The number of rescue calls received by a rescue squad in a city follows a Poisson distribution with an average of 2.83 rescues every eight hours. What is the probability that the squad will have at most 2 calls in an hour? Round answer to 4 decimal places.

Answer:

A.

Step-by-step explanation:

Hello!

First a little reminder:

The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

In the example, the results of a drug test were reported, being the null hypothesis "the drug has no effect" and the alternative hypothesis "the drug decreases vision loss in people with Macular Degeneration"

After conducting the test, the researchers obtained the p-value 0.004

Taking the previous definition of the p-value, the correct answer is A.

To tell whether B. and C. are correct or incorrect there should be specified what signification level was used in the analysis. Remember, the p-value is the probability of the statistic value under the null hypothesis and to use it to make a decision ver the null hypothesis you have to compare it with the signification level. If the p-value is greater than the level of significance, then you don't reject the null hypothesis (Then you can conclude that the drug has no effect) If the p-value is equal or less than the level of significance, then you reject the null hypothesis. (Then you can conclude that the drug decreases the vision loss)

Using a level of signification of 0.01 then the decision is to not reject the null hypothesis but with levels of 0.05 or 0.1 then the decision is to reject it. This is why it is important to know what level of significance was used in the test when interpreting the p-value.

I hope it helps!

The final answer is approximately X≈43.76.

To find the 15th percentile of a normal distribution, you can use the Z-score formula and then use the standard normal distribution table (Z-table) or a calculator to find the corresponding Z-score.

The formula to convert a value from a normal distribution to a standard normal distribution (Z-score) is:

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

Where:

X is the value in the original distribution.

μ is the mean of the original distribution.

σ is the standard deviation of the original distribution.

Z is the Z-score.

Given:

μ=50

σ=6

We want to find the 15th percentile, which corresponds to the value of

X for which 15% of the data falls below it.

First, we find the Z-score corresponding to the 15th percentile using the standard normal distribution table:

[tex]$Z_{15}=$ Z-score for 15 th percentile[/tex]

Then, we rearrange the formula to solve for X:

[tex]$X=\mu+Z_{15} \times \sigma$[/tex]

Let's calculate:

Find the Z-score for the 15th percentile using the Z-table. The closest value is approximately -1.04.

Substitute the values into the formula to find X:

[tex]$\begin{aligned} & X=50+(-1.04) \times 6 \\ & X=50-6.24 \\ & X \approx 43.76\end{aligned}$[/tex]

So, the 15th percentile of the normal distribution with mean μ=50 and standard deviation σ=6 is approximately 43.76.

Answer:

There is a probability of 90.49% of needing more than 80 rolls to reach 300.

Step-by-step explanation:

The Central Limit Theorem tells us that the sampling distribution, as the size of the sample gets larger, approaches the normal distribution. This is independent of the population distribution of the random variable.

Then we can use the normal distribution parameters to model this problem.

Let Y be the sum of 80 dices.

[tex]Y=\sum_{i=1}^{80}X_i\\\\E(Y)=80*E(X_i)=80*[(1/6)*(1+2+3+4+5+6)]=80*3.5=280\\\\V(Y)=80V(X_i)\\\\V(Y)=80*(1/6)[(1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2]\\\\V(Y)=80*(1/6)*17.5=233.33[/tex]

[tex]\sigma=\sqrt{V(Y)}=\sqrt{233.33}=15.28[/tex]

We can use the z-value to calculate the probability of getting 300 or more in 80 rolls.

[tex]z=(X-\mu)/\sigma=(300-280)/15.28=1.31\\\\\\P(X>300)=P(z>1.31)=0.0951\\\\P(X<300)=1-0.0951=0.9049[/tex]

There is a probabilitity P=0.9049 that the 80 rolls will not reach a 300 score. Thus we can conclude that there is a probability of 90.49% of needing more than 80 rolls to reach 300.

Answer:

0.9430

Step-by-step explanation:

The other answer is almost right but answers the wrong question. The question asks for the probability that at least 80 rolls are necessary not more than 80. So we can use the same process but use where we are using the sum of 79 die. The other answer also misses continuity correction. We must use continuity correction since we're approximating a discrete variable with a continuous one through CLT so P(X <= 300) -> P(X < 300.5). With only these changes, we get 0.943.

The fireworks display is 7621.17 feet  north of the person at point A

As we know 1 mile = 5280 feet

So, 6 miles = 6 × 5280

= 31,680 feet.

Calculate the time it takes for sound to travel from the fireworks display to each person.

Person at point A hears the burst after 4 seconds.

Person at point B hears the burst after 4 + 4 = 8 seconds.

Now, let's find the distance each person is from the fireworks display.

Distance = Speed × Time

For person A:

Distance of A = Speed of sound × Time_A

Distance of A = 1100 feet/second × 4 seconds

Distance of A = 4400 feet

For person B:

Distance of B = Speed of sound × Time_B

Distance of B = 1100 feet/second × 8 seconds

Distance of B = 8800 feet

Use the Pythagorean theorem to find the distance north of person A where the fireworks display occurred.

Let the distance north be x feet.

According to the problem, we have a right-angled triangle formed by points A, B, and the fireworks display location.

The distance between A and B (6 miles or 31,680 feet) is the hypotenuse of the triangle.

Using the Pythagorean theorem:

Distance of A² + x² = Distance of B²

(4400)² + x² = (8800)²

19360000 + x² = 77440000

x² = 77440000 - 19360000

x² = 58080000

x = √58080000

x = 7621.17 feet

So, the fireworks display occurred approximately 7621.17 feet north of the person at point A.

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The fireworks display took place approximately 30,762 feet north of the person at point A. This was calculated by using the information about the extra distance the sound traveled to reach Person B and applying the Pythagorean theorem.

It seems like this question is about the speed of sound and distances of objects based on perception. We are given that the sound from the fireworks travels at 1100 feet per second and the time difference in hearing it between two people is four seconds. Therefore, the sound traveled an extra 4400 feet (4 seconds * 1100 feet/second) to reach the second person.

Let's consider the triangle formed by Person A, Person B, and the fireworks. Since Person A and Person B are 6 miles apart (which is 31680 feet), and we know the extra distance the sound traveled to reach Person B, we can calculate how far the fireworks occurred north of Person A (using Pythagorean theorem). Thus, the distance of the fireworks display north of person A is: sqrt((31680 feet)^2 - (4400 feet)^2), which equals approximately 30,762 feet.

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The test statistic for a chi-square test with a sample size of 14, sample standard deviation of 20, and null hypothesis variance of 500 should be 10.40. However, since this number does not match any of the options provided, there may be an error in the question or the options.

The subject in question involves performing a hypothesis test to determine whether the variance of a population is greater than or equal to a specified value. Given a sample size (n = 14), sample standard deviation (s = 20), and the null hypothesis being (σ² ≤ 500), we need to calculate the test statistic for a chi-square distribution with degrees of freedom df = n - 1, which in this case is df = 13. The test statistic is calculated using the formula:

chi-square test statistic = χ² = (n - 1)s² / σ²

Plugging in the values, we get:

chi-square test statistic = (14 - 1) * 20² / 500 = 13 * 400 / 500 = 5200 / 500 = 10.40

However, since this value does not match any of the answer options provided in the initial question, it appears that there may be an error in the question itself or in the provided options. It's important to carefully review the calculation and ensure the values and hypotheses given are as intended.

Answer:

[tex]P(\bar X_C -\bar X_B > 7)=P(Z> \frac{(15-9)-7}{0.5})=P(Z>-2) =1-P(Z<-2) = 1-0.02275= 0.97725[/tex]

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

Solution to the problem

From the central limit theorem since the sample size for both cases are >30 we can assume that the average follows a normal distribution.

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

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

And let's put some notation

B= represent the bulldogs, C= Chihuahuas

[tex]\bar X_B \sim N(\mu_B =9, \sigma_{\bar x_B}=\frac{3}{\sqrt{100}}=0.3)[/tex]

[tex]\bar X_C \sim N(\mu_C =15, \sigma_{\bar x_C}=\frac{4}{\sqrt{100}}=0.4)[/tex]

And the distribution for the difference of averages would be given by:

[tex]\bar X_C -\bar X_B \sim N(\mu_D = 15-9=6, \sigma_D=\sqrt{\frac{3^2}{100}+\frac{4^2}{100}}=0.5)[/tex]

And for this case we want this probability:

[tex]P(\bar X_C -\bar X_B > 7)[/tex]

And for this can use the z score given by:

[tex] z=\frac{\bar X_D - \mu_D}{\sigma_D}[/tex]

And after replace we got:

[tex]P(Z> \frac{(15-9)-7}{0.5})=P(Z>-2)[/tex]

And we can use the complement rule and we got this:

[tex]P(Z>-2) =1-P(Z<-2) = 1-0.02275= 0.97725[/tex]

Final answer:

To approximate the probability that the average Chihuahua lives at least 7 years longer than the average Bulldog in a kennel of 100 Bulldogs and 100 Chihuahuas, calculate the difference in means and the standard deviation of the difference. Then, calculate the z-score to find the probability using a z-table or calculator. The approximate probability is less than 0.001.

Explanation:

To approximate the probability that the average Chihuahua lives at least 7 years longer than the average Bulldog in a kennel of 100 Bulldogs and 100 Chihuahuas, we need to compare the means of the two populations and determine the difference in years. The mean for Bulldogs is 9 years and the mean for Chihuahuas is 15 years, so the difference in means is 15 - 9 = 6 years.

Now, we need to find the standard deviation of the difference in means. Since we're dealing with independent samples, we can use the formula:

Standard deviation of the difference in means = sqrt((standard deviation of Bulldogs^2) / 100 + (standard deviation of Chihuahuas^2) / 100) = sqrt((3^2) / 100 + (4^2) / 100) = sqrt(0.09 + 0.16) = sqrt(0.25) = 0.5.

Next, we calculate the z-score using the formula:

z-score = (difference in means - 7) / standard deviation of the difference in means = (6 - 7) / 0.5 = -2 / 0.5 = -4.

Finally, we can find the probability using a z-table or calculator. The probability of the average Chihuahua living at least 7 years longer than the average Bulldog is extremely small, as the z-score of -4 corresponds to a very low probability value. Therefore, the approximate probability is less than 0.001.

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:

The air in the room at 0.01% carbon monoxide at 43.8 min

Step-by-step explanation:

Let be the volume of CO in the room at time t, be v(t)  and  the total volume of the room be V. The volume percent of CO in the room at a given time is then given by:

[tex]p(t) = \frac{100\times v(t)}{V}[/tex]

Volume percent is the measure of concentration used in this problem. The "Amount" of CO in the room is then measured in terms of the volume of CO in the room.

Let the rate at which fresh air enters the room be f, which is the same as the rate at which air exits the room. We assume that the air in the room mixes instantaneously with the air entering the room, so that the concentration of CO is uniform throughout the room.

As you wrote, the rate at which the volume of CO in the room changes with time is given by

[tex]\frac{dvt}{dt} = 0 \times f -\frac{f}{v} \times v(t) = -\frac{f}{v} \times v(t)[/tex]

This is a simple first-order equation:

[tex]\frac{dv}{v} = -\frac{f}{v} dt[/tex]

[tex]ln(v) - ln(c) = -\frac{f}{v} \times t[/tex]

where ln(c) is the constant of integration.

ln [tex]\frac{v}{c} = -\frac{f}{v} \times t[/tex]

[tex]v(t) = c \times e^{(-f*\frac{t}{V})}[/tex]

In terms of volume percent,

[tex]p(t) = \frac{100*v(t)}{V}= (\frac{C}{V})*exp(\frac{-f \times t}{v})[/tex]

where C = 100*\frac{c}{V} is just another way of writing the constant.

Plugging in the values for the constants, we get:

[tex]p(t) = (\frac{C}{768 cu.ft.})* exp(\frac{-t}{7.68 min})[/tex]

Now use the initial condition (p(0) = 3% at t = 0) to solve for C:

3% = C

[tex]p(t) = (3\%)\times exp(\frac{t}{7.68 min})[/tex]

To find the time when the air in the room reaches a certain value, it is easier to rewrite this solution as:

[tex]\frac{p(t)}{3\%} = exp(\frac{-t}{7.68 min})[/tex]

[tex]t(p) = -(7.68 min)ln(\frac{p}{3\%} )[/tex]

[tex]= (7.68 min)*ln(\frac{3\%}{p})[/tex]

The question asks when p(t) = 0.01%. Plugging this into the above equation, we get:

[tex]t(0.01\%) = (7.68 min)*ln(\frac{3}{0.01}) = 43.8 min[/tex]

Answer:

Part 1

a)[tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{100}}=0.0956[/tex]  

b) [tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{363}}=0.0502[/tex]

c) [tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{1551}}=0.0243[/tex]

Part 2

We can see that if we increase the sample size the margin of error decrease and that makes sense since n is on the denominator in the formula for the margin of error and if we increase the denominator the result needs to decrease.

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 means 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.  

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 population proportion have the following distribution

[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]

Part 1

For this case in order to find the margin of error we need to assume a confidence level fixed, let's assume 95% for example. The Margin of error is given by this formula:

[tex]ME= z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}[/tex]

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]    (a)  

And on this case we have that [tex]\hat p = 0.61[/tex] and we are interested in order to find the value of ME.

So we can replace for each case and see what we got:

a)[tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{100}}=0.0956[/tex]  

b) [tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{363}}=0.0502[/tex]

c) [tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{1551}}=0.0243[/tex]

Part 2

We can see that if we increase the sample size the margin of error decrease and that makes sense since n is on the denominator in the formula for the margin of error and if we increase the denominator the result needs to decrease.

Final answer:

The margin of error decreases as the sample size increases, demonstrating the importance of a larger sample size for more accurate polling results.

Explanation:

A 2002 poll reported that 61% of people worried that they would be exposed to SARS. To find the approximate margin of error for different sample sizes (n), we use the formula for the margin of error at the 95% confidence level, which conveniently simplifies to roughly 1 over the square root of the sample size (n) for large sample sizes. However, a more precise calculation involves the formula ME = Z * √(p(1-p)/n), where Z is the Z-score corresponding to the desired confidence level, p is the proportion, and n is the sample size.

As n increases, the margin of error decreases, illustrating the inverse relationship between sample size and the margin of error. This is because with larger sample sizes, we get a more accurate representation of the population, thus increasing the reliability of our poll's results.

Answer: fail to reject the seven-minute average waiting time claim.

Step-by-step explanation:

As per given ,

Objective for test : the average waiting time is seven minutes or more.

Then ,

[tex]H_0: \mu=7\\\\H_a: \mu>7[/tex]

Since alternative hypothesis is right-tailed thus the test is an right-tailed test.

In a right tailed test , the rejection area lies on the right side of the critical value.

It means that if the observed z-value is greater than the critical value then it will fall into the rejection region other wise not.

i.e. If the value of your test statistic is less than the critical value, the correct decision is we fail to reject null hypothesis.

i.e. fail to reject the seven-minute average waiting time claim.

The complete statement would become:

If the value of your test statistic is less than the critical value, the correct decision is to fail to reject the seven-minute average waiting time claim.

The correct decision is to fail to reject the seven-minute average waiting time claim.

The correct decision is to fail to reject the seven-minute average waiting time claim.

To determine whether the average waiting time exceeds seven minutes, we conduct a hypothesis test using the sample mean and standard deviation. We can use a t-test to compare the sample mean of 8.15 minutes to the claimed average of seven minutes.

If the value of the test statistic is less than the critical value for the chosen level of significance (such as 0.05), we fail to reject the null hypothesis and conclude that there is not enough evidence to support that the average waiting time exceeds seven minutes.

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Answer:

[tex]P(T_A < T_B) = P(T_A -T_B<0)=P(Z<0.608) =0.728[/tex]

Step-by-step explanation:

Assuming this problem: "Ann has 30 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. Bob has 30 jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. Ann's and Bob's times are independent. Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs".

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

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".

Solution to the problem

For this case we can create some notation.

Let A the values for Ann we know that n1 = 30 jobs solved in sequence and we can assume that the random variable [tex]X_i[/tex] the time in order to do the ith job for [tex]i=1,2,....,n_1[/tex]. We will have the following parameters for A.

[tex]\mu_A = 50, \sigma_A =10[/tex]

W can assume that B represent Bob we know that n2 = 30 jobs solved in sequence and we can assume that the random variable [tex[X_i[/tex] the time in order to do the ith job for [tex]i=1,2,....,n_2[/tex]. We will have the following parameters for A

[tex]\mu_B = 52, \sigma_B =15[/tex]

And we can find the distribution for the total, if we remember the definition of mean we have:

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

And [tex]T =n \bar X[/tex]

And the [tex]E(T) = n \mu[/tex]

[tex]Var(T) = n^2 \frac{\sigma^2}{n}=n\sigma^2 [/tex]

So then we have:

[tex]E(T_A)=30*50 =1500 , Var(T_A) = 30*10^2 =3000[/tex]

[tex]E(T_B)=30*52 =1560 , Var(T_B) = 30 *15^2 =6750[/tex]

Since we want this probability "Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs" we can express like this:

[tex]P(T_A < T_B) = P(T_A -T_B<0)[/tex]

Since we have independence (condition given by the problem) we can find the parameters for the random variable [tex]T_A -T_B [/tex]

[tex]E[T_A -T_B] = E(T_A) -E(T_B)=1500-1560=-60[/tex]

[tex]Var[T_A -T_B]= Var(T_A)+Var(T_B) =3000+6750=9750[/tex]

And now we can find the probability like this:

[tex]P(T_A < T_B) = P(T_A -T_B<0)[/tex]

[tex]P(\frac{(T_A -T_B)-(-60)}{\sqrt{9750}}< \frac{60}{\sqrt{9750}})[/tex]

[tex]P(Z<0.608) =0.728[/tex]

Answer:

[tex]\frac{37}{389}[/tex]

Step-by-step explanation:

Given that a  medical study investigated the effect of calcium and vitamin supplements on the risk of older Americans for broken bones. A total of 389 older Americans who lived at home and were in good health were studied over a three-year period. While all of the 389 people took in at least 700 milligrams of calcium and 200 units of vitamin D through their normal diet, 187 of them were given additional supplements containing 500 milligrams of calcium citrate and 70 units of vitamin D daily. Of the 187 who took additional supplements, 11 of them suffered broken bones over the three-year period. Of the 202 older Americans who did not take the additional supplement, 26 of them suffered broken bones over the study period.

                     Group I           Group II           Total

n                      187                  202                389

favour x             11                    26                  37

The fraction of older Americans who were included in the study suffered broken bones during the three-year period

=Total x/total n

= [tex]\frac{37}{389}[/tex]

The research firm should use the following procedure to analyze the data for the mean difference in salary within married couples:

a) One-sample t procedure, matched pair

It is because, The "matched pair" aspect indicates that each husband's salary is paired with his respective wife's salary. This pairing is essential because the focus is on comparing the salaries within each couple.

The one-sample t procedure is appropriate in this scenario because it compares the mean salary difference within each couple to determine if there is a significant difference between what husbands and wives earn.

This procedure is suitable when the same sample is measured twice (in this case, the salaries of husbands and wives in each couple) and the goal is to compare the means of the differences within the pairs.

By using the one-sample t procedure with matched pairs, the research firm can effectively analyze the data and draw conclusions regarding the mean salary difference within married couples.

The complete question: A research firm wants to determine whether there is a difference in married couples between what the husband earns and what the wife earns. The firm takesa random sample of married couples and measures the annual salary of each husband and wife. What procedure should the firm use to analyze the data for the mean difference in salary within married couples?

a)One-sample t procedure, matched pair

b)Two-sample t procedure

c)One-sample z procedure, matched pair

d)Two-sample z procedure

e)Not enough information to

The difference of proportions of high-school and college students reading newspapers regularly can be assessed using a 95% confidence interval, and the significance of this difference can be verified using a z-test considering appropriately formulated null and alternative hypotheses. The questions involve application of fundamental concepts of statistics and hypothesis testing.

Given the data, the proportions of high-school students (p₁) and college students (p₂) who read newspapers regularly are 287/500 (0.574) and 252/420 (0.6) respectively.

A: The 95% confidence interval for the difference between the proportions is calculated by [p₁-p₂ ± Z*√((p₁*(1-p₁)/n₁) + (p₂*(1-p₂)/n₂))], where Z is the z-value (1.96 for 95% confidence), n₁ and n₂ are the sample sizes. Plug in the given numbers to get the interval.

B: Based on the 95% confidence interval, we can judge whether 0 is in this interval. If so, we can't conclude that there's a significant difference between the two proportions. If not, there is a significant difference.

C: The null hypothesis (H₀) is p₁ - p₂ = 0, indicating no difference. And the alternative hypothesis (Hₐ) is p₁ - p₂ < 0, suggesting there's a significant difference and p₁ is less than p₂.

D: To find p ˆ, the pooled estimate is (x₁+x₂) / (n₁+n₂), where x₁ and x₂ are the counts of successes (those who read newspapers) in each group. Plug in the numbers to do the math.

E: Preconditions for a z-procedure: 1. The samples are random. 2. Both sample sizes are sufficiently large (n>30) to apply the Central Limit Theorem. 3. The events are independent.

F: SEp ˆ, the pooled estimate of the standard errors, is √(p ˆ*(1-p ˆ)*(1/n₁+1/n₂))

G: The test statistic is (p₁-p₂) / SEp ˆ, and this z-value can be used to find the P-value in a standard normal distribution table. If the P-value is small, we reject the null hypothesis in favor of the alternative.

H: If the P-value<0.05, then the conclusion would be that the proportion of high school students who read the newspaper regularly is significantly less than the proportion of college students.

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The confidence interval and hypothesis testing don't provide enough evidence to suggest a significant difference in the proportions of high-school and college students who read newspapers regularly. The requirements for using a z-interval and z-procedure were satisfied, suggesting that the statistical methods used were appropriate. The test statistic and P-value confirm that the null hypothesis cannot be rejected.

This question pertains to comparing proportions using hypothesis testing and confidence interval estimation. Before we perform these statistical analyses, we need to ensure that the conditions for each are satisfied.

A. To construct a 95% confidence interval, and to confirm that the conditions for a z-interval are satisfied, we consider the following:

Calculations work out to a confidence interval of ±3.83%. Therefore, according to our analysis, we are 95% confident that the true difference in proportions of high-school students and college students who read newspapers regularly lies in this range.

B. Given our confidence interval, we don't have enough evidence to suggest a significant difference between the two proportions of high-school and college students who read newspapers regularly.

C. The null hypothesis (H0) would be: p1=p2, indicating no difference between the proportions. The alternative hypothesis (Ha), would be: p1

D. To calculate pooled estimate of the population proportions (denoted as p ˆ), the formulas and calculations result in p ˆ = 0.574.

E. Sample size for both samples are larger than 30, implying that we can safely use a z-procedure for the hypothesis test. Also, the samples are independent and randomly selected, satisfying the necessary conditions.

F. The pooled estimate of the standard errors (SEp ˆ), works out to 0.028.

G. The calculated test statistic (z) is 0.83 and the P-value is 0.20.

H. Given a significance level, α = .05, the P-value > α. Therefore, we fail to reject the null hypothesis. This means we do not have enough evidence to suggest that the proportion of high-school students who read newspapers regularly is less than the proportion of college students who do.

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Answer:

[tex]z=\frac{0.540-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53[/tex]  

[tex]p_v =P(Z>2.53)=0.0057[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the true proportion is not significantly higher than 0.5.  

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=540 represent the people indicated that they would be willing to give up some personal time in order to make more money

[tex]\hat p=\frac{540}{1000}=0.54[/tex] estimated proportion of people indicated that they would be willing to give up some personal time in order to make more money

[tex]p_o=0.5[/tex] is the value that we want to test

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

Confidence=99% or 0.99

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 majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money :  

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/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].

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

[tex]np_o =100*0.5=500>10[/tex]

[tex]n(1-p_o)=1000*(1-0.5)=500>10[/tex]

3) Calculate the statistic  

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

[tex]z=\frac{0.540-0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53[/tex]  

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 significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

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

[tex]p_v =P(Z>2.53)=0.0057[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the true proportion is not significantly higher than 0.5.  

The sample data provides convincing evidence that the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money.

To test whether the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money, we need to conduct a hypothesis test. The null hypothesis is that the proportion of women willing to give up personal time is 0.50, and the alternative hypothesis is that it is greater than 0.50. We can use a one-sample proportion test.

In this case, the sample proportion is 540/1000 = 0.54. The sample size is large enough for conducting a test as the expected count for both categories is greater than 10. The test statistic can be calculated as (sample proportion - hypothesized proportion) / sqrt((hypothesized proportion * (1 - hypothesized proportion)) / sample size). The Z-score can then be compared to the critical Z-value at a significance level of 0.01 to determine if we reject or fail to reject the null hypothesis. The p-value can also be calculated using the Z-score.

In this specific case, the test statistic is (0.54 - 0.50) / sqrt((0.50 * (1 - 0.50)) / 1000) = 7.21. The corresponding p-value is less than 0.0001, which is smaller than the significance level of 0.01. Therefore, we reject the null hypothesis and conclude that the majority of women age 22 to 35 who work full-time would be willing to give up some personal time for more money.

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Answer:

[tex]\S^2_p =\frac{(18-1)(2.5)^2 +(18 -1)(2.3)^2}{18 +18 -2}=5.77[/tex]

[tex]S_p=2.402[/tex]

[tex]t=\frac{(12 -13)-(0)}{2.402\sqrt{\frac{1}{18}+\frac{1}{18}}}=-1.249[/tex]

[tex]p_v =2*P(t_{34}<-1.249) =0.2201[/tex]

Step-by-step explanation:

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

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

And the statistic is given by this formula:

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

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

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

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

The system of hypothesis on this case are:

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

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

Or equivalently:

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

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

Our notation on this case :

[tex]n_1 =18[/tex] represent the sample size for group 1

[tex]n_2 =18[/tex] represent the sample size for group 2

[tex]\bar X_1 =12[/tex] represent the sample mean for the group 1

[tex]\bar X_2 =13[/tex] represent the sample mean for the group 2

[tex]s_1=2.5[/tex] represent the sample standard deviation for group 1

[tex]s_2=2.3[/tex] represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

[tex]\S^2_p =\frac{(18-1)(2.5)^2 +(18 -1)(2.3)^2}{18 +18 -2}=5.77[/tex]

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

[tex]S_p=2.402[/tex]

And now we can calculate the statistic:

[tex]t=\frac{(12 -13)-(0)}{2.402\sqrt{\frac{1}{18}+\frac{1}{18}}}=-1.249[/tex]

Now we can calculate the degrees of freedom given by:

[tex]df=18+18-2=34[/tex]

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

[tex]p_v =2*P(t_{34}<-1.249) =0.2201[/tex]

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

To find the test statistic for comparing the sales of two varieties of truffles, we use the formula for a two-sample t-test with assumed equal population variances. The test statistic is calculated using the sample means, sample standard deviations, and sample sizes for both types of truffles.

The student is asking how to find the value of the test statistic when comparing the means of two independent samples with unknown but equal population variances. Using the provided sample means and standard deviations for the two varieties of milk chocolate truffles—one plain and one filled with mint—we can apply the formula for the test statistic in a two-sample t-test where population variances are assumed to be equal.

The formula is given by:
t = (X₁ - X₂) / S_p * sqrt(1/n₁ + 1/n₂)
where:

First, calculate the pooled standard deviation (S_p) using the formula:
S_p = sqrt(((n₁-1) * S₁² + (n₂-1) * S₂²) / (n₁ + n₂ - 2))
Then, plug the values into the above formula to find the t-statistic.

Answer:

1a) Plot on the graph.

1b) Its proportional if y (weight) varies directly with variation in x(age) and fits the form y=k*x

Step-by-step explanation:

1a)

draw graph with x axis in increments of 1 month

draw y-axis in increments of 1 pound.

put 1 dot at (1, 4)

put 1 dot at (2,7)

connect the 2 dots, and continue the line to hit one of the axes.

It will not hit the origin.

1b) show that y=kx

4=k*1 , so k=4

7=k*2 , so k=3.5

Since we have 2 different values for k (4 & 3.5), it's not proportional.

The random variable X represents the sum of numbers on the balls chosen. The probability distribution for X is X=3 with 1/3 probability, X=4 with 1/3 probability and X=5 with 1/3 probability. The probability that the sum is at least 4 is 2/3. The expected value or mean of X is 4.

In this problem, the random variable X represents the sum of the numbers on the two balls we pick from the box. The possible sum values, excluding the order in which we pick the balls, could be (1+2)=3, (2+3)=5, or (1+3)=4.

(a) Thus, the probability distribution for X would be:
X = 3 with probability 1/3 (occurs when we pick balls 1,2)
X = 4 with probability 1/3 (occurs when we pick balls 1,3)
X = 5 with probability 1/3 (occurs when we pick balls 2,3)

(b) A sum of at least 4 can occur either when X=4 or X=5. Because each of these two outcomes is equally likely with a probability of 1/3, we sum these probabilities to find that the probability that the sum is at least 4 is 2/3.

(c) The mean of X is the expected value, which is calculated by multiplying each outcome by its probability and summing these products. In this case, that would be (3*(1/3)) + (4*(1/3)) + (5*(1/3)) = 4

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Answer: Standard error of the average annual salary SE = $597.6

Step-by-step explanation:

Given;

Standard deviation = r = $5,000

Number of samples = n = 70

Mean = $50,000

To derive the standard error of mean SE. It is given as

SE = r/√n

SE = $5,000/√70

SE = $5,000/8.3666

SE = $597.6

Answer:

D) the decrease in the prey population due to interactions with the predator

Step-by-step explanation:

We have that:

x is the predator population

y is the prey population.

The coefficient c appears in the equation of dy/dt. So this coefficient is related to the population of the prey. It appears with a minus sign, this means that the prey population decreases. It also multiplies x, so it means that it is related to the predator population.

The correct answer is:

D) the decrease in the prey population due to interactions with the predator

Answer:

Step-by-step explanation:

Given that a university found that 20% of its students withdraw without completing the introductory statistics course.

Each student is independent of the other and there are only two outcomes

X no of students in the registered 20 is binomial with p = 0.2

a)the probability that two or fewer will withdraw.

=[tex]P(X\leq 2)\\=0.2061[/tex]

b. Compute the probability that exactly four will withdraw.

=[tex]P(X=4) = 0.2182[/tex]

c. Compute the probability that more than three will withdraw.

[tex]=P(X>3)\\\\=1-F(2)\\= 1-0.4115\\=0.5885[/tex]

d. Compute the expected number of withdrawals.

E(x) = np = 4

Answer:

u.v = $12,731.

Step-by-step explanation:

The dot product between two vectors, a and b, in which

a = (c,d,e)

b = (f,g,h)

Is given by the following formula

a.b = (c,d,e).(f,g,h) = cf + dg + eh.

In this problem, we have that:

u = (3224, 1429, 2275)

v = (1.50, 2.50, 1.90)

So

u.v = (3224, 1429, 2275).(1.50, 2.50, 1.90) = 3224*1.50 + 1429*2.50 + 2275*1.90 = $12,731.

Pseudocode is below

Step-by-step explanation:

random_number = genRandomInt[1, 100]

get_input = input(“Select a number between 1 and 100: ")  

while get_input<100

if get_input >random_number:

print(“the number you selected is high”)

else if get_input < random_number

print(“the number you selected is low”)

else:

print(“correct number!”)

end

The pseudocode provides a structured process for the task of guessing a number between 1 and 100. It directs the user through guessing, receiving feedback and adjustment for subsequent guesses, until correctly guessing the number.

The process of guessing a number can be represented in a structured flowchart or pseudocode as follows:

Step 1: Start

Step 2: Define a random number between 1 and 100

Step 3: The player enters a guess

Step 4: Check if the guess is equal to, greater than or less than the random number. If equal, go to Step 6. If the guess is greater, then output 'Guess is too high.'  If the guess is lower, output 'Guess is too low.'

Step 5: The player enters a new guess, then return to Step 4.

Step 6: Output 'Congratulations! Correct guess.'

Step 7: End

This pseudocode explains a structured process of guessing a number. The user is given feedback after each guess, allowing them to make an informed guess for the next round. This sequence continues until the player guesses the correct number, finally resulting in a congratulatory message and termination of the process.

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Answer:

0.2638,0.5902,0.1250,0.5902

Step-by-step explanation:

Given that in a certain ​ state, 32.2​% of all community college students belong to ethnic minorities.

i.e. probability for any random college student to belong to ethnic minorities

=0.322

This is constant as each student is independent of the other

X no of college students  belong to ethnic minorities in the sample of 10 is

Bin (10, 0.322)

a) Exactly 3 belong to an ethnic minority.

[tex]=P(X=3)\\=0.2638[/tex]

b. Three or fewer belong to an ethnic minority.

=[tex]P(X\leq 3)\\=0.5902[/tex]

c. Exactly 5 do not belong to an ethnic minority.

[tex]P(X'=5)=P(X=5)\=0.1250[/tex]

d. Six or more do not belong to an ethnic minority.

[tex]P(X'\geq 6)\\=P(X<4)\\=0.5902[/tex]

Answer:

0% probability of a guest arriving by taxi and requesting parking for a car.

Step-by-step explanation:

Two events are said to be mutually exclusive if one event precludes the other. That means that if A and B are mutually exclusive, the probability of A and B happening at the same time is 0%.

In this problem, we have that a guest arriving via taxi and requesting parking at the hotel are mutually exclusive.

So there is a 0% probability of a guest arriving by taxi and requesting parking for a car.

The range within which a student's typing speed will be in two months with a 95.45% probability is 49.66 to 65.34 words per minute.

To find the range within which a student's typing speed will be in two months with a 95.45% probability, we need to calculate the prediction interval. The regression model equation is given as Y' = 7x3 + 5x2 + 3x + 11, where x3 represents hours of instruction per week, x2 represents hours of practice per week, and x represents hours of typing per week necessary for school or work.

Since the student is taking 2 hrs of typing instruction per week (x3 = 2), practicing 5 hrs per week (x2 = 5), and typing 2.5 hours per week for work (x = 2.5), we can substitute these values into the regression equation to find the predicted typing speed (Y').

Using the given equation and substituting the values, we get:

Y' = 7(2) + 5(5) + 3(2.5) + 11

Y' = 14 + 25 + 7.5 + 11 = 57.5 words per minute

Since the standard error of the estimate is 4, the prediction interval can be calculated by adding or subtracting 1.96 times the standard error from the predicted value. Therefore, the range for a 95.45% probability is:

57.5 - (1.96 x 4) to 57.5 + (1.96 x 4) = 57.5 - 7.84 to 57.5 + 7.84 = 49.66 to 65.34 words per minute.

The independent variable is annual tuition. The average salary for a graduate with an annual tuition of $30,000 is $138.7 thousand. The slope of the regression model is -0.91.

The independent variable of a regression model is the variable that is being manipulated or controlled by the researcher. In this case, the independent variable is the annual tuition of the universities and colleges.

According to the given linear regression model, when the annual tuition is $30,000, the average mid-career salary for a graduate is calculated as follows: y = -0.91(30) + 161 = $138.7 thousand.

The slope of the regression model is the coefficient of the independent variable, which is -0.91.

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The independent variable is the annual tuition, and the dependent variable is the average mid-career salary of graduates. The average salary for a graduate with an annual tuition of $30,000 is estimated to be $137.9 thousand. The slope of the regression model is -0.91, indicating a decrease in salary for every increase in tuition.

The independent variable in this linear regression model is annual tuition, denoted by x. The average mid-career salary of graduates, denoted by y, is the dependent variable, meaning it depends on the value of the independent variable.

To find the average salary for a graduate of a college or university with an annual tuition of $30,000, we substitute x = 30 into the regression model. Plugging in the value, we get: ˆy = -0.91(30) + 161. Solving this equation, we find ˆy = $137.9 thousand.

The slope of the regression model is the coefficient of the independent variable, which is -0.91. This means that for every increase of $1,000 in annual tuition, the average mid-career salary of graduates is estimated to decrease by $910.

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Answer:

We can conclude that new fertilizer appears to be more effective on apple trees.

Step-by-step explanation:

Mean and standard deviation parameters are missing in the question, I will assume amounts as below:

At Frucht Orchards, apple trees produce an average of 840 pounds of fruit per tree with a standard deviation of 120 pounds. Their mango trees produce an average of 350 pounds of fruit per tree with a standard deviation of 190 pounds. Frucht Orchards is trying two new fertilizers. An apple tree fed Amazing Apples Fertilizer yielded 940 pounds of fruit, while a mango tree fed Amazing Mangos Fertilizer yielded 400 pounds of fruit. Which new fertilizer appears to be more “amazing”?  

We need to calculate standardized values (z-scores) of fertilized apple and mango trees to decide which fertilizer appears to be more "amazing".

z score for an unique value can be calculated using the equation:

z=[tex]\frac{X-M}{s}[/tex] where

For fertilized apple

z=[tex]\frac{940-840}{120}[/tex]≈0.83

For fertilized mango

z=[tex]\frac{400-350}{190}[/tex]≈0.26

Since 0.83>0.26 we can conclude that new fertilizer appears to be more effective on apple trees.

Answer:

[tex]n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65[/tex]

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

Step-by-step explanation:

1) 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 means 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.

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".

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

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

ME represent the margin of error

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

2) Solution to the problem

Since the new Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

Assuming that the deviation is known we can express the margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

Replacing into formula (b) we got:

[tex]n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65[/tex]

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

Answer: 12

Step-by-step explanation:

Given : Students at a liberal arts college study for an average of 10 hours per week with a standard deviation of 2 hours per week.

[tex]\mu=10\text{ hours}[/tex] and [tex]\sigma=2\text{ hours}[/tex]

The distribution of their study time happens to be uni-modal, symmetric and bell shaped i.e. Normally distributed.

According to the Empirical rule , about 68% of the population lies within one standard deviation from mean .

i.e. Approximately 68% of students study between [tex]\mu-\sigma[/tex] and  [tex]\mu+\sigma[/tex]  hours a week.

i.e. Approximately 68% of students study between [tex]10-2[/tex] and  [tex]10+2[/tex]  hours a week.

i.e. Approximately 68% of students study between 8 and 12 hours a week.

Hence, the value of B = 12.

The question is about a normal distribution in statistics where 68% of data falls within one standard deviation from the mean. Given the mean study, hours is 10 per week and the standard deviation is 2 hours, the value of B representing the upper limit of the 68% range is 12 hours per week.

The subject matter of this problem is based on the principles of statistics, particularly the concept of a normal distribution which is characterized by being uni-modal, symmetric, and bell-shaped. In a normal distribution, approximately 68% of data falls within one standard deviation from the mean.

In this question, we are given that the average study hours are 10 per week (the mean), and the standard deviation is 2 hours. A study time of 8 hours a week represents one standard deviation below the mean. Therefore, one standard deviation above the mean will represent 12 hours a week (mean + standard deviation: i.e. 10 + 2).

Thus, in terms of the normal distribution of study hours, the value of B is 12 hours. That means that approximately 68% of students at this particular liberal arts college study between 8 and 12 hours a week.

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