1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ksivusya [100]
3 years ago
15

The author drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 199 times. Here are the observed

frequencies for the outcomes of​ 1,2,3,4,5, and 6​ respectively: 28​, 29​, 47​, 40​, 22​, 33. Use a 0.05 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair​ die?
Mathematics
1 answer:
Anton [14]3 years ago
8 0

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28​, 29​, 47​, 40​, 22​, 33.

Probability of an Event

      =\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911

Result is not significant.

3.

(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378

Result is not significant.

4.

(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349

Result is not significant.

5.

(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773

Result is not significant.

6.

(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair​ die.

   

You might be interested in
Vince added 0.912,0.882 ,and 1.354 liters of liquid to a bottle.About how much is in the bottle?
Levart [38]
3.148 liters of liquid. 
3 0
3 years ago
Four pencils cost 5$ , how much will 12 pencils cost, I will give brainliest.
nadya68 [22]

Answer:

the answer is $60

Step-by-step explanation:

5x12=60

3 0
3 years ago
Read 2 more answers
Can someone help me out with this?
myrzilka [38]
Where the ball ended
4 0
3 years ago
tanya purchased 4 hamners for $11.00 each and 7 scewdrivers for $6.65 each write an expresion for the total cost of the tools.th
rewona [7]
X=numbers of hammers
y=numbers of  scewdrivers
f(x,y)= total cost in $

f(x,y)=11x+6.65y

f(4,7)=4(11)+6.65(7)=44+46.55=90.55

Sol: an expresion for the total cost of the tools is f(x,y)=11x+6.65 y; 
x=number of hammers
y= number of  scewdrivers.
The total cost is $90.55
7 0
3 years ago
Read 2 more answers
Find the equation of the line passing through points (6,-3) and (-2,3)
Eduardwww [97]
3/4 I’m pretty sure hejjejwiwiwjwnnsn
3 0
3 years ago
Other questions:
  • Determine the sum: (2abc − 9abc2 − 14ab2c) + (−5abc + 4ab2c).
    11·1 answer
  • The diagonals of quadrilateral ABCD intersect at point K. Is each statement needed to prove that ABCD is a parallelogram?
    10·1 answer
  • Which expression correctly represents the product 12(8 + 5) after applying
    11·1 answer
  • Which pair of ratios is proportional?
    11·1 answer
  • What is (2i)^2.. explain
    7·2 answers
  • If angle A =90 degree , b=1 cm and a =2 cm then solve the following right angled triangle ABC ​
    12·1 answer
  • 17 + 15 / 3 - 2⁴ and 12⁰ + 3 x 4 - 5¹
    14·2 answers
  • Choose a factor pair of 36.Show how this factor pair can be found in the prime factorization of 36
    15·1 answer
  • I need help on my question hurry plsss
    14·1 answer
  • Find the missing lengths of the sides
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!