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Vilka [71]
3 years ago
9

Mr. Taylor's 4th grade class uses Skittles to learn about probability. They open several randomly selected bags of Skittles and

sort and count the different colors and want to determine if Skittles are evenly distributed by color.
Col1 Color Red Orange Yellow Purple Green
Col2 Count 107 101 87 115 10

A) Choose the appropriate null and alternate hypotheses.B)What is the p-value?C)What is your conclusion?
Mathematics
1 answer:
kap26 [50]3 years ago
3 0

Answer:

The skitties are not evenly distributed by colour

Step-by-step explanation:

Given that Mr. T  aylor's 4th grade class uses Skittles to learn about probability. They open several randomly selected bags of Skittles and sort and count the different colors and want to determine if Skittles are evenly distributed by color.

H_0: Skitties are equally distributed\\H_a: atleast two are not equally distributed

(Two tailed chi square test)

If all are equally distributed then expected values would be equal to 420/5 =104

Observed                       Red   Orange  Yellow   Purple   Green  Total

                                       107          101      87           115           10     420

Expected                         104          104     104          104         104    420

Chi square 0.0865 0.0865 2.7788 1.16345 84.9615 89.0769

Chi square is calculated as (obs-exp)^2/exp

Total chi square = 89.0769

df = 4

p value = <0.00001

Reject null hypothesis

The skitties are not evenly distributed by colour

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AnnyKZ [126]

Hello from MrBillDoesMath!

Answer:

21

Discussion:

$350  at 6% =                              => as "percent" means per 100

350 * (6/100) =                            => as 350 * 6 = 2100

2100/100 =

21

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7 0
3 years ago
Due tomarrow please help!!!!!!!!!!!!!!!!!!!!!!!!! Identify the independent and dependent variables in the following statement. T
miskamm [114]
The independent variable is the rate that water is flowing into the pool, and the dependent variable is the depth of water in the pool.

I believe this would be the case due to the fact that we are not in control of the flow yet the flow impacts the depth. So flow would be independent while the pool depth would be dependent on the rate of flow.
8 0
3 years ago
Use the information to answer the following questions.
lyudmila [28]

If a manager at a large company wants to understand whether the employees are feeling. The best way to collect data is: B. survey.

<h3>What is survey?</h3>

Survey can be defined as the data or information collected from group of people so as to draw or reach a conclusion.

Based on the information Survey would be the best method that would be  suited to understand whether the employees are feeing overworked or not.

The manager can send an online survey to the employee so as to collect that data that will enables the manager to reach a conclusion.

Therefore the correct option is B.

Learn more about survey here:brainly.com/question/968894

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8 0
2 years ago
The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large sh
MAXImum [283]

Answer:

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. CI [6583.336 ,6816.336]

d.  The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

Step-by-step explanation:

Population mean = u= 6500 hours.

Population standard deviation = σ=500 hours.

Sample size =n= 50

Sample mean =x`=  6,700 hours

Sample standard deviation=s=  600 hours.

Critical values, where P(Z > Z) =∝ and P(t >) =∝

Z(0.10)=1.282  

Z(0.05)=1.645  

Z(0.025)=1.960  

t(0.01)(49)= 1.299

t(0.05)= 1.677  

t(0.025,49)=2.010

Let the null and alternate hypotheses be

H0: u = 6500 against the claim Ha: u ≠ 6500

Applying Z test

Z= x`- u/ s/√n

z= 6700-6500/500/√50

Z= 200/70.7113

z= 2.82=2.82

Applying  t test

t= x`- u /s/√n

t= 6700-6500/600/√50

t= 2.82

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

Yes we reject H0  for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645

For t test  we reject H0   as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. The 95 % confidence interval of the population mean life is estimated by

x` ±  z∝/2  (σ/√n )

6700± 1.645 (500/√50)

6700±116.336

6583.336 ,6816.336

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

6 0
3 years ago
1) work out the value of (1.7x10^4)x (8.5x10^-2) give your answer in standard form.
Mashutka [201]

ANSWER TO QUESTION 1

(1.7\times 10^4)\times(8.5\times 10^{-2})


We rewrite to obtain;


(1.7\times 10^4)\times(8.5\times 10^{-2})=(1.7\times 8.5)\times(10^4\times 10^{-2})


Recall this product law of indices


a^m\times a^n=a^{m+n}


we apply this law to obtain,


(1.7\times 10^4)\times(8.5\times 10^{-2})=14.45\times10^{4+-2}



This simplifies to


(1.7\times 10^4)\times(8.5\times 10^{-2})=14.45\times10^{2}


We need to rewrite this in standard form;


(1.7\times 10^4)\times(8.5\times 10^{-2})=1.445\times 10^1\times10^{2}


We apply the product law again to get


(1.7\times 10^4)\times(8.5\times 10^{-2})=1.445\times 10^{1+2}


This simplifies to

(1.7\times 10^4)\times(8.5\times 10^{-2})=1.445\times 10^{3}



ANSWER TO QUESTION 2


(6.8\times 10^2)\times(1.3\times 10^{-3})


We rewrite to obtain;


(6.8\times 10^2)\times(1.3\times 10^-3)=(6.8\times 1.3)\times(10^2\times 10^{-3})


Recall this product law of indices


a^m\times a^n=a^{m+n}


we apply this law to obtain,


(6.8\times 10^2)\times(1.3\times 10^-3)=8.84\times10^{2+-3}



This simplifies to


(6.8\times 10^2)\times(1.3\times 10^-3)=8.84\times10^{-1}


This is already in standard form.







8 0
3 years ago
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