AT&T would like to test the hypothesis that the proportion of 18- to 34-year-old Americans that own a cell phone is less than the proportion of 35- to 49-year-old Americans. A random sample of 200 18- to 34-year-old Americans found that 126 owned a smartphone. A random sample of 175 35- to 49-year-old Americans found that 119 owned a smartphone. If Population 1 is defined as 18- to 34-year-old Americans and Population 2 is defined as 35- to 49-year-old Americans, the correct hypothesis statement for this hypothesis test would be

Answers

Answer 1

Answer:

The null and alternative hypothesis can be written as:

[tex]H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0[/tex]

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion of 18- to 34-year-old Americans that own a cell phone is less than the proportion of 35- to 49-year-old Americans.

This claim will be reflected in the alternnative hypothesis, that will state that the population proportion 1 (18 to 34) is significantly smaller than the population proportion 2 (35 to 49).

On the contrary, the null hypothesis will state that the population proportion 1 is ot significantly smaller than the population proportion 2.

Then, the null and alternative hypothesis can be written as:

[tex]H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0[/tex]

The significance level is assumed to be 0.05.

The sample 1, of size n1=200 has a proportion of p1=0.63.

[tex]p_1=X_1/n_1=126/200=0.63[/tex]

The sample 2, of size n2=175 has a proportion of p2=0.68.

[tex]p_2=X_2/n_2=119/175=0.68[/tex]

The difference between proportions is (p1-p2)=-0.05.

[tex]p_d=p_1-p_2=0.63-0.68=-0.05[/tex]

The pooled proportion, needed to calculate the standard error, is:

[tex]p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{126+119}{200+175}=\dfrac{245}{375}=0.653[/tex]

The estimated standard error of the difference between means is computed using the formula:

[tex]s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.653*0.347}{200}+\dfrac{0.653*0.347}{175}}\\\\\\s_{p1-p2}=\sqrt{0.001132+0.001294}=\sqrt{0.002427}=0.049[/tex]

Then, we can calculate the z-statistic as:

[tex]z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.05-0}{0.049}=\dfrac{-0.05}{0.049}=-1.01[/tex]

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

[tex]\text{P-value}=P(z<-1.01)=0.1554[/tex]

As the P-value (0.1554) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the proportion of 18- to 34-year-old Americans that own a cell phone is less than the proportion of 35- to 49-year-old Americans.


Related Questions

The problem is: On a Map, 3 inches represents 40 miles, How many inches represents 480 miles?

Answers

Answer: 36

480/40=12
12x3=36

find the circumference of a circle with a diameter of 6 cm

Answers

Circumference = πd

~substitute (π)(6 cm)

~simplify → 6π cm.

So the circumference of the circle shown here is 6π cm.

Answer:

18.85 cm

Step-by-step explanation:

The circumference of a circle has a formula.

Circumference = π × diameter

The diameter is 6 centimeters.

Circumference = π × 6

Circumference ≈ 18.85

The circumference of the circle is 18.85 centimeters.

The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of 1/8 inch. The width of a door is normally distributed with a mean of 23 7/8 inches and a standard deviation of 1/16 inch. Assume independence. a. Determine the mean and standard deviation of the difference between the width of the casing and the width of the door. b. What is the probability that the width of the casing minus the width of the door exceeds 1/4 inch? c. What is the probability that the door does not fit in the casing?

Answers

Answer:

a) Mean = 0.125 inch

Standard deviation = 0.13975 inch

b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25) = 0.18673

c) Probability that the door does not fit in the casing = P(X < 0) = 0.18673

Step-by-step explanation:

Let the distribution of the width of the casing be X₁ (μ₁, σ₁²)

Let the distribution of the width of the door be X₂ (μ₂, σ₂²)

The distribution of the difference between the width of the casing and the width of the door = X = X₁ - X₂

when two independent normal distributions are combined in any manner, the resulting distribution is also a normal distribution with

Mean = Σλᵢμᵢ

λᵢ = coefficient of each disteibution in the manner that they are combined

μᵢ = Mean of each distribution

Combined variance = σ² = Σλᵢ²σᵢ²

λ₁ = 1, λ₂ = -1

μ₁ = 24 inches

μ₂ = 23 7/8 inches = 23.875 inches

σ₁² = (1/8)² = (1/64) = 0.015625

σ₂ ² = (1/16)² = (1/256) = 0.00390625

Combined mean = μ = 24 - 23.875 = 0.125 inch

Combined variance = σ² = (1² × 0.015625) + [(-1)² × 0.00390625] = 0.01953125

Standard deviation = √(Variance) = √(0.01953125) = 0.1397542486 = 0.13975 inch

b) Probability that the width of the casing minus the width of the door exceeds 1/4 inch = P(X > 0.25)

This is a normal distribution problem

Mean = μ = 0.125 inch

Standard deviation = σ = 0.13975 inch

We first normalize/standardize 0.25 inch

The standardized score of any value is that value minus the mean divided by the standard deviation.

z = (x - μ)/σ = (0.25 - 0.125)/0.13975 = 0.89

P(X > 0.25) = P(z > 0.89)

Checking the tables

P(x > 0.25) = P(z > 0.89) = 1 - P(z ≤ 0.89) = 1 - 0.81327 = 0.18673

c) Probability that the door does not fit in the casing

If X₂ > X₁, X < 0

P(X < 0)

We first normalize/standardize 0 inch

z = (x - μ)/σ = (0 - 0.125)/0.13975 = -0.89

P(X < 0) = P(z < -0.89)

Checking the tables

P(X < 0) = P(z < -0.89) = 0.18673

Hope this Helps!!!

I NEED HELP PLEASE, THANKS! :)
A rock is tossed from a height of 2 meters at an initial velocity of 30 m/s at an angle of 20° with the ground. Write parametric equations to represent the path of the rock. (Show work)

Answers

Answer:

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

Step-by-step explanation:

If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -

x = ( 30 cos 20° )( time ),

y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2

To determine " ( 30 cos 20° )( time ) " you would do the following calculations -

( x = 30 * 0.93... = ( About ) 28.01t

This represents our horizontal distance, respectively the vertical distance should be the following -

y = 30 * 0.34 - 4.9t^2,

( y = ( About ) 10.26t - 4.9t^2 + 2

In other words, our solution should be,

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

These are are parametric equations

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of six per hour.
(a) What is the probability that exactly three arrivals occur during a particular hour? (Round your answer to three decimal places.)
(b) What Is the probability that at least three people arrive during a particular hour? (Round your answer to three decimal places.)
(c) How many people do you expect to arrive during a 15-min period?

Answers

Answer:

a) P(x=3)=0.089

b) P(x≥3)=0.938

c) 1.5 arrivals

Step-by-step explanation:

Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.

The variable X is modeled by a Poisson process with a rate parameter of λ=6.

The probability of exactly k arrivals in a particular hour can be written as:

[tex]P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k![/tex]

a) The probability that exactly 3 arrivals occur during a particular hour is:

[tex]P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\[/tex]

b) The probability that at least 3 people arrive during a particular hour is:

[tex]P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938[/tex]

c) In this case, t=0.25, so we recalculate the parameter as:

[tex]\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5[/tex]

The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.

[tex]E(x)=\lambda=1.5[/tex]

Consider the function represented by 9x + 3y = 12 with x as the independent variable. How can this function be
written using function notation?
Of) = -
O F(x) = - 3x + 4
Of(x) = -x +
O fb) = - 3y+ 4

Answers

Answer:

f(x) = -3x + 4

Step-by-step explanation:

Step 1: Move the 9x over

3y = 12 - 9x

Step 2: Divide everything by 3

y = 4 - 3x

Step 3: Rearrange

y = -3x + 4

Step 4: Change y to f(x)

f(x) = -3x + 4

i am stuck on this please help!

Answers

Answer:

[tex]20 {x}^{3} - 36 {x}^{2} + 7x + 3[/tex]

Solution,

[tex](5x + 1)(2x - 1)(2x - 3)[/tex]

[tex] = 5x(2x - 1) + 1(2x - 1) \times (2x - 3) \\ = (10 {x}^{2} - 5x + 2x - 1)(2x - 3) \\ = (10 {x}^{2} - 3x - 1)(2 x - 3) \\ = 10 {x}^{2} (2x - 3) - 3x(2 x - 3) - 1(2x - 3) \\ = 20 {x}^{3} - 30 {x}^{2} - 6 {x }^{2} + 9x - 2x + 3 \\ = 20 {x}^{3} - 36 {x}^{2} + 7x + 3[/tex]

Hope this helps..

Good luck on your assignment...

Five thousand tickets are sold at​ $1 each for a charity raffle. Tickets are to be drawn at random and monetary prizes awarded as​ follows: 1 prize of ​$800​, 3 prizes of ​$200​, 5 prizes of ​$50​, and 20 prizes of​ $5. What is the expected value of this raffle if you buy 1​ ticket?

Answers

Answer:

The expected value of this raffle if you buy 1​ ticket is $0.41.

Step-by-step explanation:

The expected value of the raffle if we buy one ticket is the sum of the prizes multiplied by each of its probabilities.

This can be written as:

[tex]E(X)=\sum p_iX_i[/tex]

For example, the first prize is $800 and we have only 1 prize, that divided by the number of tickets gives us a probability of 1/5000.

If we do this with all the prizes, we can calculate the expected value of a ticket.

[tex]E(X)=\sum p_iX_i\\\\\\E(X)=\dfrac{1\cdot800+3\cdot200+5\cdot50+20\cdot20}{5000}\\\\\\E(X)=\dfrac{800+600+250+400}{5000}=\dfrac{2050}{5000}=0.41[/tex]

I need help pls pls pls pls​

Answers

Answer:

D.  4

Step-by-step explanation:

If he leaves the science assignments for the next day, he will spend zero hours on science assignments.  This means that y is equal to 0.  Plug this into the given equation and solve for x.

2x + y = 8

2x + 0 = 8

2x = 8

x = 4

Gerald can complete 4 math assignments.

M/J Grade 8 Pre-Algebra-PT-FL-1205070-003

Answers

Answer:

Following are the description of the given course code:

Step-by-step explanation:

The given course code is Pre-Algebra, which is just an introduction arithmetic course programs to train high school in the Algebra 1. This course aims to strengthen required problem solving skills, datatypes, equations, as well as graphing.

In this course students start to see the "big picture" of maths but also understand that mathematical, algorithmic, and angular principles are intertwined to form a basis for higher mathematics education.The duration of this code is in year and it is divided into two levels. In this, code it includes PreK to 12 Education Courses , with the general mathematics .

Answer:

A

Step-by-step explanation:

11/n = 8/5 solve for n

Answers

Answer:

n = 55/8

Step-by-step explanation:

You can solve it by cross multiplying. Where you multiply the denominator of the fraction on the left side with the numerator on the right side, and vice versa.

11/n = 8/5

n x 8 = 11 x 5

8n = 55

n = 55/8

(or 6.875)

Answer:

[tex]\boxed{\pink{n = 7 \frac{3}{8} }}[/tex]

Step-by-step explanation:

[tex] \frac{11}{n} = \frac{8}{5} \\ [/tex]

Use cross multiplication

[tex]11 \times 5 = 8 \times n \\ 55 = 8n \\ \frac{55}{8} = \frac{8n}{8} \\ n = 7 \frac{3}{8} [/tex]

[!] Urgent [!] Find the domain of the graphed function.

Answers

There is no way I can answer this without the graph

Which of the following statements about feasible solutions to a linear programming problem is true?A. Min 4x + 3y + (2/3)z
B. Max 5x2 + 6y2
C. Max 5xy
D. Min (x1+x2)/3

Answers

Answer:

The answer is "Option A"

Step-by-step explanation:

The valid linear programming language equation can be defined as follows:

Equation:

[tex]\Rightarrow \ Min\ 4x + 3y + (\frac{2}{3})z[/tex]

The description of a linear equation can be defined as follows:

It is an algebraic expression whereby each term contains a single exponent, and a single direction consists in the linear interpolation of the equation.

Formula:

[tex]\to \boxed{y= mx+c}[/tex]

The average life a manufacturer's blender is 5 years, with a standard deviation of 1 year. Assuming that the lives of these blenders follow approximately a normal distribution, find the probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.

Answers

Answer:

55.11% probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question:

[tex]\mu = 5, \sigma = 1, n = 9, s = \frac{1}{\sqrt{9}} = 0.3333[/tex]

Find the probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.

This is the pvalue of Z when X = 5.1 subtracted by the pvalue of Z when X = 4.5. So

X = 5.1

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

By the Central Limit Theorem

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

[tex]Z = \frac{5.1 - 5}{0.3333}[/tex]

[tex]Z = 0.3[/tex]

[tex]Z = 0.3[/tex] has a pvalue of 0.6179

X = 4.5

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

[tex]Z = \frac{4.5 - 5}{0.3333}[/tex]

[tex]Z = -1.5[/tex]

[tex]Z = -1.5[/tex] has a pvalue of 0.0668

0.6179 - 0.0668 = 0.5511

55.11% probability that the mean life a random sample of 9 such blenders falls between 4.5 and 5.1 years.

how many solution does this equation have LOOK AT SCREENSHOT ATTACHED

Answers

Answer:

One solution

Step-by-step explanation:

99% of the time, linear equations (equations that have the first degree) have only one solution. However, it's always good to check.

6 - 3x = 12 - 6x

6 = 12 - 3x

-3x = -6

x = 2

As you can see, only one solution. Hope this helps!

An instructor asks students to rate their anxiety level on a scale of 1 to 100 (1 being low anxiety and 100 being high anxiety) just before the students take their final exam. The responses are shown below. Construct a relative frequency table for the instructor using five classes. Use the minimum value from the data set as the lower class limit for the first row, and use the lowest possible whole-number class width that will allow the table to account for all of the responses. Use integers or decimals for all answers.
48,50,71,58,56,55,53,70,63,74,64,33,34,39,49,60,65,84,54,58
Provide your answer below:
Lower Class Limit Upper Class Limit Relative Frequency

Answers

Answer:

The frequency table is shown below.

Step-by-step explanation:

The data set arranged ascending order is:

S = {33 , 34 , 39 , 48 , 49 , 50 , 53 , 54 , 55 , 56 , 58 , 58,  60 , 63 , 64 , 65 , 70 , 71 , 74 , 84}

It is asked to use the minimum value from the data set as the lower class limit for the first row.

So, the lower class limit for the first class interval is 33.

To determine the class width compute the range as follows:

[tex]\text{Range}=\text{Maximum}-\text{Minimum}[/tex]

          [tex]=84-33\\=51[/tex]

The number of classes requires is 5.

The class width is:

[tex]\text{Class width}=\frac{Range}{5}=\frac{51}{2}=10.2\approx 10[/tex]

So, the class width is 10.

The classes are:

33 - 42

43 - 52

53 - 62

63 - 72

73 - 82

83 - 92

Compute the frequencies of each class as follows:

Class Interval                  Values                        Frequency

   33 - 42                      33 , 34 , 39                             3

   43 - 52                      48 , 49 , 50                            3

   53 - 62          53 , 54 , 55 , 56 , 58 , 58,  60              7

   63 - 72                 63 , 64 , 65 , 70 , 71                      5

   73 - 82                              74                                  1

   83 - 92                             84                                   1

   TOTAL                                                                   20

Compute the relative frequencies as follows:

Class Interval          Frequency        Relative Frequency

   33 - 42                        3                   [tex]\frac{3}{20}\times 100\%=15\%[/tex]

   43 - 52                        3                   [tex]\frac{3}{20}\times 100\%=15\%[/tex]

   53 - 62                        7                   [tex]\frac{7}{20}\times 100\%=35\%[/tex]

   63 - 72                        5                   [tex]\frac{5}{20}\times 100\%=25\%[/tex]

   73 - 82                         1                   [tex]\frac{1}{20}\times 100\%=5\%[/tex]

   83 - 92                         1                   [tex]\frac{1}{20}\times 100\%=5\%[/tex]

   TOTAL                        20                          100%

Determine the logarithmic regression of the data below using either a calculator or spreadsheet program. Then, estimate the x−value when the y−value is 5.2. Round your answer to one decimal place. (4.7,10.7),(7.8,20.6),(10.5,30.2),(15.6,41),(20.8,56.1),(22,65.1). Please help right away! Thank you so much!

Answers

Answer:

y ≈ 33.7·ln(x) -45.94.6

Step-by-step explanation:

A graphing calculator can perform logarithmic regression, as can a spreadsheet. The least-squares best fit log curve is about ...

  y ≈ 33.7·ln(x) -45.9

The value of x estimated to make y = 5.2 is about 4.6.

Mia, Maya, and Maria are sisters. Mia's age is twice Maya's age and Maria is seven years younger than Mia. If Maria is 3 years old, how old are Mia and Maya?

Answers

Answer:

Mia:10 Maya:5 Maria:3

Step-by-step explanation:

3+7= 10= Mia's age

10÷2=5= Maya's age

Answer:

Mia - 10

Maya - 5

Maria - 3

The graphs below are the same shape what is the equation of the blue graph

Answers

Answer:

B. g(x) = (x-2)^2 +1

Step-by-step explanation:

When you see this type of equation your get the variables H and K in a quadratic equation. In this case the (x-2)^2 +1  is your H. The (x-2)^2 +1 is your K.

For the H you always do the opposite so in this case instead of going to the left 2 times you go to the right 2 times (affects your x)

For the K you go up or down which in this case you go up one (affects your y)

And that's how you got your (2,1) as the center of the parabola

-Hope this helps :)

Kylie and miranda began arguing about who did better on their tests, but they couln't decide who did better given that they took different tests, kylie took a test in Art History and earned a 77.3, and Tan took a test in English and earned a 62.9. Use the fact that all the students' test grades in the Art History class had a mean of 73 and a standard deviation of 10.7, and all the students' test grades in English had a mean of 66.8 and a standard deviation of 10.8 to answer the following questions.
a) Calculate the Z-score for Isaac's test grade.
b) Calculate the 2-score for lan's test grade.
c) Which person did relatively better?
A. Kylie
B. miranda
C. They did equally well.

Answers

Answer:

a) 77.3-73/10.7= 0.40187

b) 62.9-66.8/10.8= -0.36111

c) Kylie did relatively better

Step-by-step explanation:

Conde Nast Traveler publishes a Gold List of the top hotels all over the world. The Broadmoor Hotel in Colorado Springs contains 700 rooms and is on the 2004 Gold List (Conde Nast Traveler, January 2004). Suppose Broadmoor's marketing group forecasts a demand of 670 rooms for the coming weekend. Assume that demand for the upcoming weekend is normally distributed with a standard deviation of 30.

a.What is the probability all the hotel's rooms will be rented (to 4 decimals)?

b. What is the probability 50 or more rooms will not be rented (to 4 decimals)?

Answers

Answer:

(a) The probability that all the hotel's rooms will be rented is 0.1587.

(b) The probability that 50 or more rooms will not be rented is 0.2514.

Step-by-step explanation:

We are given that the Broadmoor Hotel in Colorado Springs contains 700 rooms and is on the 2004 Gold List.

Suppose Broadmoor's marketing group forecasts a mean demand of 670 rooms for the coming weekend. Assume that demand for the upcoming weekend is normally distributed with a standard deviation of 30.

Let X = demand for rooms in the hotel

So, X ~ Normal([tex]\mu=670,\sigma^{2} =30^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                           Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = mean demand for the rooms = 670

            [tex]\sigma[/tex] = standard deviation = 30

(a) The probability that all the hotel's rooms will be rented means that the demand is at least 700 = P(X [tex]\geq[/tex] 700)

          P(X [tex]\geq[/tex] 700) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\geq[/tex] [tex]\frac{700-670}{30}[/tex] ) = P(Z [tex]\geq[/tex] 1) = 1 - P(Z < 1)

                                                             = 1 - 0.8413 = 0.1587

The above probability is calculated by looking at the value of x = 1 in the z table which has an area of 0.8413.

(b) The probability that 50 or more rooms will not be rented is given by = P(X [tex]\leq[/tex] 650)

         P(X [tex]\leq[/tex] 650) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{650-670}{30}[/tex] ) = P(Z [tex]\leq[/tex] -0.67) = 1 - P(Z < 0.67)

                                                             = 1 - 0.7486 = 0.2514

The above probability is calculated by looking at the value of x = 0.67 in the z table which has an area of 0.7486.

If -5(x+8) =-25, then x=-3

Answers

Answer:

Correct!

Step-by-step explanation:

-5(x+8)=-25

x+8=5

x=-3

Answer:

here, -5(x+8)=-25

or, -5x +(-40)= -25

or, -5x=-25+40

or, x= 15/-5

therefore the value of x is -3....ans..

hope u understood..

7. The mean age at first marriage for respondents in a survey is 23.33,
with a standard deviation of 6.13. For an age at first marriage of 33.44,
the proportion of area beyond the Z score associated with this age is
.05. What is the percentile rank for this score?

Answers

Answer:

[tex] \mu = 23.33, \sigma =6.13[/tex]

And for this case we are analyzing the value os 33.44 and we can use the z score formula given by:

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

And replacing we got:

[tex] z=\frac{33.44 -23.33}{6.13}= 1.649[/tex]

We know that the proportion of area beyond the Z score associated with this age is  .05 so then the percentile would be: 95

Step-by-step explanation:

For this case we have the following parameters:

[tex] \mu = 23.33, \sigma =6.13[/tex]

And for this case we are analyzing the value os 33.44 and we can use the z score formula given by:

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

And replacing we got:

[tex] z=\frac{33.44 -23.33}{6.13}= 1.649[/tex]

We know that the proportion of area beyond the Z score associated with this age is  .05 so then the percentile would be: 95

The chi-square value for a one-tailed (lower tail) test when the level of significance is .1 and the sample size is 15 is a. 23.685. b. 6.571. c. 7.790. d. 21.064.

Answers

Answer:

The degrees of freedom are given by:

[tex] df =n-1= 15-1=14[/tex]

And if we look in the chi square distribution with 14 degrees of freedom and if we find a quantile who accumulates 0.1 of the area in the left we got:

[tex] \chi^2 = 7.790[/tex]

And then the best answer would be:

c. 7.790

Step-by-step explanation:

For this case we know that we are using a one tailed (lower tail) critical value using a significance level of [tex]\alpha=0.1[/tex] and for this case we know that the ample size is n=15. The degrees of freedom are given by:

[tex] df =n-1= 15-1=14[/tex]

And if we look in the chi square distribution with 14 degrees of freedom and if we find a quantile who accumulates 0.1 of the area in the left we got:

[tex] \chi^2 = 7.790[/tex]

And then the best answer would be:

c. 7.790

Please answer this correctly without making mistakes

Answers

Answer:

Question 2

Step-by-step explanation:

2) The time when she woke up was -  3° C

During nature walk, temperature got 3° C warmer than when she woke up.

So, temperature during nature walk = - 3 + 3 = 0° C

a) Al usar un microscopio el microscopio se amplía una célula 400 veces. Escribe el factor de ampliación como cociente o como escala.
b) La imagen de una célula usando dicho microscopio mide 1,5 mm ¿ Cuánto mide la célula en la realidad?

Answers

Answer:

x = 0,00375 mm

Step-by-step explanation:

a) El factor de ampliación es 400/1   es decir el tamaño real se verá ampliado 400 veces mediante el uso del microscopio

b) De acuerdo a lo establecido en la respuesta a la pregunta referida en a (anterior) podemos establecer una regla de tres, según:

Si al microscopio el tamaño de la célula es 1,5 mm, cual será el tamaño verdadero ( que es reducido 400 en relación al que veo en el microscopio)

Es decir     1,5 mm      ⇒    400

                    x (mm)    ⇒       1 (tamaño real de la célula)

Entonces

x  =  1,5 /400

x = 0,00375 mm

A cardboard box without a lid is to have a volume of 8,788 cm3. Find the dimensions that minimize the amount of cardboard used.

Answers

Answer:

x = y = 26 cm; z = 13 cm

Step-by-step explanation:

We can calculate the dimensions of the square base as

∛(2·8788) = 26 cm

the height of the box will be half of 26/2 which is 13 cm.

x = y = 26 cm; z = 13 cm

then the minimum area for the given volume can be calculated using what we call Lagrange multipliers, this makes it easier

area = xy +2(xz +yz)

But we were given the volume as 8788

Now we will make the partial derivatives of L to be in respect to the cordinates x, y, z, as well as λ to be equal to zero, then

L = xy +2(xz +yz) +λ(xyz -8788)

For x: we have

y+2z +λyz=0

For y we have

y: x +2z +λxz=0

For z we have 2x+2y +λxy=0............eqn(*)

For we have xyz -8788=0

If we simplify the partial derivative equation of y and x above then we have

λ = (y +2z)/(yz).

= 1/z +2/y............eqn(1)

λ = (x +2z)/(xz)

= 1/z +2/x.............eqn(2)

Set eqn(1 and 2) to equate we have

1/z +2/y = 1/z +2/x

x = y

From eqn(*) we can get z

λ = (2x +2y)/(xy) = 2/y +2/x

If we simplify we have

1/z +2y = 2/x +2/y

Then z = x/2

26/2 =13

Therefore,

x = y = 2z = ∛(2·8788)

X= 26

y = 26 cm

z = 13 cm

Joe hypothesizes that the students of an elite school will score higher than the general population. He records a sample mean equal to 568 and states the hypothesis as μ = 568 vs μ > 568. What type of test should Joe do?

Answers

Answer:

The test to be used is the right tailed test.

Step-by-step explanation:

The type of test joe should do would be a right tailed test. This is because;

A right tailed test which we sometimes call an upper test is where the hypothesis statement contains the greater than (>) symbol. This means that, the inequality points to the right. For example, we want to compare the the life of batteries before and after a manufacturing change.

If we want to know if the battery life of maybe 90 hours would be greater than the original, then our hypothesis statements might be:

Null hypothesis: (H0 = 90).

Alternative hypothesis: (H1) > 90.

In the null hypothesis, there are no changes, but in the alternative hypothesis, the battery life in hours has increased.

So, the most important factor here is that the alternative hypothesis (H1) is what determines if we have a right tailed test, not the null hypothesis.

Thus, the test to be used is the right tailed test.

Answer:

right tailed test.

Step-by-step explanation:

HELP ASAP WILL MARK BRAINIEST IF YOU ARE RIGHT !Which of the following represents a function?

Answers

Answer:

Option C.

Step-by-step explanation:

This is a function because all of the numbers have a partner, and none of them have more than one.

                                    Example of Not a Function

Function                                Not a Function

-4 to 5                                       -4 to 5                             <

9 to 7                                       -4 to 3                              <

13 to 3                                       13 to 3                              ^

-7 to 5                                        9 to 7                               ^

                                                 -7 to 5                               ^

                                           Not a Function because of this

will give brainliest Evaluate 15/k when k is 3

Answers

Answer:

Hey there!

15/k, when k=3

15/3=5

Answer:

5

Step-by-step explanation:

its a simple as 15/3 = 5

have fun

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