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Maksim231197 [3]
3 years ago
8

In a study conducted in the United Kingdom about sleeping positions, 1000 adults in the UK were asked their starting position wh

en they fall asleep at night. The most common answer was the fetal position (on the side, with legs pulled up), with 41% of the participants saying they start in this position. Use a normal distribution to find a 95% confidence interval for the proportion of all UK adults who start sleep in this position. Use the fact that the standard error of the estimate is 0.016.
Mathematics
1 answer:
Lady_Fox [76]3 years ago
4 0

Answer: = ( 0.411, 0.409)

Therefore at 95% confidence interval (a,b) = (0.411, 0.409)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean gain x = 0.41

Standard deviation r = 0.016

Number of samples n = 1000

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

0.41+/-1.96(0.016/√1000)

0.41+/-1.96(0.000506)

0.41+/-0.00099

0.41+/-0.001

= ( 0.411, 0.409)

Therefore at 95% confidence interval (a,b) = (0.411, 0.409)

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2 years ago
Can someone give me an answer for this?
stealth61 [152]

Answer:

444 adult ticket

763 student ticket

Step-by-step explanation:

let a = number of adult ticket

let s = number of student ticket

s+a=1207

5a+s=2983

get s by itself in the 1st equation so you can plug it into the 2nd

s+a=1207

-a      -a

s=1207-a

5a+(1207-a)=2983

solve for a

4a+1207=2983

-1207      -1207

4a= 1776

444

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8 0
3 years ago
A Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer; a sample
8090 [49]

Answer:

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given first random sample size n₁ = 500</em>

Given  Roper survey reported that 65 out of 500 women ages 18-29 said that they had the most say when purchasing a computer.

<em>First sample proportion </em>

<em>                              </em>p^{-} _{1} = \frac{65}{500} = 0.13

<em>Given second sample size n₂ = 700</em>

<em>Given a sample of 700 men (unrelated to the women) ages 18-29 found that 133 men said that they had the most say when purchasing a computer.</em>

<em>second sample proportion </em>

<em>                              </em>p^{-} _{2} = \frac{133}{700} = 0.19

<em>Level of significance = α = 0.05</em>

<em>critical value = 1.96</em>

<u><em>Step(ii)</em></u><em>:-</em>

<em>Null hypothesis : H₀: There  is no significance difference between these proportions</em>

<em>Alternative Hypothesis :H₁: There  is significance difference between these proportions</em>

<em>Test statistic </em>

<em></em>Z = \frac{p_{1} ^{-}-p^{-} _{2}  }{\sqrt{PQ(\frac{1}{n_{1} } +\frac{1}{n_{2} } )} }<em></em>

<em>where </em>

<em>         </em>P = \frac{n_{1} p^{-} _{1}+n_{2} p^{-} _{2}  }{n_{1}+ n_{2}  } = \frac{500 X 0.13+700 X0.19  }{500 + 700 } = 0.165<em></em>

<em>        Q = 1 - P = 1 - 0.165 = 0.835</em>

<em></em>Z = \frac{0.13-0.19  }{\sqrt{0.165 X0.835(\frac{1}{500 } +\frac{1}{700 } )} }<em></em>

<em>Z =  -2.76</em>

<em>|Z| = |-2.76| = 2.76 > 1.96 at 0.05 level of significance</em>

<em>Null hypothesis is rejected at 0.05 level of significance</em>

<em>Alternative hypothesis is accepted at 0.05 level of significance</em>

<u><em>Conclusion:</em></u><em>-</em>

<em>There is there is a difference between these proportions at α = 0.05</em>

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Answer:

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Answer:

Step-by-step explanation: My variables are

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My equations are 6=3q2d+3=qI

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