Question

Please help! see attached

Answer 1

Answer: 0.5262 - 0.5538

Explanation:

1) Find the standard deviation with the given information:

n=5000

p=54% ⇒ 0.54

1-p = 1 - 0.54 = 0.46

`\sigma =\sqrt{(p(1-p))/(n)}\\\\\\.\ =\sqrt{(0.54(0.46))/(5000)}\\\\\\.\ =\sqrt{(0.2484)/(5000)}\\\\\\.\ =√(0.00004968)\\\\\\.\ =0.007048`

2) Find the margin of error (ME) with the given information:

C=95% ⇒ Z = 1.960

σ=0.007048

ME = Z × σ

= 1.96 (0.007048)

= 0.01381

3) Find the confidence interval with the given information:

p = 0.54

ME = 0.01381

C = p ± ME

= 0.54 ± 0.01381

= (0.5262, 0.5538)

Answer 2

1. The lowest value of the confidence interval is 0.5262 or 52.62%
2. The highest value of the confidence interval is 0.5538 or 55.38%

Explanation:

Here you estimate the proportion of people in the population that said did not have children under 18 living at home.It can also be given as a percentage.

The general expression to apply here is;

`C.I=p+-z*\sqrt{(p(1-p))/(n) }`

where ;

p=sample proportion

n=sample size

z*=value of z* from the standard normal distribution for 95% confidence level

Given;

n=5000

Find p

From the question 54% of people chosen said they did not have children under 18 living at home

`(54)/(100) *5000 = 2700\\\\p=(2700)/(5000) =0.54`

To calculate the 95% confidence interval, follow the steps below;

• Find the value of z* from the z*-value table

The value of z* from the table is 1.96

• calculate the sample proportion p

The value of p=0.54 as calculated above

• Find p(1-p)

`0.54(1-0.54)=0.2484`

• Find p(1-p)/n

Divide the value of p(1-p) with the sample size, n

`(0.2448)/(5000) =0.00004968`

• Find the square-root of p(1-p)/n

`=√(0.00004968) =0.007048`

• Find the margin of error

Here multiply the square-root of p(1-p)/n by the z*

`=0.007048*1.96=0.0138`

The 95% confidence interval for the lower end value is p-margin of error

`=0.54-0.0138=0.5262`

The 95% confidence interval for the upper end value is p+margin of error

`0.54+0.0138=0.5538`