. In a survey of 600 community college students, 470 indicated that they have read a book for personal enjoyment during the school year. Determine a 95% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year.

C.I. at 95% is as C.I.[0.7504 to 0.8162]

Step-by-step explanation:

Given:

Total number n=600

and Required No of students=x=470

To Find:

Determine a 95% confidence interval .

Solution:

Using normal distribution table for Z,

when 95 % C.I. Z=1.96

Now calculate ,

Population proportion,p=x/n

p=470/600

p=0.7833

Now calculate margin of error

MOE=Z*Sqrt[p(1-p)/n]

MOE=1.96*Sqrt[(0.7833*0.21667)/600]

=1.96*Sqrt(0.00028286)

=1.96*0.01681

=0.0329

So

The Boundaries will be as follows:

1)p-MOE

=0.7833-0.0329

=0.7504

2)p+MOE

=0.7833+0.0329

=0.8162

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;