppm - 18.6 ppm) / 5.9 ppm = 1.119.

The area to the right of 1.119 under the standard normal curve is 0.1316.

Therefore, the probability of someone riding a motorbike for 5 km on a highway in this city experiencing a carbon monoxide exposure of more than 25 ppm is 0.1316.

Complete question:

According to a research paper, the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in a particular city is approximately normally distributed with a mean of 18.6 ppm. Suppose that the standard deviation of carbon monoxide exposure is 5.9 ppm (a) Approximately what proportion of those who ride a motorbike for 5 km on a highway in this city will experience a carbon monexide exposure of more than 20 ppm? (Round your answer to fou decimal places.) (b) Approximately what proportion of those who ride a motorbike for 5 km on a highway in this city will experience a carbon monoxide exposure of more than 25 ppm?

Answer:

[tex]P(X>140)=1-P(X\leq 140)=1-P(\frac{X-\mu}{\sigma}<\frac{140-\mu}{\sigma})=1-P(Z<\frac{140-100}{19})=1-P(Z<2.105)=0.0176[/tex]

Number =0.0176*280000000=4928000 approximately Americans would have a IQ score more than higher than 140

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 IQ scores of the population of interest, and for this case we know the distribution for X is given by:

[tex]X \sim N(100,19)[/tex]  

Where [tex]\mu=100[/tex] and [tex]\sigma=19[/tex]

We are interested on this probability

[tex]P(X>140)[/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]

If we apply this formula and the complement rule to our probability we got this:

[tex]P(X>140)=1-P(X\leq 140)=1-P(\frac{X-\mu}{\sigma}<\frac{140-\mu}{\sigma})=1-P(Z<\frac{140-100}{19})=1-P(Z<2.105)=0.0176[/tex]

And we can find this probability with the following excel code:

"=1-NORM.DIST(2.105,0,1,TRUE)"

This number 0.0176 represent the proportion of Americans that present a score higher than 140.

And now since we ar einterested on the approximate number of people in the United States (assuming a total population of 280,000,000) with an IQ higher than 140, we just need to do this:

Number =0.0176*280000000=4928000 approximately Americans would have a IQ score more than higher than 140

This equation is separable, as

[tex]\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx[/tex]

Integrate both sides; on the left, expand the fraction as

[tex]\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)[/tex]

Then

[tex]\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C[/tex]

[tex]\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C[/tex]

Since [tex]y(0)=1[/tex], we get

[tex]\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1[/tex]

so that the particular solution is

[tex]\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}[/tex]

Final answer:

The 95% confidence interval for the PCB level in the fish is (11.4228, 11.5372) ppm. The 95% lower confidence interval is 11.4228 ppm and the 95% upper confidence interval is 11.5372 ppm, based on both the t-distribution and the provided sample data and standard deviation.

Explanation:

To find the 95% confidence interval, lower confidence interval, and upper confidence interval of the PCB level in a fish from Lake Michigan, based on 10 measurements and a standard deviation of 0.08 ppm, we first need to calculate the sample mean and then apply the appropriate formulas.

To calculate the mean PCB concentration (μ), we sum all the values and divide by the number of measurements (n=10):
μ = (11.2 + 12.4 + 10.8 + 11.6 + 12.5 + 10.1 + 11.0 + 12.2 + 12.4 + 10.6) / 10 = 114.8 / 10 = 11.48 ppm.

For calculating the confidence intervals, we use the t-distribution since the sample size is small. We need the t-value for 9 degrees of freedom (n-1) at the 95% confidence level which, assuming it is approximately 2.262 (values differ slightly depending on the t-distribution table used).

The standard error (SE) is calculated using the sample standard deviation (s) and the square root of the number of measurements: SE = s/sqrt(n) = 0.08/sqrt(10) = 0.0253 ppm.

The 95% confidence interval is given by:
CI = μ ± (t-value * SE)
CI = 11.48 ± (2.262 * 0.0253)
CI = 11.48 ± 0.0572
CI = (11.4228, 11.5372) ppm

The 95% lower confidence interval is the mean minus the product of the t-value and SE:
LCI = μ - (t-value * SE)
LCI = 11.48 - (2.262 * 0.0253)
LCI = 11.4228 ppm

The 95% upper confidence interval is the mean plus the product of the t-value and SE:
UCI = μ + (t-value * SE)
UCI = 11.48 + (2.262 * 0.0253)
UCI = 11.5372 ppm