In a study of automobile collision insurance costs, a random sample of n = 35 repair costs of front-end damage caused by hitting a wall at a specified speed had a mean of $1,438. (a) Given that σ = $269 for such data, what can be said with 98% confidence about the maximum error if x = $1, 438 is used as an estimate of the average cost of such repairs.

Answer:

A

Step-by-step explanation:

Having an inexpensive wedding helps young couples avoid financial burdens that may strain their marriage. This is true as it will ensure the young couples spend wisely. They understand that if they splurge, the returns will not be as great as that from a long-lasting marriage.

Answer:

There is a significant linear correlation between the two variables.

Step-by-step explanation:

Given that part time weekly earnings are

x y

10 93

15 171

20 204

20 156

35 261

Correlation 0.9199

H0: r=correlatin =0

Ha: r ≠0

(two tailed test)

r difference = 0.9199

t = [tex]r*\sqrt{(n-2)/(1-r^2)} \\=4.063[/tex]

p value < 0.0001

Since p <0.05 we reject null hypothesis

There is a significant linear correlation between the two variables.

This answer demonstrates how to create a scatter plot in Excel, interpret the correlation, calculate SSxx, SSyy, and SSxy, find t.05 from a t-distribution table, and perform a t-test to determine if the correlation coefficient is significantly different from zero.

In Excel, you can create a scatter plot by selecting the appropriate data and choosing Scatter Plot under the Insert tab. The scatter plot suggests a positive correlation between hours worked (X) and weekly pay (Y), indicating that as hours worked increase, so does weekly pay.

To calculate SSxx, SSyy, and SSxy, first, calculate the average of X and Y. Next, subtract the average from each individual data point and square the result to find SSxx and SSyy. For SSxy, subtract the average of X from each X data point and the average of Y from each Y data point, multiply these together, and take the sum.

With regards to finding t.05, look at the t-distribution table available with your study materials. Lastly, calculate the t test statistics using the formula t = r * sqrt((n-2)/(1-r^2)), where n is the number of pairs, r is the correlation coefficient obtained from your Excel scatter plot, and sqrt is the square root function. If the calculated t-value exceeds the t.05 value, you reject the null hypothesis (p = 0).

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Answer:Bradley practices for 328500 minutes more than Jonas

Step-by-step explanation:

Jonas practices guitar 2 hours a day for 2 years. 24 hours make a day. Assuming there are 365 days in a year.

If he plays 2 hours for 2 days

Number of hours for which he practiced for 365 days would be (365×2)/2 = 365 hours.

Over the course of 3 years, number of hours for which he practiced would be 365×3 = 1095 hours

Bradley practices the guitar 5 hours a day more then Jonas. Jonas practiced for 1 hour a day. This means that Bradley practices the guitar for 5+1 = 6 hours a day. In a year, it would be 6×365 = 2190 hours. In 3 years, it would be 3×2190 = 6570 hours

Difference in number of hours between Bradley and Jonas is

6570 - 1095 = 5475 hours. Since 60 minutes = 1 hour,

5475 hours would be

5475×60 = 328500 minutes.

Answer:the linear equation for the fundraising goal is 8x + 12y = 240

Step-by-step explanation:

Let x represent the number of water bottles that the Environmental Club would sell.

Let x represent the number of tote bags that the Environmental Club would sell.

The Environmental Club is selling water bottles for $8 and tote bags for $12 to raise $240 to donate to charity. This means that

8x + 12y = 240

Final answer:

The probability of getting a red card or a face card from a deck of playing cards is 8/13.

Explanation:

To determine P(E1 or E2), we need to find the probability of either event E1 (outcome is a red card) or event E2 (outcome is a face card) occurring.

To find P(E1), we can count the number of red cards in the deck, which is 26 (there are 2 red suits and each suit has 13 cards).

To find P(E2), we can count the number of face cards in the deck, which is 12 (there are 3 face cards in each suit and 4 suits).

Since E1 and E2 are not mutually exclusive (some cards are both red and face cards), we need to subtract the number of cards that satisfy both events to avoid double counting. In this case, there are 6 red face cards in the deck (2 red suits each with 3 face cards).

Therefore, P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2) = 26/52 + 12/52 - 6/52 = 32/52 = 8/13.

Final answer:

The probability of drawing either a red card or a face card from a standard deck of 52 playing cards is 8/13.

Explanation:

To calculate P(E1 or E2), where E1 is the event of drawing a red card and E2 is the event of drawing a face card from a standard deck of 52 playing cards, we can use the formula for the probability of the union of two events: P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2). Since there are 13 red cards in each of the two red suits (hearts and diamonds), P(E1) = 26/52. There are 3 face cards (jack, queen, king) in each of the four suits, so P(E2) = 12/52. However, among these face cards, 6 are red (3 in hearts and 3 in diamonds), so P(E1 and E2) = 6/52. Thus, the probability of drawing either a red card or a face card is P(E1 or E2) = 26/52 + 12/52 - 6/52.

In conclusion: P(E1 or E2) = (26 + 12 - 6) / 52

Simplifying that, we get: P(E1 or E2) = 32/52, which can be reduced to P(E1 or E2) = 8/13.

Answer:

The value of the test statistic z for this hypothesis test is -2.79

Step-by-step explanation:

Consider the provided information.

To calculate the test statistic use the formula:

[tex]z=\frac{\hat p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}[/tex]

Where, z  is Test statistics, n is Sample size, p₀ = Null hypothesized value  and [tex]\hat p[/tex] = Observed proportion.

p₀ = 0.50

Thus 1-p₀= 0.50

42% responded that if they had a cell phone, thus [tex]\hat p=0.42[/tex]

The sample size is 305.

Substitute the respective values in the above formula.

[tex]z=\frac{0.42-0.50}{\sqrt{\frac{0.50(0.50)}{305}}}[/tex]

[tex]z=\frac{-0.08}{\sqrt{0.00082}}[/tex]

[tex]z=-2.79[/tex]

Hence, the value of the test statistic z for this hypothesis test is -2.79

The value of the test statistic [tex]\( z \)[/tex] for this hypothesis test is approximately -2.797.

To calculate the test statistic [tex]\( z \)[/tex], we follow these steps:

1. State the null hypothesis [tex]\( H_0: p = 0.50 \)[/tex] and the alternative hypothesis [tex]\( H_A: p < 0.50 \)[/tex].2. Identify the sample proportion [tex]\( \hat{p} \)[/tex], which is the proportion of the sample that has the characteristic of interest.

In this case, [tex]\( \hat{p} = \frac{128}{305} \approx 0.42 \)[/tex]

3. Calculate the standard error of the proportion, which is given by [tex]\( SE = \sqrt{\frac{p(1-p)}{n}} \)[/tex], where [tex]\( p \)[/tex] is the population proportion under the null hypothesis and [tex]\( n \)[/tex] is the sample size

[tex]\( SE = \sqrt{\frac{0.50(1-0.50)}{305}} \approx \sqrt{\frac{0.25}{305}} \)[/tex]

4. Compute the test statistic [tex]\( z \)[/tex] using the formula [tex]\( z = \frac{\hat{p} - p}{SE} \)[/tex], where [tex]\( \hat{p} \)[/tex] is the sample proportion, [tex]\( p \)[/tex] is the population proportion under the null hypothesis, and [tex]\( SE \)[/tex] is the standard error.

5. Plug in the values to get [tex]\( z = \frac{0.42 - 0.50}{\sqrt{\frac{0.25}{305}}} \)[/tex].

6. Simplify the expression to find the value of [tex]\( z \)[/tex].

[tex]\[ z = \frac{0.42 - 0.50}{\sqrt{\frac{0.25}{305}}} = \frac{-0.08}{\sqrt{\frac{1}{1220}}} = \frac{-0.08}{\sqrt{0.0008197}} = \frac{-0.08}{0.0286} \approx -2.797 \][/tex]

To estimate the difference between the proportion of elementary school teachers and high school teachers who are satisfied, we need to calculate a 95% confidence interval. The CI is (0.0181, 0.1589).

To estimate the difference between the proportion of all elementary school teachers who are satisfied and all high school teachers who are satisfied, we need to calculate a 95% confidence interval (CI). First, we calculate the sample proportions for both groups. The sample proportion of elementary school teachers who are very satisfied is 226/397 = 0.5698, and the sample proportion of high school teachers who are very satisfied is 129/268 = 0.4813. Next, we calculate the standard error using the formula: √(phat1*(1-phat1)/n1 + phat2*(1-phat2)/n2), where phat1 and phat2 are the sample proportions and n1 and n2 are the sample sizes. Substituting the values, we get √((0.5698*(1-0.5698)/397 + 0.4813*(1-0.4813)/268)). This gives us a standard error of 0.0359.

Next, we need to calculate the margin of error (ME) by multiplying the standard error by 1.96 (which corresponds to a 95% confidence level). ME = 1.96 * 0.0359 = 0.0704.

Finally, we can calculate the confidence interval by subtracting the margin of error from the difference in sample proportions and adding the margin of error to the difference in sample proportions. The difference in sample proportions is 0.5698 - 0.4813 = 0.0885. So, the 95% confidence interval is (0.0885 - 0.0704, 0.0885 + 0.0704), which simplifies to (0.0181, 0.1589).

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Answer:

a) We would expect to see 500*0.88=440

b) [tex]z=\frac{0.9 -0.88}{\sqrt{\frac{0.88(1-0.88)}{500}}}=1.376[/tex]  

[tex]p_v =2*P(Z>1.376)=0.167[/tex]  

So the p value obtained was a very high value 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, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when [tex]p_v <\alpha[/tex] we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".  

Step-by-step explanation:

Data given and notation

n=500 represent the random sample taken

X=450 represent the people that have the seat belt fastened

[tex]\hat p=\frac{450}{500}=0.9[/tex] estimated proportion of people that have the seat belt fastened

[tex]p_o=0.88[/tex] is the value that we want to test

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

z would represent the statistic (variable of interest)

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

Part a

We would expect to see 500*0.88=440

Part b

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion changes fro m 0.88.:  

Null hypothesis:[tex]p=0.88[/tex]  

Alternative hypothesis:[tex]p \neq 0.88[/tex]  

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

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

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

Calculate the statistic  

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

[tex]z=\frac{0.9 -0.88}{\sqrt{\frac{0.88(1-0.88)}{500}}}=1.376[/tex]  

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 assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(Z>1.376)=0.167[/tex]  

So the p value obtained was a very high value 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, and we can said that at 5% of significance the true proportion is not significant different from 0.9.

The p value is a criterion to decide if we reject or not the null hypothesis, when [tex]p_v <\alpha[/tex] we reject the null hypothesis in other case we FAIL to reject the null hypothesis. And represent the "probability of obtaining the observed results of a test, assuming that the null hypothesis is correct".  

The probability of drawing a 2 and then drawing another 2 in the board game is 1/72.

To calculate the probability of drawing a 2 and then drawing another 2 in the board game, we first need to understand the concept of probability. Probability represents the likelihood of an event occurring, ranging from 0 (impossible) to 1 (certain). In this case, we're interested in the probability of two specific events happening sequentially: first drawing a 2, and then drawing another 2.

In the board game scenario, there are 9 numbers in total. So, the probability of drawing a 2 on the first draw is 1 out of 9 since there's one 2 among the nine numbers. After drawing the first number and not replacing it, there are now 8 numbers left. If the first draw was a 2, then there's one less 2 available among the remaining 8 numbers. So, the probability of drawing another 2 on the second draw is 1 out of 8.

To find the overall probability of both events happening in sequence (drawing a 2 and then drawing another 2), we multiply the probabilities of each event. Thus, the probability of drawing a 2 and then drawing another 2 is (1/9) * (1/8) = 1/72.

Therefore, the probability of drawing a 2 and then drawing another 2 in the board game is indeed 1/72. This means that out of all the possible sequences of two draws without replacement, only 1 out of 72 sequences will result in drawing a 2 followed by another 2.

Complete question :- In a board game, students draw a number do not replace it and then draw a second number

Determine the probability of each event occurring,

1

6

6

9

21

62

What is the probability of drawing a 2, then drawing another 2?

Answer:

b) ( N c ∩ D ) ∪ ( N ∩ D c )

Step-by-step explanation:

Let's see each option:

a) N ∪ D

This includes those who likes both, so this is not correc.t

b) ( N c ∩ D ) ∪ ( N ∩ D c )

Nc is the complement of N. That is those who do not subscribe to news magazine. Intensected with D, those are who do not subscribe to the news magazine but do to the daily informer. By the same logic, the second intersection are those who do subscribe to news magazine but not to the daily informer. This is the correct answer

c) N c ∩ D

This includes only those who subscribe to the daily informer but not to the news maganize. It also needs those who subscribe to the news magazine but not to the daily informer.

d) ( N ∩ D )c

This also involves those who do not subscribe to any of these news sources.

e) ( N c ∪ D ) ∩ ( N ∪ D c )

Wrong... Intersection of those who subscribe to the daily informer but not the news magazine and those who subscribe to the news magazine but not to the daily informer. This is the empty set.

Answer:

We are confident at 99% that the difference between the two proportions is between [tex]0.380 \leq p_{Republicans} -p_{Democrats} \leq 0.420[/tex]

Step-by-step explanation:

Part a

Data given and notation  

[tex]X_{D}=3266[/tex] represent the number people registered as Democrats

[tex]X_{R}=2137[/tex] represent the number of people registered as Republicans

[tex]n=7525[/tex] sampleselcted

[tex]\hat p_{D}=\frac{3266}{7525}=0.434[/tex] represent the proportion of people registered as Democrats

[tex]\hat p_{R}=\frac{2137}{7525}=0.284[/tex] represent the proportion of people registered as Republicans

The standard error is given by this formula:

[tex]SE=\sqrt{\frac{\hat p_D (1-\hat p_D)}{n_{D}}+\frac{\hat p_R (1-\hat p_R)}{n_{R}}}[/tex]

And the standard error estimated given by the problem is 0.008

Part b

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]p_A[/tex] represent the real population proportion of Democrats that approve of the way the California Legislature is handling its job  

[tex]\hat p_A =\frac{1894}{3266}=0.580[/tex] represent the estimated proportion of Democrats that approve of the way the California Legislature is handling its job  

[tex]n_A=3266[/tex] is the sample size for Democrats

[tex]p_B[/tex] represent the real population proportion of Republicans that approve of the way the California Legislature is handling its job  

[tex]\hat p_B =\frac{385}{2137}=0.180[/tex] represent the estimated proportion of Republicans that approve of the way the California Legislature is handling its job

[tex]n_B=2137[/tex] is the sample for Republicans

[tex]z[/tex] represent the critical value for the margin of error  

The population proportion have the following distribution  

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

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]  

For the 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.64[/tex]  

And replacing into the confidence interval formula we got:  

[tex](0.580-0.180) - 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.380[/tex]  

[tex](0.580-0.180) + 1.64 \sqrt{\frac{0.580(1-0.580)}{3266} +\frac{0.180(1-0.180)}{2137}}=0.420[/tex]  

And the 99% confidence interval would be given (0.380;0.420).  

We are confident at 99% that the difference between the two proportions is between [tex]0.380 \leq p_{Republicans} -p_{Democrats} \leq 0.420[/tex]

Answer:

C) Yes, because the joint probability of P and M is equal to 0.

Step-by-step explanation:

Given that, P represents the event that the selected person preferred pepperoni only.

And M represents the event that the selected person preferred mushroom only.

so, P and M are two mutually exclusive events because two events are mutually exclusive if they do not occur together.

Here, P and M cannot occur at same time.

so, the joint probability of P and M i.e) probability of both P and M to occur at same time is zero.

In the given problem we have 55 percent preferred pepperoni only, 30 percent preferred mushroom only, and 15 percent preferred something other than pepperoni and mushroom. The probability of event can be identified by selecting how many people can chose pepperoni and mushroom.

The correct option is (c)

Given:

Pepperoni preferred by 55 percent.

Mushroom preferred by 30 percent.

The people who preferred other than Mushroom and Pepperoni are 15 percent.

In the given problem, A person can not select both pepperoni and mushroom because percentages of the pizza preferences adds to 1 in this survey. So, the mathematically,

P(P∩M)=0P(P∩M)=0

The above expression indicates the events are mutually exclusive.

Therefore, the joint probability of P and M is equal to 0.

Thus, the correct option is (c)

Learn more about what probability is here:

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Answer:

a) [tex]div F = 27 \frac{W}{km^3}[/tex]

b) [tex]\alpha=\frac{27 W/km^3}{3}= 9\frac{W}{Km^3}[/tex]

c) [tex]T(0,0,0)=-\frac{10}{2(27000)} (0) +7605.185=7605.185[/tex]

Step-by-step explanation:

(a) Suppose that the actual heat generation is 27W/km3 What is the value of div F? div F- Include units)

For this case the value for div F correspond to the generation of heat.

[tex]div F = 27 \frac{W}{km^3}[/tex]

(b) Assume the heat flows outward symmetrically. Verify that [tex] F= \alpha r[/tex] where [tex]r=xi +yj+zk[/tex]. Find a α, (Include units.)

For this case we can satisfy this condition:

[tex]div[\alpha (xi +yj +z k)]]=\alpha(1+1+1)=3\alpha[/tex]

And since we have the value for the [tex]div F[/tex] we can find the value of [tex]\alpha[/tex] like this:

[tex]\alpha=\frac{27 W/km^3}{3}= 9\frac{W}{Km^3}[/tex]

(c) Let T (x,y,z) denote the temperature inside the earth. Heat flows according to the equation F= -k grad T where k is a constant. If T is in °C then k=27000 C/km. Assuming the earth is a sphere with radius 6400 km and surface temperature 20°C, what is the temperature at the center? 27,0 C/km.

For this case we have this:

[tex] F =-k grad T[/tex]

And grad T represent the direction of the greatest decrease related to the temperature.

So we have this equation:

[tex] 10(xi +yj+zk)=-27000 grad T[/tex]

And we can solve for grad T like this:

[tex] grad T = -\frac{10}{(27000)} (xi+yj+zk)[/tex]

Andif we integrate in order so remove the gradient on both sides we got:

[tex]T=-\frac{10}{2(27000)} (x^2 +y^2 +z^2) +C[/tex]

For our case we have the following condition:

[tex]x^2 +y^2 +z^2 = 6400 , T=20 C[/tex]

[tex] T=-\frac{1}{54000} (6400^2)+C =20[/tex]

And we can solve for C like this:

[tex] C= 20+\frac{6400^2}{5400}= 7605.185 [/tex]

So then our equation would be given by:

[tex]T=-\frac{10}{2(27000)} (x^2 +y^2 +z^2) +7605.185[/tex]

And for our case at the center we have that [tex]x^2+ y^2+ z^2 =0[/tex]

And we got:

[tex]T(0,0,0)=-\frac{10}{2(27000)} (0) +7605.185=7605.185[/tex]

Answer: D. 60100

Step-by-step explanation:

The formula for determining the sum of n terms of an arithmetic sequence is expressed as

Sn = n/2[2a + (n - 1)d]

Where

n represents the number of terms in the arithmetic sequence.

d represents the common difference of the terms in the arithmetic sequence.

a represents the first term of the arithmetic sequence.

From the information given,

n = 200 terms

a = 2

d = 3

Therefore, the sum of the first 200 terms, S200 would be

S200 = 200/2[2 × 2 + (200 - 1)3]

S200 = 100[4 + 597)

S200 = 100 × 601 = 60100

Answer:

The answer is 0.2865

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

In this problem, we have that:

The mean number of accidents on any given day in Coralville is 5. Of those, 25% are with an uninsured drive.

So [tex]\mu = 5*0.25 = 1.25[/tex]

Calculate the probability that on a given day in Coralville there are no trafficaccidents that involve an uninsured driver.

This is P(X = 0). So

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-1.25}*(1.25)^{0}}{(0)!} = 02865[/tex]

To calculate the probability of no traffic accidents involving uninsured drivers in Coralville, we can use the Poisson distribution formula with a mean of 5 accidents. The probability is approximately 0.00674, or 0.674%.

To calculate the probability that on a given day in Coralville there are no traffic accidents that involve an uninsured driver, we can use the Poisson distribution formula. The Poisson distribution is used to model the number of events occurring in a fixed interval of time or space, given the average rate of occurrence. In this case, the mean number of traffic accidents is 5, and the probability of an accident involving an uninsured driver is 0.25.

The formula for the Poisson distribution is P(X=k) = (e^(-λ) * λ^k) / k!, where λ is the mean number of events and k is the number of events we are interested in.

To find the probability of no traffic accidents involving uninsured drivers, we can plug in λ=5 and k=0 into the formula:

P(X=0) = (e^(-5) * 5^0) / 0!

P(X=0) = e^(-5) / 1 = 0.006737947

Therefore, the probability that on a given day in Coralville there are no traffic accidents that involve an uninsured driver is approximately 0.00674, or 0.674%.

Answer:

The 95% confidence interval would be given by [tex]-23.44 \leq \mu_1 -\mu_2 \leq -8.16[/tex]  

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".  

[tex]\bar X_1 =114.1[/tex] represent the sample mean 1

[tex]\bar X_2 =129.9[/tex] represent the sample mean 2

n1=5 represent the sample 1 size  

n2=2 represent the sample 2 size  

[tex]s_1 =5.08[/tex] sample standard deviation for sample 1

[tex]s_2 =5.37[/tex] sample standard deviation for sample 2

[tex]\mu_1 -\mu_2[/tex] parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:

[tex]\bar X_1 -\bar X_2 =114.1-129.9=-15.8[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:  

[tex]df=n_1 +n_2 -1=5+5-2=8[/tex]  

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that [tex]t_{\alpha/2}=\pm 2.31[/tex]  

The standard error is given by the following formula:

[tex]SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex]

And replacing we have:

[tex]SE=\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=3.306[/tex]

Confidence interval

Now we have everything in order to replace into formula (1):  

[tex]-15.8-2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-23.437[/tex]  

[tex]-15.8+2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-8.163[/tex]  

So on this case the 95% confidence interval would be given by [tex]-23.44 \leq \mu_1 -\mu_2 \leq -8.16[/tex]  

To calculate a 95% confidence interval for the difference between the true average stopping distances for cars equipped with system 1 and system 2, use the formula CI = (x1 - x2) ± (c * SE) where x1 and x2 are the sample means, and SE is the standard error of the difference.

To calculate a 95% confidence interval for the difference between the true average stopping distances for cars equipped with system 1 and system 2, we can use the formula:


CI = (x1 - x2) ± (c * SE)


where x1 and x2 are the sample means, and SE is the standard error of the difference. The critical value c can be found using the t-distribution table for the given sample sizes n1 and n2.

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Answer:

SAT scores account for 4 % of the fluctuation in GPA

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:

[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]  

The determinarion coefficient represent "the proportion of the variance in the dependent variable that is predictable or explained from the independent variable".

And is just defined as [tex]r^2[/tex]

Solution to the problem

The % of variation is given by the determination coefficient given by [tex]r^2[/tex] and on this case [tex]0.2^2 =0.04[/tex], so then the % of variation explained by the linear model is 4%.

And we can conclude that: SAT scores account for 4 % of the fluctuation in GPA

Answer:

n=601

Step-by-step explanation:

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{\hat p(1-\hat 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 confidence interval for the mean is given by the following formula:  

[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/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]ME =\pm 0.04[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex]   (b)  

Since we don't have a prior estimation for the proportion we can use 0.5 as estimation. And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.96})^2}=600.25[/tex]  

And rounded up we have that n=601

Answer:

[tex]\mu_{p_E -p_H} = 0.4-0.6 =-0.2[/tex]

[tex]SE_{p_E -p_H}=\sqrt{\frac{0.4(1-0.4)}{20} +\frac{0.6 (1-0.6)}{20}}=0.1549[/tex]

b. Normal with mean μ= -0.2 and standard deviation σ=0.1549

Step-by-step explanation:

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]p_E[/tex] represent the real population proportion for Engineering

[tex]\hat p_E =\frac{8}{20}=0.4[/tex] represent the estimated proportion for Engineering

[tex]n_E=20[/tex] is the sample size required for Engineering

[tex]p_H[/tex] represent the real population proportion for Humanities  

[tex]\hat p_H =\frac{12}{20}=0.6[/tex] represent the estimated proportion for Humanities

[tex]n_H=20[/tex] is the sample size required for Humanities

We assume that the population proportions follows a normal distribution since we satisfy these conditions:

[tex]np\geq 5[/tex] ,[tex]n(1-p)\geq 5[/tex]

[tex]20*0.4= 8\geq 5 , 20(1-0.4)=12\geq 5[/tex]

The population proportion have the following distribution  

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

And we are interested on the distribution for [tex]p_E-p_H[/tex]

For this case we know that the distribution for the differences of proportions is also normal and given by:

[tex]p_E -p_H \sim N(\hat p_E -\hat p_H, \sqrt{\frac{\hat p_E(1-\hat p_E)}{n_E} +\frac{\hat p_H (1-\hat p_H)}{n_H}}[/tex]

So then we can find the mean and the deviation like this:

[tex]\mu_{p_E -p_H} = 0.4-0.6 =-0.2[/tex]

[tex]SE_{p_E -p_H}=\sqrt{\frac{0.4(1-0.4)}{20} +\frac{0.6 (1-0.6)}{20}}=0.1549[/tex]

So then the best option is :

b. Normal with mean μ= -0.2 and standard deviation σ=0.1549

The shape is unknown since there is no knowledge about it and the representative sample is limited.

Total number of interview in engineering and humanities = 20

Number of interview in engineering = 8

Number of interview in humanities = 12

⇒ Probability of interview in engineering - Probability of interview in humanities

⇒ 8/20 - 12/20

⇒ -0.20

The shape is unknown since there is no knowledge about it and the representative sample is limited.

Therefore, Option D is the right answer.

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Answer: All the statements can be proved using definitions from group theory.

Step-by-step explanation:

We will use the fact that the set [tex]\{0,1,2\}:=\mathbb{Z}_3[/tex] is a group under addition modulo 3 (denoted by +). This is because addition modulo 3 is associative, 0 is the identity element for this group and -0=0, -1=2, -2=1.

Let [tex]ax^2+bx+c,dx^2+ex+f, gx^2+hx+i \in G[/tex].

Addition in G is well defined, because [tex]ax^2+bx+c+dx^2+ex+f=(a+d)x^2+(b+e)x+(c+f)[/tex] and the coefficients of this last polynomial are computed using addition modulo 3, which is well defined.

Because addition modulo 3 is associative, we have that [tex](ax^2+bx+c+dx^2+ex+f)+gx^2+hx+i=((a+d)x^2+(b+e)x+(c+f))+gx^2+hx+i=((a+d)+g)x^2+((b+e)+h)x+((c+f)+i)=(a+(d+g))x^2+(b+(e+h))x+(c+(f+i))=ax^2+bx+c+dx^2+((d+g)x^2+(e+h)x+c+(f+i))=ax^2+bx+c+dx^2+(dx^2+ex+f+gx^2+hx+i)[/tex] therefore the addition defined in G is associative.

The identity element for this group is the polynomial [tex]e=0:=0x^2+0x+0\in G[/tex]. Indeed, we have that  [tex]ax^2+bx+c+0=(a+0)x^2+(b+0)x+(c+0)=ax^2+bx+c=(0+a)x^2+(0+b)x+(0+c)=0+ax^2+bx+c[/tex]

The inverse element of a polynomial [tex]ax^2+bx+c \in G[/tex] is the polynomial [tex](-a)x^2+(-b)x+(-c) \in G[/tex], because [tex]ax^2+bx+c+(-a)x^2+(-b)x+(-c)=(a-a)x^2+(b-b)x+(c-c)=0=(-a)x^2+(-b)x+(-c)+]ax^2+bx+c [/tex]

This group has order 27: to count all the polynomials in G we have to count the number of ways to choose a, b and c and construct a polynomial. a,b and c can be either 0,1 or 2 so each one can be chosen in 3 ways. In total, there are 3×3×3=27 possible choices that result in a element of G.

This group is not cyclic. Take any polynomial [tex]ax^2+bx+c \in G[/tex], then its generated group is the set [tex]<ax^2+bx+c>=\{0,ax^2+bx+c,2ax^2+2bx+2c\}[/tex]. This is because any element of the form[tex]n(ax^2+bx+c)=nax^2+nbx+nc, n\in \mathbb{Z}[/tex] can be reduced to [tex](n\mod 3)ax^2+(n\mod 3)bx+(n\mod 3)c[/tex] and [tex]n\mod 3[/tex] only takes the values 0,1,2. The generated group of any element of G has less than 27 elements, so G can't be cyclic.  

The set G of all polynomials ax^2+bx+c with coefficients {0, 1, 2} forms a group under addition with modulo 3 and has order 27, but it is not cyclic, because no element of G, when raised to each power from 0 to the group's order, can generate all elements of G.

Let G be the set of all polynomials of the form ax^2+bx+c, with coefficients {0, 1, 2}. Each coefficient has 3 possibilities (0, 1, 2) so there are 3^3 = 27 total polynomials, hence the order of G is 27.

To show that G is a group under addition with modulo 3, we need to show four things: closure, associativity, identity element, and inverses. Polynomials form a ring under normal addition and multiplication, and since mod 3 addition follows the same rules, G is closed and associative. The identity element is the polynomial where all coefficients are 0. Finally, regarding inverses, every element in G has an additive inverse in G (for example, the inverse of 2 is 1 mod 3).

To show that G is not cyclic, note that a group is cyclic if and only if it has a generator which, when raised to every power from 0 to the order of the group, generates every element of the group. But in G there is no such polynomial, hence G is not cyclic.

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The age of the artifact is approximately 3378 years.

To determine the age of the artifact, we can use the concept of radioactive decay and the half-life of carbon-14 (14C), which is 5715 years.

The formula for the decay of a radioactive substance is given by:

[tex]A = A_0 \times (0.5)^{(t / T_{1/2})[/tex]

where:

A = Final activity of the sample (in counts per minute, cpm)

A₀ = Initial activity of the sample (in counts per minute, cpm)

t = Time elapsed (in years)

T₁/₂ = Half-life of the radioactive substance (in years)

We are given:

A₀ (activity of the artifact) = 38.0 cpm

A (activity of the standard of zero age) = 58.2 cpm

T₁/₂ (half-life of carbon-14) = 5715 years

Let's rearrange the formula to solve for t (age of the artifact):

[tex]t = T_{1/2} \times log_2(A / A_0)[/tex]

Now, plug in the given values:

t = 5715 × log₂(58.2 / 38.0)

Using a calculator:

t ≈ 5715 × log₂(1.53263158)

t ≈ 5715 × 0.591170846

t ≈ 3378.115647

Hence, the age of the artifact is approximately 3378 years.

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Answer:

(a) z-value = -2.12

(b) p-value = 0.0170

(c) Reject H0

Step-by-step explanation:

(a)

[tex]std-err=\frac{std-dev}{\sqrt{n}}=\frac{2}{\sqrt{50}}=0.2828[/tex]

[tex]z-value=\frac{X-mean}{std-err}=\frac{19.4-20}{0.2828}=-2.12[/tex]

(b)

with

z-value = -2.12

significance level = 0.05

one-tail hypothesis (H0: μ ≥ 20)

We can see on the normal distribution table (Z-Score table) that

p-value = 0.0170

(c)

Since p value (0.0170) is less than α=0.05 we reject H0

Hope this helps!

The sample of 100 horses is too small to obtain an unbiased estimate.

Applied research frequently focuses more on finding solutions to particular issues that directly impact people in the present. As an illustration, a social psychologist conducting fundamental research on violence would consider the various potential causes of violence in general.

Given, A study is being planned by a group of veterinary researchers to determine how many enteroliths are present on average in affected horses. When the study is financed, the researchers will sign up the first 100 horses they observe.

A study plan, which is considered to be the most important piece of information describing the projected research sample of an investigator, is the main part of an application. An important analyst is also given the opportunity to speak on the planned study, highlighting its benefits and explaining the implementation and conduct of the research. Because it covers your whole research project from start to finish and identifies and explains your emphasis, technique, and goals, a research plan is crucial.

Therefore, The sample size of 100 horses is insufficient to provide a fair assessment.

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Answer:

[tex] 6P6 = \frac{6!}{(6-6)!}=\frac{6!}{0!}= 6! =6*5*4*3*2*1=720[/tex]

D. 720

Step-by-step explanation:

For this case we have 6 different cities that needs to be ordered in different ways. And the best way to solve this problem is using permutations since we can't repeat the route.

We need to remember the concept of permutation.

A permutation, known as "arrangement number"  "is a rearrangement of the elements of an ordered list S into a one-to-one correspondence with S itself". Where S correspond to the sample space. And the formula is given by:

[tex]nPX = \frac{n!}{(n-x)!}[/tex]

And for this case we have a total of 6 cities and we want to know how many routes with these 6 cities we can create, so then n=6 and k=6 and if we replace we got:

[tex] 6P6 = \frac{6!}{(6-6)!}=\frac{6!}{0!}= 6! =6*5*4*3*2*1=720[/tex]

And then we have 720 ways to visit six different cities. So the best option is:

D. 720

The number of different routes possible for a tourist visiting six cities in Ireland is 720, calculated by the formula for permutations of a set (6!).

The question is asking for how many different routes are possible if a tourist wants to visit six different cities in Ireland. This is an application of the mathematical principle of permutations, which is a way of counting or ordering objects. In this case, for 6 cities, we use the formula for permutations of a set, which is represented as n!, where n is the number of cities to visit.

For 6 cities, this would be 6!, which is calculated as 6 * 5 * 4 * 3 * 2 * 1 = 720. So, there are 720 different routes that the tourist can take to visit the six different cities.

Hence, option 'D. 720' is the correct answer.

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Answer:

10000

Step-by-step explanation:

Answer:

a) 0.0019

b) 0.0913

c) 9.13%

Step-by-step explanation:

We are given the following information in the question:

[tex]Q(t)=0.0019(2.22)^{-3t}[/tex]

The standard form of exponential function is

[tex]f(t) = ab^{t}[/tex]

where a is the initial amount and b is the base.

Rewriting the the given function, we have:

[tex]Q(t)=0.0019(2.22)^{-3t}\\Q(t)=0.0019((2.22)^{-3})^t\\Q(t)=0.0019(0.0913)^t[/tex]

a) Initial Value

Putting t = 0, we get,

[tex]Q(0)=0.0019(0.0913)^0 = 0.0019[/tex]

a = 0.0019

b) Growth factor

Comparing, we get, b = 0.0913

c) Growth rate

[tex]\text{Growth factor}\times 100\% = 0.0913\times 100\% = 9.13\%[/tex]

Answer:

[tex]t=\frac{85.8-113.4}{\sqrt{\frac{(50)^2}{145}+\frac{(75)^2}{141}}}}=-3.65[/tex]  

[tex]p_v =2*P(t_{(284)}<-3.65)=0.000312[/tex]

Using Excel we can use the following code: "=2*T.DIST(-3.65;284;TRUE)"

So the p value is a very low value and using any significance level giveen [tex]\alpha=0.05[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the means are different.

Step-by-step explanation:

1) Data given and notation

[tex]\bar X_{M}=85.8[/tex] represent the mean for the males

[tex]\bar X_{F}=113.4[/tex] represent the mean for females

[tex]s_{M}=50[/tex] represent the sample standard deviation for males

[tex]s_{F}=75[/tex] represent the sample standard deviation for female

[tex]n_{M}=145[/tex] sample size for the group poisoned

[tex]n_{F}=141[/tex] sample size for the group unpoisoned

t would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for the two groups are equal or not , the system of hypothesis would be:

Null hypothesis:[tex]\mu_{M} = \mu_{F}[/tex]

Alternative hypothesis:[tex]\mu_{M} \neq \mu_{F}[/tex]

If we analyze the size for the samples both are higher than 30 but the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

[tex]t=\frac{\bar X_{M}-\bar X_{F}}{\sqrt{\frac{s^2_{M}}{n_{M}}+\frac{s^2_{F}}{n_{F}}}}[/tex] (1)

t-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.

3) Calculate the statistic

We can replace in formula (1) the results obtained like this:

[tex]t=\frac{85.8-113.4}{\sqrt{\frac{(50)^2}{145}+\frac{(75)^2}{141}}}}=-3.65[/tex]  

4) Statistical decision

The first step is calculate the degrees of freedom, on this case:

[tex]df=n_{M}+n_{F}-2=145+141-2=284[/tex]

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

[tex]p_v =2*P(t_{(284)}<-3.65)=0.000312[/tex]

Using Excel we can use the following code: "=2*T.DIST(-3.65;284;TRUE)"

So the p value is a very low value and using any significance level giveen [tex]\alpha=0.05[/tex] always [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the means are different.

Answer:

We will need about 3.3 gallons of red paint.

Step-by-step explanation:

Let

The amount of red paint is half the amount of white paint.

r = 1/2 w

w = 2r [1]

We need to make a total of 10 gallons. Then,

r + w = 10 [2]

Replacing [1] in [2]

r + (2r) = 10

3r = 10

r = 3.3

We will need about 3.3 gallons of red paint.

To make 10 gallons of pink paint with red paint being half the amount of white paint, you would need 5 gallons of red paint.

The amount of red paint needed to make 10 gallons of pink paint when the amount of red paint is half that of white paint:

Answer:

17x² - 5x + 11 = 0

Step-by-step explanation:

To add or subtract a polynomial do the operation with like - terms.

i.e., Two terms of x² gets added or subtracted. Similarly terms of x and constant.

Here, we have to subtract:

[tex]$ 2x^2 + 9x - 3 - 7x^2 + 4x - 8 - 9x^2 - 13x + 5 - 9x^2 + 13x - 5 + 5x^2 - 13x - 1 + 5x^2 + 5x - 5 $[/tex]

Club all the like terms for easier simplification. We get:

[tex]$ (-2 - 7 - 9 - 9 + 5 + 5)x^2 + (9 + 4 - 13 + 13 - 13 + 5)x + (-3 - 2 + 5 - 5 - 1) $[/tex]

[tex]$ \implies - 17 x^2 + 5x - 11 = 0 $[/tex]

Multiplying by -1 throughout:

17x² - 5x + 11 = 0 is the answer.

Answer: z= -2.35

Step-by-step explanation:

When the population variation is know, then we calculate the z-statistic.

Formula to calculate the z-test statistic is given by :-

[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]

where , n= sample size

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

[tex]\mu[/tex] = Population mean

[tex]\sigma[/tex] =Population standard deviation.

As per given we have,

n= 270

[tex]\mu= 108[/tex]

[tex]\overline{x}=107[/tex]

[tex]\sigma^2=49\\\\\Rightarrow \sigma=\sqrt{49}=7[/tex]

Then, z-statistic  will be

[tex]z=\dfrac{107-108}{\dfrac{7}{\sqrt{270}}}[/tex]    (Substitute all the values )

[tex]\Rightarrow\ z=\dfrac{-1}{\dfrac{7}{16.4316767252}}[/tex]  

[tex]\Rightarrow\ z=\dfrac{-1}{0.426006433614}[/tex]  

[tex]\Rightarrow\ z=-2.34738238931\approx-2.35[/tex]  

Hence, the value of the test statistic= z= -2.35