A screening test for a newly discovered disease was evaluated in order to determine how well the test performs. Investigators administered the test to a random sample of 900 people. Of the participants with the disease, 150 tested positive and 60 tested negative. Of those without the disease, 50 tested positive. What is the sensitivity of the screening test?

A. 7.14%
B. 71.43%
C. 7.50%
D. 75.00%
E. None of the above

Answer:

Total Exercises each group will do =[tex]2\frac{x}{4}[/tex]

Total exercises individual need=   [tex]y=\frac{4x}{2}[/tex]      

                                                         [tex]y-\frac{4x}{2}[/tex]  

Step-by-step explanation:

No. of groups = 4

Each group has to do exercises = 2

Total no.  of exercises individual need = y

Total Exercises each group will do =[tex]2\frac{x}{4}[/tex]

Total exercises individual need= [tex]y=2\frac{x}{4}(4)[/tex]

                                                      [tex]y=\frac{4x}{2}[/tex]      

                                                       [tex]y-\frac{4x}{2}[/tex]  

Answer:

y-2{x/4}

Step-by-step explanation:

Answer:

Step-by-step explanation:

Matching each vocabulary word with its corresponding example:

(1). All Americans = f. Population (a group of items, units or subjects under reference of study e.g. inhabitants of a region, numbers of cars in a city, workers in a factory and so on)  

(2). The proportion of all Americans who strongly agree that prisoners with life sentences who have Alzheimer's should be released = c. Parameter ( the number that summarizes some characteristic of a population)  

(3). The proportion of the 85 Americans surveyed who strongly agree that prisoners with life sentences who have Alzheimer's should be released = d. Statistic (the sample characteristic corresponding to a population parameter used when a sample is used to make statistical inference about a population)

(4). The answer-Strongly Agree' or 'Agree, or Disagree, or Strongly Disagree = b. Variable (any quality, characteristic, quantity or number that can be counted or measured)

(5). The 85 Americans who participated in the survey = e. Sample (a part or fraction of a population selected on some basis)

(6). The list of answers that the 85 Americans gave = a. Data (the pieces of information collected to be used for the analysis)

Final answer:

In the research question provided, the key terms like Population, Parameter, Statistic, Sample, Variable, and Data are matched with examples based on a Likert-Response Scale survey on the release of prisoners with Alzheimer's who have life sentences.

Explanation:

A researcher inquiring about the public's view on releasing prisoners with Alzheimer's who have a life sentence used a Likert-Response Scale to understand the level of agreement with a statement. This method has several components we can define using the given options:

Population - (B) All Americans

Parameter - (C) The proportion of all Americans who strongly agree that prisoners with life sentences who have Alzheimer's should be released

Statistic - (D) The proportion of the 85 Americans surveyed who strongly agree that prisoners with life sentences who have Alzheimer's should be released

Variable - (CV) The answer 'Strongly Agree' or 'Agree, or Disagree, or Strongly Disagree'

Sample - (E) The 85 Americans who participated in the survey

Data - (F) The list of answers that the 85 Americans gave

In this study, the variable constitutes the different levels of agreement, while the data consists of the actual responses collected from the survey participants. The sample involves the subset of the population, which is the 85 Americans who were surveyed, and the statistic is the result derived from this sample. The population, on the other hand, includes all Americans, and the parameter is the value that would be obtained if all Americans could be surveyed.

Answer:

Policeman is correct.

Step-by-step explanation:

Consider the provided information.

The tires left three skid marks of 69 ft, 70 ft, and 74 ft.

The road had a drag factor of 0.95. Your brakes were operating at 98% efficiency.

Thus, drag factor (f) = 0.95 and brakes efficiency (n) = 0.98

Compute the skid distance by finding the mean of length of 3 skids.

[tex]D=\frac{69+70+74}{3}\\D=\frac{213}{3}\\D=71[/tex]

Therefore, the skid distance is 71 ft.

Now find the speed of car using skid speed formula:

[tex]S=\sqrt{30Dfn}[/tex]

Where, S is the speed of car, D is the skid distance, f is the drag factor and n is the beak efficiency.

Substitute the respective values in above formula.

[tex]S=\sqrt{30\times 71\times0.95 \times0.98}[/tex]

[tex]S=\sqrt{1983.03}[/tex]

[tex]S\approx 44.5312[/tex]

Your speed was 44.5312 mi/h which is greater than 40 mi/hr.  So policeman is correct.

Answer:

[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]  

[tex]p_v =P(t_{12}>3.036) =0.0051[/tex]  

So with the p value obtained and using the significance level given [tex]\alpha=0.05[/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 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).  

Step-by-step explanation:

The system of hypothesis on this case are:  

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

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

Or equivalently:  

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

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

Our notation on this case :  

[tex]n_1 =8[/tex] represent the sample size for group 1 (Orno)  

[tex]n_2 =6[/tex] represent the sample size for group 2  (Edne)

We can calculate the sampel means and deviations with the following formulas:

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

[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

[tex]\bar X_1 =10[/tex] represent the sample mean for the group 1  

[tex]\bar X_2 =7[/tex] represent the sample mean for the group 2  

[tex]s_1=2.268[/tex] represent the sample standard deviation for group 1  

[tex]s_2=1.414[/tex] represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a right tailed test.

On this case since the significance assumed is 0.05 and we are conducting a bilateral test we have one critica value, and we need on the right tail of the distribution [tex]\alpha/2 = 0.05[/tex] of the area.  

The distribution on this case since we don't know the population deviation for both samples is the t distribution with [tex]df=8+6 -2=12[/tex] degrees of freedom.

We can use the following excel code in order to find the critical value:

"=T.INV(1-0.05,12)"

And the rejection zone is: (1.78,infinity)

The statistic is given by this formula:  

[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{S^2_2}{n_2}}}[/tex]  

And now we can calculate the statistic:  

[tex]t=\frac{(10 -7)-(0)}{\sqrt{\frac{2.268^2}{8}+\frac{1.414^2}{6}}}=3.036[/tex]  

The degrees of freedom are given by:  

[tex]df=8+6-2=12[/tex]

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

[tex]p_v =P(t_{12}>3.036) =0.0051[/tex]  

So with the p value obtained and using the significance level given [tex]\alpha=0.05[/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 5% of significance the mean of the group 1 (Orno) is significantly higher than the mean for the group 2 (Edne).  

Answer:

-0.4

B. The answer is the same irrespective of the sample sizes.

Step-by-step explanation:

Given that an article in the November 1983 Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline AA batteries and Eveready Energizer Alkaline AA batteries were given as 4.1 hours and 4.5 hours, respectively

Let X be the sample average lifetime of 81 Duracell and Y be the sample average lifetime of 81 Eveready Energizer batteries.

Now we have sample mean difference does not depend about the sample sizes.

This is because

E(X-Y) = E(X)-E(Y) for all X and Y

Mean of X-Y = [tex]4.1-4.5=-0.4[/tex]

This does not depend on the specified sample sizes

B. The answer is the same irrespective of the sample sizes.

The mean value of X - Y is 0.4 hours. The answer does not depend on the sample sizes.

The mean value of X - Y, where X is the sample average lifetime of 81 Duracell batteries and Y is the sample average lifetime of 81 Eveready Energizer batteries, is given by the difference in the population average lifetimes of the two types of batteries. In this case, the difference is 4.5 - 4.1 = 0.4 hours. So the distribution of X - Y is centered around 0.4 hours.

The answer does not depend on the specified sample sizes in this case. The mean value of X - Y remains the same regardless of the sample sizes, as long as the samples are representative of the populations and the population parameters do not change. So the answer is B. The answer is the same irrespective of the sample sizes.

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Answer:

B. H0: p1 − p2 ≤ 0; HA: p1 − p2 > 0

[tex]z=\frac{0.447-0.414}{\sqrt{0.431(1-0.431)(\frac{1}{150}+\frac{1}{140})}}=0.57[/tex]    

[tex]z_{crit}=1.64[/tex]

A. Do not reject H0; there is no increase in the proportion of people using LinkedIn

Step-by-step explanation:

1) Data given and notation  

[tex]X_{1}=67[/tex] represent the number of recent jobseekers

[tex]X_{2}=58[/tex] represent the number of job seekers three years ago.

[tex]n_{1}=150[/tex] sample of recent jobseekers selected  

[tex]n_{2}=140[/tex] sample of job seekers three years ago selected  

[tex]p_{1}=\frac{67}{150}=0.4468[/tex] represent the proportion of recent jobseekers

[tex]p_{2}=\frac{58}{140}=0.4143[/tex] represent the proportion of job seekers three years ago

z would represent the statistic (variable of interest)  

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

[tex]\alpha=0.05[/tex] significance level given

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if "More people are using social media to network, rather than phone calls or e-mails", the system of hypothesis would be:  

Null hypothesis:[tex]p_{1} - p_{2} \leq 0[/tex]  

Alternative hypothesis:[tex]p_{1} - p_{2} > 0[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex]   (1)  

Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{67+58}{150+140}=0.4310[/tex]  

3) Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.4468-0.4143}{\sqrt{0.4310(1-0.4310)(\frac{1}{150}+\frac{1}{140})}}=0.5671[/tex]    

In order to find the critical value since we have a right tailed test the we need to find a value on the z distribution that accumulates 0.05 of the area on the right tail, and this value is[tex]z_{crit}=1.64[/tex].

4) Statistical decision

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

[tex]p_v =P(Z>0.5671)= 0.285[/tex]  

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

So the correct conclusion would be:

A. Do not reject H0; there is no increase in the proportion of people using LinkedIn

Answer:

Maximum profit is $87 when 3 blenders and 11 mixers are produced.

Step-by-step explanation:

let blender is represented by [tex]x_{1}[/tex] and and mixer by [tex]x_{2}[/tex].

total time to deliver parts = 24 hrs

total time to assemble = 30 hrs

time taken by each blender to deliver parts = 1 hr

time taken by each mixer to deliver parts = 2 hr

time taken by blenders in final assembling= 2 hr

time taken by mixers in final assembling = 3 hr

Each blender produced nets the firm=  $7

Each mixer produced nets the firm=  $6

Using this all data linear system of equation will be:

[tex]x_{1} + 2x_{2} =24  ----- (1)\\2x_{1} + 2x_{2} = 30 ----- (2)\\[/tex]

profit function:

[tex]z= 7x_{1} +6x_{2} --- (3)[/tex]

[tex]from (1)\\x_{1} = 0 \implies x_{2}= 12\\x_{2}= 0 \implies x_{1}= 24\\[/tex]

Coordinate points obtained from (1) are (0,12) and (24,0)

[tex]from (2)\\x_{1}=0 \implies x_{2}=10\\x_{2}=0 \implies x_{1}=15\\[/tex]

Coordinate points obtained from (2) are (0,10) and (15,0)

plotting these on graph

points lying in feasible region are:

A(0,0)

B(0,10)

C(3,11)

D(12,0)

substituting these points in (3) to find the maximum profit:

for A (0,0)

z = 0

for B (0,10)

z = 60

for C (3,11)

z =  87

for D (12,0)

z=84

So maximum profit is $87 when 3 blenders and 11 mixers are produced.

Answer:

Reject the null hypothesis. There is sufficient evidence to prove that the mean life is different from 7463 hours.

95% confidence interval also supports this result.

Step-by-step explanation:

Let mu be the population mean life of a large shipment of CFLs.

The hypotheses are:

[tex]H_{0}[/tex]: mu=7463 hours

[tex]H_{a}[/tex]: mu≠7463 hours

Test statistic can be calculated using the equation:

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

Then z=[tex]\frac{7163-7463}{\frac{1080}{\sqrt{81} } }[/tex] = -2.5

p-value is  0.0124, critical values at 0.05 significance are ±1.96

At the 0.05 level of​ significance, the the result is significant because 0.0124<0.05. There is significant evidence that mean life of light bulbs is different than 7463 hours.

95% Confidence Interval can be calculated using M±ME where

margin of error (ME) from the mean can be calculated using the formula

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

Then ME=[tex]\frac{1.96*1080}{\sqrt{81} }[/tex] =235.2

Thus 95% confidence interval estimate of the population mean life of the light bulbs is 7163±235.2 hours. That is between 6927.8 and 7398.2 hours.

Answer:

a) [tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]  

[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]  

b) The 90% confidence interval would be given by (63.330;81.070)

[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]  

Step-by-step explanation:

1) Previous concepts  and notation

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".

Let put some notation  

x=left arm , y = right arm  

x: 175 169 182 146 144  

y: 102 101 94 79 79

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

d:  73, 68, 88, 67, 65

The second step is calculate the mean difference  

[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2[/tex]  

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

[tex]s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311[/tex]  

2) Confidence interval

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

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

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=5-1=4[/tex]  

Since the 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: "=-T.INV(0.05,4)".And we see that [tex]t_{\alpha/2}=2.13[/tex]  

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

[tex]72.2-2.13\frac{9.311}{\sqrt{5}}=63.330[/tex]  

[tex]72.2+2.13\frac{9.311}{\sqrt{5}}=81.070[/tex]  

So on this case the 90% confidence interval would be given by (63.330;81.070)

[tex]63.330 < \mu_{left}- \mu_{right} <81.070[/tex]  

Answer:

0.0561,0.0673,0.999997

Step-by-step explanation:

Given that a sign on  the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 8 cars needs to have oil added.

X- no of cars that need oil in a sample of 8 cars is binomial with p = 1/8

since each car is independent of the other and there are only two outcomes.

a) Exactly three out of the next eight cars needs oil.

Probability = [tex]P(X=3) =0.0561[/tex]

b) At least three out of the next eight cars needs oil.

Probability = [tex]P(X\geq 3)\\= 0.0673[/tex]

c) At most 6 out of the next 8 cars needs oil.

Probability =[tex]P(X\leq 6)\\=0.999997[/tex]

Answer:

Probability=1-[tex]\frac{(5!)^{3} }{15!}[/tex]

Step-by-step explanation:

We have to solve this question with the help of complementary method. Observe the statement "Probability that every family has at least two of its children in the same tent". The complementary of this statement will be every family does not have atleast two of its children in the same tent ie. Each child of a family is not in the same tent as one from the same family.

P(A)=1-[tex]P(A)^{c}[/tex]

Therefore, we can get P(A) if we just take out the value of [tex]P(A)^{c}[/tex]

Probability=[tex]\frac{TotalNo.OfFavourableOutcomes}{TotalNo.ofOutcomes}[/tex]

Imagine that we have 15 locations to fill and 15 people for it, so the total no. of cases= 15!

Bernstein's children can be arranged in 5 different tents in 5! ways.

Similarly Henderson's and Smith's children can be arranged in 5 different tents in 5! ways only.

Therefore, [tex]P(A)^{c}[/tex]= [tex]\frac{(5!)^{3} }{15!}[/tex]

P(A)=1- [tex]\frac{(5!)^{3} }{15!}[/tex]

Answer: the amount of money invested in the first property is $140000

the amount of money invested in the second is property is $62353

the amount of money invested in the third is property is $20235

Step-by-step explanation:

Let x represent the amount of money invested in the first property.

Let y represent the amount of money invested in the second property.

Let z represent the amount of money invested in the third property.

The man has S400,000 invested in three rental properties. This means that

x + y + z = 400000 - - - - - - - - -1

The first property earns 7.5% per year on the investment, the second earns 8%, and the third earns 9%. The total annual earnings from the three properties is S33,700. This means that

7.5/100 × x + 8/100×y + 9/100×z = 33700

0.075x + 0.08y + 0.09z = 33700 - - - - - 2

The amount invested at 9% equals the sum of the first two investments. This means that

z = x + y - - - - - - - - - - -3

Substituting equation 3 into equation 1 and equation 2, it becomes

x + y + x + y = 400000

2x + 2y = 400000 - - - - - - - - 4

0.075x + 0.08y + 0.09(x + y) = 33700

0.075x + 0.08y + 0.09x + 0.09y = 33700

0.165x + 0.17y = 33700 - - - - - - - - 5

Multiplying equation 4 by 0.165 and equation by 2, it becomes

0.33x + 0.33y = 66000

0.33x + 0.34y = 67400

Subtracting

- 0.01x= - 1400

x = $140000

Substituting x = 140000 into equation 5, it becomes

0. 165(140000) + 0.17y = 33700

23100 + 0.17y = 33700

0.17y = 33700 - 23100 = 10600

y = 10600/0.17 = $62353

z = x + y = 140000 + 62353 = $202353

Answer:

a.the goodness of fit for the estimated multiple regression equation increases.

Step-by-step explanation:

As the value of the multiple coefficient of determination increases,

a. the goodness of fit for the estimated multiple regression equation increases.

As we know that the coefficient of determination measures the variability of response variable with the help of regressor. As we know that if the value of the coefficient of determination increases strength of fit also increases.  

To determine if the mean hourly wage of employees in the manufacturing industry differs from the reported mean for the goods-producing industries, a hypothesis test can be conducted. Steps include setting up null and alternative hypotheses, selecting a significance level, collecting a sample, calculating a test statistic, determining critical values or p-values, and making a conclusion based on the results.

To determine if the mean hourly wage of employees in the manufacturing industry differs from the reported mean of $24.57 for the goods-producing industries, we can conduct a hypothesis test.

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Answer:

Option A) 0.0127  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 16 weeks

Standard Deviation, σ = 2 weeks

Sample size = 20

We are given that the distribution of time taken to find a job is a bell shaped distribution that is a normal distribution.

Formula:  

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

P(20 young workers average less than 15 weeks)  

P(x < 15)  

[tex]P( x < 15) = P( z < \displaystyle\frac{15-16}{\frac{2}{\sqrt{20}}}) = P(z < -2.236)[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]P(x < 15) =0.0127= 1.27\%[/tex]

Thus, 0.0127 is the probability that 20 young workers average less than 15 weeks to find a job.

The probability of a group of 20 workers finding jobs in less than 15 weeks, given the specific mean and standard deviation, was calculated using z-scores and the normal distribution. The result was approximately 0.0571, but this was an interpretation that wasn't presented as an answer choice.

In this question, we're asked to calculate the probability that a group of 20 young workers on average find a job in less than 15 weeks, given that the mean is 16 weeks and standard deviation is 2 weeks. This is a problem of statistics, specifically involving the

normal distribution

and

z-scores

. The z-score is calculated using the formula Z = (X - μ) / (σ/√n), where X is the random variable (15 weeks in this case), μ is the population mean (16 weeks), σ is the standard deviation (2 weeks) and n is the sample size (20 workers). This gives us Z = (15 - 16) / (2/√20), which is approximately -1.58. Looking up this z-score in a standard normal distribution table gives us a probability of 0.0571, which isn't an option provided in the question. However, we may have made an error in calculation or interpretation. It's important to thoroughly understand and practice these statistical concepts.

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Answer:

n=500 represent the random sample taken    

[tex]\hat p=0.12[/tex] estimated proportion of people that chose chocolate​ pie

[tex]\hat q =1-\hat p=1-0.12=0.88[/tex] represent the people that NOT chose chocolate​ pie

E=0.05 represent the error or margin of error given by the following formula:

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

p= true population proportion of people that chose chocolate​ pie

If the confidence level is 90 ​%, what is the value of alpha ​?

[tex]\alpha=1-0.9 =0.1[/tex] and the value of [tex]\alpha/2 =0.05[/tex],

[tex]z_{\alpha/2}=-1.64[/tex] and [tex]z_{1-\alpha/2}=1.64[/tex]

[tex]ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238[/tex]

Step-by-step explanation:

Data given and notation

What values do ModifyingAbove p with caret ​, ModifyingAbove q with caret ​, ​n, E, and p​ represent?

n=500 represent the random sample taken    

X represent the people that chose chocolate​ pie

[tex]\hat p=0.12[/tex] estimated proportion of people that chose chocolate​ pie

[tex]\hat q =1-\hat p=1-0.12=0.88[/tex] represent the people that NOT chose chocolate​ pie

E=0.05 represent the error or margin of error given by the following formula:

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

z would represent the quantile of the normal standard distribution

p= true population proportion of people that chose chocolate​ pie

The confidence interval for the population proportion is given by this formula :

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

If the confidence level is 90 ​%, what is the value of alpha ​?

On this case the value for the significance would be [tex]\alpha=1-0.9 =0.1[/tex] and the value of [tex]\alpha/2 =0.05[/tex], we can find the quantiles of the normal standard distribution given by:

[tex]z_{\alpha/2}=-1.64[/tex] and [tex]z_{1-\alpha/2}=1.64[/tex]

And with the following excel codes:

"=NORM.INV(0.05,0,1)" "=NORM.INV(1-0.05,0,1)"

And we can find the margin of error like this:

[tex]ME=1.64 \sqrt{\frac{0.12(1-0.12)}{500}}=0.0238[/tex]

Final answer:

In the poll,
ModifyingAbove p with caret and
ModifyingAbove q with caret represent the sample proportions of respondents who chose and did not choose chocolate pie, respectively. The symbol n represents the sample size, E denotes the margin of error, and p is the population proportion. If the confidence level is 90%, the value of alpha is 0.10.

Explanation:

In the context of the provided poll scenario, the various symbols represent the following statistical terms:
ModifyingAbove p with caret (
ModifyingAbove p with caret) represents the sample proportion, which is the observed percentage in the sample that chose chocolate pie. In this case, it would be 0.12 or 12%.
ModifyingAbove q with caret (
ModifyingAbove q with caret) is the sample proportion of respondents who did not choose chocolate pie, which would be 1 - 0.12 = 0.88 or 88%.

n represents the sample size, which is the number of respondents in the poll, 500 in this scenario.

E denotes the margin of error, which is
plus or minus 5 percentage points in this case.

p represents the population proportion, which is the true percentage of all adults who would choose chocolate pie as their favorite if everyone was surveyed.

If the confidence level is 90%, the value of alpha (
alpha) is the probability that the true population parameter will not be contained in the confidence interval. For a 90% confidence level, alpha would be 1 - 0.90 = 0.10 or 10%. This is often split into two tails of the normal distribution for a two-tailed test, thus each tail would have an area of 0.05.

Answer:

The required formula for y, the number of cars sold is: [tex]y = 0.008x +200[/tex].

Step-by-step explanation:

Consider the provided information.

Let x represents the amount of money spend on advertising .

y is number of cars sold,

For each additional 5000 dollars spent, they sold 40 more cars.

[tex]Slope = m = \frac{Rise}{Run}=\frac{40}{5000}=0.008[/tex]

The y-intercept form is: [tex]y = mx + b[/tex]

Substitute x=60,000, y=680 and m=0.008 in the above equation.

[tex]680= 0.008(60,000)+b[/tex]

[tex]b=680-480[/tex]

[tex]b=200[/tex]

Thus, the required formula for y, the number of cars sold is: [tex]y = 0.008x +200[/tex].

To create a formula for the number of cars sold (y) as a function of advertising expenditure (x, in thousands of dollars), we use the data points provided to establish a linear equation: y = 8x + 200, where x represents the advertising spending and y the number of cars sold.

To find a formula for y, the number of cars sold, given the linear relationship between advertising expenditure (x, in thousands of dollars) and sales, we first establish the given data points:

An additional $5,000 in advertising (or an increase of 5 in x since it's measured in thousands) results in selling 40 more cars.

We can express this relationship with the linear equation y = mx + b, where m is the slope (change in y over change in x) and b is the y-intercept (the number of cars sold when no money is spent on advertising).

Let's calculate the slope, m:

For every increase of 5 in x, y increases by 40, so m = 40/5 = 8.

We can then use this slope and one of the data points to find the y-intercept, b:

680 = 8(60) + b, so b = 680 - (8*60) = 680 - 480 = 200.

The linear equation representing the relationship between advertising spend and cars sold is:

y = 8x + 200

Answer: 33 bananas

Step-by-step explanation:

Since the monkey can only carry 100 bananas at a time.

And a total of 200 bananas is to be transferred through 100yards.

With the condition that he eats 1 banana per yard .

If he decides to carry 100 banana straight through the whole 100yards. He would be left with no banana at the end of the journey and will not have enough banana to be able to come back and carry the rest.

Therefore, the best way is to transfer all the banana to a destination where he will be left with 100bananas that he can then move at once to the final destination.

To move 200 bananas to a particular destination given that he can carry only 100 banana each.

He would need to travel twice that means ( three trips i.e leave 100 return with 0 leave 0)

For three trips through a particular length in yards to exhaust 100 yards. The length in yards is given as

100 = 3 × l

l = 100/3

l = 33 yards

Travelling 33 yards he would be left with 100 bananas that he can carry in just one trip through the rest of the journey which is 100 -33 yards = 67yards

Carrying 100 bananas through 67 yards in one trip

He will exhaust;

1 banana per yard × 67yards = 67 bananas

Therefore, he would be left with

100 - 67 bananas

= 33 bananas

Answer:

a) The 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

b) Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% of significance.  

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]p[/tex] represent the real population proportion of interest  

[tex]\hat p =0.09[/tex] represent the estimated proportion for the sample  

n=800 is the sample size required  

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

Confidence =0.95 or 95%

Part a

The population proportion have the following distribution  

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

The confidence interval would be given by this formula  

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

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

[tex]z_{\alpha/2}=1.96[/tex]  

The margin of error is given by :

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

[tex]Me=1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0198[/tex]

And replacing into the confidence interval formula we got:  

[tex]0.09 - 1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0702[/tex]  

[tex]0.09 + 1.96 \sqrt{\frac{0.108(1-0.09)}{800}}=0.1098[/tex]  

And the 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

Part b

Since the confidence interval NOT contains the value 0.06 we  have anough evidence to reject the claim at 5% of significance.

The 95% confidence interval for the true percentage of men taking on second jobs is approximately 7.02% to 10.98%. The pundit's claim of 6% is not plausible.

To construct a 95% confidence interval for the true percentage of men taking on second jobs, we'll use the sample proportion (\( \hat{p} \)) and the margin of error formula. Then, we'll use this interval to test the pundit's claim.

Given:

- Sample size ( n ) = 800

- Sample proportion [tex](\( \hat{p} \))[/tex] = 9% = 0.09

a) Constructing a 95% confidence interval:

The margin of error ( E ) for a 95% confidence interval can be calculated using the formula:

[tex]\[ E = Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \][/tex]

Where:

-  Z  is the Z-score corresponding to the desired confidence level (95%)

- [tex]\( \hat{p} \)[/tex] is the sample proportion

-  n  is the sample size

We can find the Z-score corresponding to a 95% confidence level, which is approximately 1.96.

Now, let's calculate the margin of error and construct the confidence interval.

b) Testing the pundit's claim:

We'll compare the pundit's claim (6%) with the confidence interval constructed in part (a). If the pundit's claim falls within the confidence interval, it is plausible. Otherwise, it is not plausible.

Let's proceed with the calculations.

a) Constructing a 95% confidence interval:

First, let's calculate the margin of error ( E ):

[tex]\[ E = 1.96 \times \sqrt{\frac{0.09 \times (1 - 0.09)}{800}} \][/tex]

[tex]\[ E \approx 1.96 \times \sqrt{\frac{0.09 \times 0.91}{800}} \][/tex]

[tex]\[ E \approx 1.96 \times \sqrt{\frac{0.0819}{800}} \][/tex]

[tex]\[ E \approx 1.96 \times \sqrt{0.000102375} \][/tex]

[tex]\[ E \approx 1.96 \times 0.010118 \][/tex]

[tex]\[ E \approx 0.0198 \][/tex]

Now, let's construct the confidence interval:

Lower bound: [tex]\( \hat{p} - E = 0.09 - 0.0198 = 0.0702 \)[/tex]

Upper bound: [tex]\( \hat{p} + E = 0.09 + 0.0198 = 0.1098 \)[/tex]

So, the 95% confidence interval for the true percentage of men taking on second jobs is approximately [tex]\( (7.02\%, 10.98\%) \).[/tex]

b) Testing the pundit's claim:

The pundit's claim is 6%, which falls below the lower bound of the confidence interval (7.02%). Therefore, it is not plausible given the poll data.

Answer:

b) 36

Step-by-step explanation:

We can use combinations to solve this problem.

The binomial coefficient [tex]\binom{n}{k}=\frac{n!}{k!(n-k)!}[/tex] counts the number of ways of choose k elements from a set of n elements.

The product rule from combinatorics says that if there are N ways of doing something and M ways of doing another thing, the number of ways of doing both things is equal to NM.

First, we choose the blue balls. The urn contains 4 blue balls and we select 2 so there are [tex]N=\binom{4}{2}=6[/tex] ways of doing this. Similarly, we choose the 5 orange balls from the set of 6 in the urn, which can be done in [tex]M=\binom{6}{5}=6[/tex] ways. By the product rule, there are MN=6(6)=36 ways of selecting all the balls.

The number of ways 2 blue balls and 5 orange balls can be selected from the urn is 36 number of ways: Option C is correct

Combination has to do with selection.

If r number is selected from n number, this is expressed using the formula:

[tex]nC_r=\frac{n!}{(n-r)!r!}\\[/tex]

If 2 blue balls are selected from 4 blue balls, this is expressed as:

[tex]4C_2=\frac{4!}{(4-2)!2!}\\4C_2=\frac{4\times 3 \times 2!!}{2!2!}\\4C_2=\frac{12}{2} = 6 ways[/tex]

Similarly, if 5 orange balls are selected from 5 orange balls, this is expressed as:

[tex]6C_5=\frac{6!}{(6-5)!5!}\\6C_5=\frac{6\times 5!}{1!5!}\\6C_5=\frac{6}{1} = 6 ways[/tex]

The number of ways 2 blue balls and 5 orange balls can be selected from the urn is 36 number of ways

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Answer:

x + 3y - 16 = 0

Step-by-step explanation:

When two points, say [tex]$ (x_1, y_1) $[/tex] and [tex]$ (x_2, y_2) $[/tex] are given, the equation is determined using Two - point form.

The two - point form is as follows:

[tex]$ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} $[/tex]

Here: [tex]$ (x_1, y_1) = (-6, 7) $[/tex] and [tex]$(x_2, y_2) = (-3, 6) $[/tex].

Substituting in the formula we get the equation of the line as:

[tex]$ \frac{y - 7}{6 - 7} = \frac{x + 6}{- 3 + 6}[/tex]

[tex]$ \implies \frac{y - 7}{- 1} = \frac{x + 6}{ 3} $[/tex]

[tex]$ \implies 3y - 21 = - x - 6 $[/tex]

Rearranging we get: x + 3y - 15 = 0

This is the required equation of the line.

Answer:

Step-by-step explanation:

Final answer:

The decision rule for the hypothesis test at a 0.10 significance level involves rejecting the null hypothesis if the test statistic is greater than the critical value. The calculated Z-value is approximately 1.8233, and the corresponding p-value will determine our decision to reject or not reject the null hypothesis.

Explanation:

Decision Rule and Test Statistic

For a hypothesis test comparing two population means with known standard deviations, we use a Z-test for two samples. The decision rule at a significance level of 0.10 involves rejecting the null hypothesis (H0: μ1 = μ2) if the absolute value of the test statistic exceeds the critical value from the standard normal distribution.

Calculation of Test Statistic

The test statistic (Z) is calculated using the formula: Z = (101.0 - 99.3)/ √[ (4.6²/40) + (4.0²/47) ]

Plugging in the numbers: Z = (1.7)/ √[ (21.16/40) + (16/47) ] Z = (1.7)/ √[ 0.529 + 0.3404255 ] Z = (1.7)/√[0.8694255] Z = (1.7)/0.932447 Z ≈ 1.8233

Decision Based on P-value

The p-value associated with this Z-value can be found using standard normal distribution tables or software. If the p-value is less than 0.10, we reject the null hypothesis. Otherwise, we fail to reject it.

Answer:

a) The mean of the sampling distribution of the sample means is 92.

b) The variance of the sampling distribution of the sample mean is 3.24.

c) The standard error of the sampling distribution of the sample mean is 1.8.

d) 28.77% probability that the sample mean ex-ceeds 93.0 units.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 92, \sigma = 3.6[/tex].

a. Find the mean of the sampling distribution of thesample means.

The mean is the same as the mean of the population. So the mean of the sampling distribution of the sample means is 92.

b. Find the variance of the sampling distribution ofthe sample mean.

The standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{4}} = 1.8[/tex]

The variance is [tex]s^{2} = 1.8^{2} = 3.24[/tex]

c. Find the standard error of the sampling distribution of the sample mean.

This is the same as the standard deviation of the sample. So the standard error of the sampling distribution of the sample mean is 1.8.

d. What is the probability that the sample mean ex-ceeds 93.0 units?

This is 1 subtracted by the pvalue of Z when X = 93. So

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

[tex]Z = \frac{93 - 92}{1.8}[/tex]

[tex]Z = 0.56[/tex]

[tex]Z = 0.56[/tex] has a pvalue of 0.7123.

So there is a 1-0.7123 = 0.2877 = 28.77% probability that the sample mean ex-ceeds 93.0 units.

The mean of the sampling distribution of the sample means is 92.0. The variance is 0.81. The standard error is 0.9.

a. The mean of the sampling distribution of the sample means can be calculated by finding the mean of the original population, which is 92.0.

b. The variance of the sampling distribution of the sample mean can be calculated using the formula (standard deviation/original population size)^2. In this case, it is (3.6/4)^2 = 0.81.

c. The standard error of the sampling distribution of the sample mean can be calculated by taking the square root of the variance. So, in this case, it is √0.81 = 0.9.

d. To find the probability that the sample mean exceeds 93.0 units, you can calculate the z-score using the formula (sample mean - population mean)/standard error. Then, you can find the probability of the z-score using a standard normal distribution table or calculator.

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Answer: 176.

Step-by-step explanation:

Formula to find the sample size is given by :-

[tex]n=(\dfrac{z^*\cdot \sigma}{E})^2[/tex]

, where [tex]\sigma[/tex] = Population standard deviation from prior study.

E = margin of error.

z* = Critical value.

As per given , we have

[tex]\sigma=\$675000[/tex]

E= $100,000

We take 95% confidence interval.

Critical value (Two tailed)=[tex]z^*=1.96[/tex]

The required sample size = [tex]n=(\dfrac{(1.96)\cdot 675000}{100000})^2[/tex]

[tex]n=(13.23)^2\\\\ n=175.03299\approx176[/tex] [Round to next integer]

Hence, the required sample size = 176.

Answer:

19

Step-by-step explanation:

Given:

measuring cup size = 1/4 cup

total amount to be measured,

= 4 3/4 cups (convert to improper fraction)

= 19/4 cups.

number of 1/4 cups

= total amount measured ÷ 1/4 cup

= 19/4 ÷ 1/4

= 19/4 x 4/1

= 19/4 x 4

= 19         (1/4-cups) used

To figure out how many 1/4 cups Mel will need, we need to convert 4 3/4 cups to quarters. As 4 cups equal 16 quarters and 3/4 equals 3 quarters, we have a sum of 19 quarters. Therefore, Mel will need 19 measures of his 1/4 cup.

To solve this problem, you need to understand how to convert measurements using fractions. We know that Mel needs 4 3/4 cups of flour and the only measuring cup he has measures 1/4 cup.

The first thing we can do is convert 4 3/4 into an improper fraction. One whole cup is 4 quarters, so 4 cups are 4 x 4 = 16 quarters. Additionally, there are 3 more quarter cups, totaling 16 + 3 = 19 quarter cups.

So, Mel needs 19 quarter cups of flour.

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Answer:

[tex]0.0864 < p< 0.1358[/tex]

We are confident (95%) that the true proportion of people admitted they did not buy a ticket is betwen 0.0864 and 0.1358.  

Step-by-step explanation:

1) Data given and notation  

n=621 represent the random sample taken    

X=69 represent the people admitted they did not buy a ticket

[tex]\hat p=\frac{69}{621}=0.111[/tex] estimated proportion of people admitted they did not buy a ticket

[tex]\alpha=0.05[/tex] represent the significance level (no given, but is assumed)    

z would represent the statistic

p= population proportion of people admitted they did not buy a ticket

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".

2) Confidence interval

The confidence interval would be given by this formula

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

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

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

And replacing into the confidence interval formula we got:

[tex]0.111 - 1.96 \sqrt{\frac{0.111(1-0.111)}{621}}=0.0864[/tex]

[tex]0.111 + 1.96 \sqrt{\frac{0.111(1-0.111)}{621}}=0.1358[/tex]

And the 95% confidence interval would be given (0.0864;0.1358).

[tex]0.0864 < p< 0.1358[/tex]

We are confident (95%) that the true proportion of people admitted they did not buy a ticket is betwen 0.0864 and 0.1358.  

Answer:The type of the problem described above is a Sinking Fund

Option B

Step-by-step explanation:

In order to understand the solution to this question we have to be familiar with these concepts  

Sinking Fund

A sinking fund is an account earning compound interest into which you make periodic deposits. Suppose that the account has an annual interest rate of (r) compounded (m) times per year, so that (i=r/m) is the interest rate per compounding period. If you make a payment of PMT at the end of each period, then the future value after (t) years, or (n = mt) periods, will be  

   FV = [tex] PMT  (〖(1+i )〗^n  -1)/i [/tex]

Where FV is the amount that would be accumulated after t years    

Payment Formula for a Sinking Fund  

Suppose that an account has an annual rate of (r) compounded (m) times per year, so that is (i=r/m) is the interest rate per compounding period. If you want to accumulate a total of FV in the account after t years, or (n = mt) periods, by making payments of PMT at the end of each period, then each payment must be

                                           PMT = [tex] FV  ( i)/(〖(1+i)〗^n  -1) [/tex]

From the question  

 Rate= r = 3/100 = 0.03

Number of times it was paid (compounded) in a year = m = 4 its value is Four cause the payment is made 4 times in one year i.e. Quarterly  

The interest rate per compounding period = I = r/m = 0.03/4 = 0.0075

Number of times it was paid (compounded) t years n = 4 x 3 = 12

The amount that ted desires to be in that account  after 3 years =FV = $25,000  

   So the investment that Ted needs to make Quarterly in order to get his desired amount is  

               = [tex]25000 × (0.0075/(〖(1+0.0075)〗^12  -1 )) [/tex]

                = $2000  

The correct option is b. Sinking Fund. Ted should invest approximately $22,857.14 each quarter to have $25,000 in 3 years with an annual interest rate of 3% compounded quarterly.

To determine how much Ted should invest each quarter into an account that pays 3% per year compounded quarterly, we can use the sinking fund formula:[tex]\[ P = \frac{FV}{\left(1 + \frac{r}{m}\right)^{n \cdot m}} \][/tex]

where:

P is the payment made each period (quarterly in this case),

FV is the future value of the investment, which is $25,000,

r is the annual interest rate (3% or 0.03),

m is the number of times interest is compounded per year (4 for quarterly),

n is the number of years (3).

First, we need to adjust the interest rate for quarterly compounding:

[tex]\[ \text{Quarterly interest rate} = \frac{r}{m} = \frac{0.03}{4} = 0.0075 \][/tex]

Next, we calculate the number of total compounding periods:

[tex]\[ n \cdot m = 3 \cdot 4 = 12 \][/tex]

Now we can plug these values into the sinking fund formula to solve for P:

[tex]\[ P = \frac{25000}{\left(1 + 0.0075\right)^{12}} \][/tex]

[tex]\[ P = \frac{25000}{\left(1.0075\right)^{12}} \][/tex]

[tex]\[ P = \frac{25000}{1.0934} \][/tex]

[tex]\[ P \approx \frac{25000}{1.0934} \][/tex]

[tex]\[ P \approx 2285.71 \][/tex]

Therefore, Ted should invest approximately $22,857.14 each quarter to have $25,000 in 3 years with an annual interest rate of 3% compounded quarterly.

Answer:

V = 8π/3

Step-by-step explanation:

Haven't done calc in years so I might be wrong.

Graph out all the lines needed so you can have a better look at it.

I set y=2x = y=2 to find where the intersect each others so I can have my boundaries for integration.

You goal is to find the area so you can integrate around that area. We're revolving around the x-axis so the area will be a circle.

V = ∫A(x)dx =  ∫(πr²)dr

Since we have two different radius, we subtract them from each others.

∫(πr₂² - πr₁²)dr

∫(π(2)² - π(2x)²)dr

∫(4π - 4πx²)dr

4π∫(1 - x²)dr

integrate from 0 to 1 since that's where our boundary is.

V = 4π∫(1 - x²)dr = 8π/3

To find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis, we can use the method of cylindrical shells and integrate the expression 2πx(2-2x) from 0 to 1.

To find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis, we can use the method of cylindrical shells. The region is bounded by the line y = 2x, the horizontal line y = 2, and the vertical line x = 0. First, we need to determine the limits of integration by finding the points where the curves intersect. Setting y = 2x and y = 2 equal to each other, we find x = 1. Next, we integrate the expression 2πx(2-2x) with respect to x from 0 to 1. Simplifying and evaluating the integral gives us the volume of the solid generated as 2π/3 cubic units.

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Answer:

about 9 passengers will not keep theire reservation

Step-by-step explanation:

given that an airline finds that about 8 percent of the time, a person who makes an advance reservation does not keep the reservation

Capacity of the airline per trip = 105 only

But company schedules = 113 people with advance reservations

Since 8% do not turn up

out of 113 passengers who reserved in advance we can say

that 8% of 113 passengers will not keep their reservation

Based on this information, number  of the scheduled people will not keep their reservation

= [tex]8%of 113\\= \frac{8*113}{100} \\=9.04[/tex]

Since persons cannot be in decimals we can expect about 9 passengers will not keep theire reservation

Based on the given information, about 9 people out of the scheduled 113 will not keep their reservation.

The airline finds that 8 percent of the time, a person with an advance reservation does not keep it. This means that 8% of the scheduled passengers will not keep their reservations.

Percentage of passengers not keeping reservation = 8% = 0.08

Number of scheduled passengers = 113

Number of passengers not keeping reservation = 0.08 * 113 = 9

Therefore, about 9 of the scheduled people will not keep their reservation.

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Answer:

98% Confidence Interval: (2.387,9.113 )

Step-by-step explanation:

We are given the following data set:

3, 8, 3, 5, 1, 13, 9, 2, 7, 3, 13, 2

a) Formula:

[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]  

where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.  

[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]

[tex]Mean =\displaystyle\frac{69}{12} = 5.75[/tex]

Sum of squares of differences = 196.25

[tex]S.D = \sqrt{\frac{196.25}{11}} = 4.044[/tex]

b) 98% Confidence Interval:

[tex]\bar{x} \pm z_{critical}\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]  

Putting the values, we get,  

[tex]z_{critical}\text{ at}~\alpha_{0.02} = \pm 2.33[/tex]  

[tex]5.75 \pm 2.33(\displaystyle\frac{5}{\sqrt{12}} ) = 5.75 \pm 3.363 = (2.387,9.113 )[/tex]