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kirill [66]
3 years ago
8

"twelve diminished by six times a number

Mathematics
1 answer:
Mnenie [13.5K]3 years ago
7 0
The answer to the problem is 12-6x
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Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 ​students, she finds 2 w
irina1246 [14]

Answer:

A 95​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus is [0.012, 0.270].

Step-by-step explanation:

We are given that Jane wants to estimate the proportion of students on her campus who eat cauliflower. After surveying 24 ​students, she finds 2 who eat cauliflower.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

                              P.Q.  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of students who eat cauliflower

           n = sample of students

           p = population proportion of students who eat cauliflower

<em>Here for constructing a 95% confidence interval we have used a One-sample z-test for proportions.</em>

<u>So, 95% confidence interval for the population proportion, p is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

Now, in Agresti and​ Coull's method; the sample size and the sample proportion is calculated as;

n = n + Z^{2}__(\frac{_\alpha}{2})

n = 24 + 1.96^{2} = 27.842

\hat p = \frac{x+\frac{Z^{2}__(\frac{\alpha}{2}_)  }{2} }{n} = \hat p = \frac{2+\frac{1.96^{2}   }{2} }{27.842} = 0.141

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

 = [ 0.141 -1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } } , 0.141 +1.96 \times {\sqrt{\frac{0.141(1-0.141)}{27.842} } } ]

 = [0.012, 0.270]

Therefore, a 95​% confidence interval for the proportion of students who eat cauliflower on​ Jane's campus [0.012, 0.270].

The interpretation of the above confidence interval is that we are 95​% confident that the proportion of students who eat cauliflower on​ Jane's campus is between 0.012 and 0.270.

7 0
3 years ago
Please help me figure this out
love history [14]

-13 Faren equals -25 Cel! Give the guy above me brainliest, his was more wrote out but this is just a simple answer lol, good job btw guy above me! and goodluck OP. <3 !

6 0
1 year ago
Read 2 more answers
What are the zeros of this function?​
Yakvenalex [24]

Answer: 0 and 5

Step-by-step explanation: The zeros of the function are the points were the graph pass through x - axis

5 0
3 years ago
Randall wants to buy a pizza. He can select from 5 different sizes, 4 types of crust, and 12 toppings for his pizza. Which of th
hjlf

Answer:

240

Step-by-step explanation:

5x4x12=240

if you multiply the amount of sizes he can get times the amount of crust sizes he can get you get the pizza it's self (20)

now including the toppings (12) this means 20x12 gets you the amount of toppings on those induviduale pizza's.

Hope this helps!

Brainliest please!

Please fully rate!

Have a great day!

8 0
2 years ago
Read 2 more answers
Im bad at money :/ do please help me
Semenov [28]

The third answer

1 quarter, 6 dimes, 1 nickel, 7 pennies

6 0
2 years ago
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